School of Mathematics and Statistics
MATH 1021
Calculus of One Variable
© 2003–2021
Revised February 2021

Table of contents
Acknowledgements 1
Introduction 2
1 Real and Complex Numbers 5
1.1 Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.2 Number Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.3 The real number line – Intervals . . . . . . . . . . . . . . . . . . . . . . . . 9
1.4 Complex numbers - Cartesian form . . . . . . . . . . . . . . . . . . . . . . . 11
1.5 Arithmetic in Cartesian form . . . . . . . . . . . . . . . . . . . . . . . . . . 14
1.6 The set of complex numbers . . . . . . . . . . . . . . . . . . . . . . . . . . 17
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
2 Polar Forms of Complex Numbers 26
2.1 Standard Polar form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
2.2 Polar exponential form - Euler’s formula . . . . . . . . . . . . . . . . . . . . 33
2.3 Arithmetic in polar form . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
2.4 Roots of complex numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
2.5 Roots of polynomial equations . . . . . . . . . . . . . . . . . . . . . . . . . 42
2.6 Sine and cosine in terms of exponentials . . . . . . . . . . . . . . . . . . . . 45
2.7 Complex exponential function . . . . . . . . . . . . . . . . . . . . . . . . . 46
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
3 Functions 52
3.1 Functions – definitions and examples . . . . . . . . . . . . . . . . . . . . . . 52
3.2 Combining functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
3.3 Injective and inverse functions . . . . . . . . . . . . . . . . . . . . . . . . . 58
3.4 Inverse trigonometric functions . . . . . . . . . . . . . . . . . . . . . . . . . 62
3.5 Hyperbolic functions and their inverses . . . . . . . . . . . . . . . . . . . . 63
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67
4 Limits and Continuity 70
4.1 Informal definition of limit . . . . . . . . . . . . . . . . . . . . . . . . . . . 70
4.2 One-sided limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
4.3 The basic limit laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75
4.4 Limits at infinity – Horizontal asymptotes . . . . . . . . . . . . . . . . . . . 76
4.5 Infinite limits – Vertical asymptotes . . . . . . . . . . . . . . . . . . . . . . 78
iii
4.6 The squeeze law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80
4.7 Continuous and discontinuous functions . . . . . . . . . . . . . . . . . . . . 82
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88
5 Differentiation 90
5.1 The derivative at a point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90
5.2 The derivative as a function . . . . . . . . . . . . . . . . . . . . . . . . . . . 93
5.3 Basic rules of differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . 95
5.4 The chain rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96
5.5 Implicit differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98
5.6 The Mean Value Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 100
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101
6 Applications of Differentiation 103
6.1 Optimizing functions of one variable . . . . . . . . . . . . . . . . . . . . . . 103
6.2 Increasing and decreasing functions . . . . . . . . . . . . . . . . . . . . . . 107
6.3 Concavity and points of inflection . . . . . . . . . . . . . . . . . . . . . . . 111
6.4 Curve sketching . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115
6.5 L’Hôpital’s rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121
7 Taylor Polynomials 123
7.1 An approximation for ex . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123
7.2 Taylor polynomials about x= 0 . . . . . . . . . . . . . . . . . . . . . . . . . 126
7.3 Taylor polynomials about x= a . . . . . . . . . . . . . . . . . . . . . . . . . 129
7.4 Taylor’s formula – The remainder term . . . . . . . . . . . . . . . . . . . . . 131
7.5 How good is the Taylor polynomial approximation? . . . . . . . . . . . . . . 132
7.6 Proof of the remainder formula . . . . . . . . . . . . . . . . . . . . . . . . . 134
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135
8 Taylor Series 137
8.1 Infinite series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137
8.2 Taylor series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140
8.3 Euler’s formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144
8.4 The binomial series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145
8.5 A series for the inverse tan function . . . . . . . . . . . . . . . . . . . . . . 148
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149
9 The Riemann Integral 151
9.1 Riemann sums – The area problem . . . . . . . . . . . . . . . . . . . . . . . 151
9.2 The Riemann integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153
9.3 Calculating Riemann sums . . . . . . . . . . . . . . . . . . . . . . . . . . . 155
9.4 Properties of the Riemann integral . . . . . . . . . . . . . . . . . . . . . . . 159
iv
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162
10 Fundamental Theorem of Calculus 164
10.1 Integrals as functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164
10.2 The Fundamental Theorem of Calculus I . . . . . . . . . . . . . . . . . . . . 166
10.3 The Fundamental Theorem of Calculus II . . . . . . . . . . . . . . . . . . . 167
10.4 Leibniz Integral Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170
10.5 The natural logarithm and exponential functions . . . . . . . . . . . . . . . . 171
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176
11 Integration Techniques 178
11.1 Basic rules of integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178
11.2 Integration by substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . 179
11.3 Integration by parts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183
11.4 Partial fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191
12 Applications of Integration 193
12.1 Further integration techniques . . . . . . . . . . . . . . . . . . . . . . . . . 193
12.2 Length of a curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 198
12.3 Area between two curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200
12.4 Solids of revolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207
A Formal Definition of Limits 208
B Geometric proof that limx→0 sinx/x= 1 212
C Linear approximations and differentials 214
D The Distance Problem 219
E Growth Rates 223
F Table of Standard Integrals 225
G Answers to Selected Exercises 226
v

Acknowledgements
The material in these notes has been developed over many years by the following members
of the School of Mathematics and Statistics:
Eduardo Altmann Mary Myerscough
Sandra Britton Nigel O’Brian
Chris Durrant Sharon Stephen
Dave Galloway Fernando Viera
Jenny Henderson Haotian Wu
Andrew Mathas
1
Introduction
Calculus is one of the major achievements of the 17th century. It plays a key role in almost
every instance where mathematics is applied in the sciences, engineering, or in economics.
Without calculus, we would not have cars, computers, televisions or mobile phones; Einstein
would never have penned his theory of relativity; we would not know of the existence of
DNA; we would never have landed on the moon. The list goes on.
Whether you end up continuing in mathematics or majoring in another field it will be im-
portant for you to learn and understand the meaning of calculus. The reason for this is quite
simple. In high school you can progress simply by memorizing formulas; in university there
will be times when you need to develop formulas for yourself, and this is where a proper
understanding of calculus will be a definite asset.
These notes are intended to supplement the lectures ofMATH1021. Your lecturers will almost
certainly use different examples and they will also explain some of the material in the course
slightly differently from these notes. In some places these notes go into more detail than your
lectures; at other times your lecturer will go into more detail.
Reading mathematics is not like reading a novel; we have to think and struggle with every
sentence. We are professional mathematicians and we are not ashamed to say that in our
research there have been times when we have spent more than a day trying to understand a
single line of mathematics! You will be pleased to know that in this course you should not
have to spend this long on a single sentence; however, there will be times when you do have
to think quite hard to understand what is going on. If you do get stuck then go and ask your
lecturer or tutor to explain it to you!
In addition to thinking when you read mathematics you should also work through the calcu-
lations yourself using pen and paper.
At some places in the notes and in the Appendices we have included material which is more
“advanced” than we expect you to know or understand. You are free to either read these
sections or skip over them, as you wish.
Tutorial problems and Exercise sheets
There are plenty of problems with full solutions for you to practice.
a) Worked examples with full solutions have been included in these lecture notes
throughout all chapters.
b) Exercises are available at the end of the chapters in these notes and answers to Selected
Exercises can be found in Appendix G.
2
Introduction 3
c) Exercise sheets containing problems to be solved before the tutorial session are avail-
able on the MATH1021 web page. Full solutions will be available online at the end of
the corresponding week.
A detailed list of mathematical objectives (knowledge, understanding and skills) for a
given chapter is provided in the weekly Exercise Sheets.
d) Board tutorial sheets will be handed out during tutorials with problems to be solved
during the tutorial class. Full solutions will be available online at the end of the corre-
sponding week.
e) Solutions will be provided to assignments 1 and 2.
f) Questions and solutions to selected past exam papers will be made available near the
end of semester.
Why study mathematics
The study of mathematics enhances your ability to think logically and an-
alytically, move from the particular to the general, work quantitatively and
improve problem-solving skills. By reading and working carefully through
the material in these notes you will develop the following additional generic
skills:
• Generalise simple and familiar ideas to more complex settings.
• Use geometric/visual techniques to help understand new concepts.
• Apply simple techniques in unfamiliar situations.
• Estimate values by using suitable approximation techniques.
• Recognise that bounds on the error are an important part of any good
approximation.
A note about definitions
A mathematical definition is a precise description of some mathematical con-
cept. Historically, many concepts in mathematics have been used extensively
before a precise definition of the concept has been formulated.
While precision in definitions is certainly important, learning a definition off
by heart, without an understanding of the concept, is unlikely to be helpful.
It is important to spend some time thinking about a definition in order to gain
this understanding.
4 MATH 1021 Calculus of One Variable
C H A P T E R 1
Real and Complex Numbers
Mathematics includes not only the study of logic, structure and geometry, but also ideas about
numbers. Real numbers in particular, are fundamental to calculus and many other branches
of mathematics. In this chapter we review the concepts of sets and extend previous work on
numbers, particularly the real numbers, before introducing the set of complex numbers.
1.1 Sets
Set notation is a convenient and precise way to write about collections of numbers. We start
by talking about general sets.
Definition
A "set" is a collection of objects which are called "members" or "elements"
of the set.
Example 1.1a A set can be written as a list, for example, A= {a,b,c,d}, where
• A is the name of the set,
• a,b,c,d are the elements of the set enclosed in braces and separated by commas.
• If the list of elements is large, three dots may be used to mean ’and so on’. For example,
the set of natural numbers may be denoted by N= {0,1,2,3, . . .}.
♦
1.2 Number Sets
Our understanding of numbers, what they are and how they work, develops from simple
counting through fractions and negative numbers to an appreciation of irrational numbers
and real numbers. Mathematically, different types of numbers belong to different sets.
5
6 MATH 1021 Calculus of One Variable
The set of "natural numbers" {0,1,2,3,4, . . .}, is denoted by the symbol N. It is closed
under the operations of addition and multiplication. That is, adding two natural numbers
gives another natural number, as does multiplying them together.
The set of "integers" {. . . ,−4,−3,−2,−1,0,1,2,3,4, . . .}, denoted by Z, is the set of
whole numbers, including both positive whole numbers, negative whole numbers and zero.
The set of integers is closed under the operations of addition, subtraction and multiplication.
The set of "rational numbers", denoted by Q, is the set of all numbers of the form n/m
where n and m are integers and m 6= 0. Some examples are 1
2
,−1
4
, 4
3
. Rational numbers
include decimals which either terminate or repeat. Note that the integers are a subset of the
rational numbers, since they are of the form n/m where m = 1. The set of rational numbers
is closed under the operations of addition, subtraction, multiplication and division, provided
that division by zero is excluded.
The set of "real numbers", denoted by R, includes all rational numbers and all irrational
numbers. Irrational numbers cannot be expressed as n/m, where m and n are integers, al-
though some may be interpreted geometrically. For example,
√
2 is the length of a diagonal
of a unit square. The irrational number pi is the ratio of the circumference of a circle to the
circle’s diameter.
The set of "complex numbers", denoted by C, contains all the other number sets mentioned
above and all the imaginary numbers to be introduced in Section 1.4.
In fact we can summarise these numbers sets diagrammatically as shown in Figure 1.1.
Natural Numbers, N
0, 1, 2, 3, . . .
Integers, Z
. . . ,−2,−1,
Rational Numbers, Q
eg. 1
2
,−4
3
, 0.1,−7.2, 0.3˙
Real Numbers, R
eg. pi, e,−√5, 0.1010010001 . . .
Complex Numbers, C
eg. 1+2 i, 3 i . . . , with i2 =−1
Figure 1.1: Number Sets
Chapter 1: Real and Complex Numbers 7
Set notation
• Element of a set – The symbol ∈ means “is an element of”. For example, −3 ∈ Z is
read as “−3 is an element of the set of integers”; y ∈ B is read as “y is an element of
the set B” or “y is a member of the set B”.
• Subset of a set – The symbol⊆ should be read as “is a subset of”. For example, N⊆Z
is read as “the set of natural numbers is a subset of the set of integers” or “the set of
natural numbers is contained in the set of integers”.
• Strictly a subset – Sometimes you may see the symbol⊂which means that the smaller
set is strictly a subset of the larger; the two cannot be equal. For example, it is most
precise to write N⊂ Z as the two sets are not the same.
• Contains a set – The reversed symbol⊇means “contains”. For example, R⊇Q reads
“the set of real numbers contains the set of rational numbers”. (There is also a symbol
⊃ which means “contains, but is not equal to”.) If A⊆ B then B⊇ A.
• Not an element of a set – A forward slash through any of these symbols above means
“not”. For example, −1 6∈ N is read as “−1 is not an element of the set of natural
numbers”.
• Not a subset – Another example, R 6⊆ Z, is read as “the set of real numbers is not a
subset of the set of integers.”
There are other symbols which describe sets formed from other sets:
• Union of sets – The expression A∪B denotes the union of set Awith set B. The "union"
of two sets is the set of elements which are members of either one or both of the sets.
If an element occurs in both sets, it is only listed once in the union. For example
{1,2,3,4,}∪{3,4,5,6}= {1,2,3,4,5,6}
• Intersection of sets – The intersection of sets A and B is written A∩B. The "inter-
section" of two sets is the set of elements which are members of both of the sets. For
example
{1,2,3,4,}∩{3,4,5,6}= {3,4}
• Subtraction of sets – The symbol \ which is read “minus” or “without”, is used to
indicate the set of elements which are in one set but not in another. That is, A\B is the
set of all elements which are in A but not in B. So for example,
{1,2,3,4,}\{3,4,5,6}= {1,2}.
8 MATH 1021 Calculus of One Variable
Venn Diagrams
A set can be represented in a simple, graphical way by a "Venn diagram". Each set is drawn
as a circle, a square or some other closed shape. Shapes representing sets may overlap one
another if sets have elements in common. Sometimes, the elements of the sets are written on
the Venn diagram but often they are not. Different parts of a Venn diagram can be shaded to
illustrate different parts of the set.
Venn diagrams are a useful way to represent relations between sets. Note that A\B is not the
same as B\A.
A B
A∪B
A B
A\B
A B
A∩B
A B
B\A
Conditions on Sets
If we want to specify a set whose elements fulfil a certain condition then we do this in the
way illustrated in the following examples.
• If we want to express that “A is the set of all rational numbers x such that x is positive”,
we write
A= {x ∈Q | x> 0}.
The vertical slash should be read as such that.
• LetW be the set of words in English. Then
B= {x inW |x begins with the letter “P”}
reads “B is the set of all elements x of the set of English words such that x begins with
P” or “B is the set of all English words that begin with P.”
Chapter 1: Real and Complex Numbers 9
• If we want to say that “C is the set of all integers x such that x/2 is an integer” or “C is
the set of all even integers”, we write
C = {x ∈ Z | x
2
∈ Z}.
• If we want to say that “D is the set of all real numbers which are greater than −1 and
less or equal to 1”, we write
D= {x ∈ R | −1< x≤ 1}.
1.3 The real number line – Intervals
• The real number line – Every real number can be located on the "real number line".
For example:
0−1 3
2
pi 4
It is straightforward to sketch sets that are written using interval notation on the real
number line.
Note that an open dot is used if the end point of the interval is not included in the set.
If the endpoint is part of the set, then a closed dot is used.
• Interval notation – Sets of real numbers which lie between two end points can be
represented using "interval notation". For example
D= {x ∈ R | −1< x≤ 1}= (−1,1]
A curved bracket is used to show that an endpoint (such as −1 in this example) is not
included in the set and a square bracket is used when the endpoint is part of the set.
• Open interval – An interval where neither endpoint is part of the set is called an "open
interval".
a b
(a, b) = {x ∈ R | a< x< b}
The interval (a,b) = {x ∈R | a< x< b} and the point (a,b) in the Cartesian plane are
written in exactly the same way. They are not, however, the same thing. It is usually
clear from the context whether (a,b) represents a point or an interval.
• Closed interval – If both endpoints are part of the interval it is called a "closed inter-
val".
10 MATH 1021 Calculus of One Variable
a b
[a, b] = {x ∈ R | a≤ x≤ b}
It is also possible that one endpoint will be in the set and the other will not be. For
example,
a b
(a, b] = {x ∈ R | a< x≤ b}
a b
[a, b) = {x ∈ R | a≤ x< b}
• Semi-infinite intervals – There is special notation for sets of the number line that
extend infinitely in one direction or the other.
(a,∞) = {x ∈ R | x> a}; (−∞,a) = {x ∈ R | x< a}
[a,∞) = {x ∈ R | x≥ a}; (−∞,a] = {x ∈ R | x≤ a}
Note that ∞ is not a number, rather, it represents infinity.
Both ∞ and −∞ always take a round bracket.
Examples 1.3a
i) A= [7,29] = {x ∈ R | 7≤ x≤ 29}
0 7 29
[7, 29]
ii) S= (2,∞) = {x ∈ R | x> 2}
0 2
(2, ∞)
iii) V = (−3,−1)∪ [2,5] = {x ∈ R | −3< x<−1 or 2≤ x≤ 5}
0-3 -1 2 5
(−3,−1)∪ [2, 5]
iv) T = (−∞,0)∪ (0,∞) = {x ∈ R | x 6= 0} = R\{0}. As you can see there may be a
number of ways of writing down a set.
0
R\{0}
♦
Chapter 1: Real and Complex Numbers 11
Modulus or absolute value
The "modulus" or "absolute value" |x| of a real number x gives the distance on the real number
line from x to zero. The modulus of x is defined in this way:
|x|=
{
x if x≥ 0,
−x if x< 0.
For example |5|= 5 and |−10|= 10.
The distance between two numbers on the number line can also be expressed using modulus.
The distance between x and y is given by |x−y|= |y−x|. For example, the distance between
3 and −4 is |3− (−4)| = |3+ 4| = 7 which is what we intuitively expect to be the distance
from −4 to 3. Alternatively we could have written |−4−3|= |−7|= 7.
1.4 Complex numbers - Cartesian form
Suppose that you are asked to solve the equation
x2+1= 0.
Your first response might be to say that there will be two solutions as it is a quadratic equation.
Very quickly you might write down the line
x2 =−1.
At that point you might conclude, correctly, that there are no real solutions to the equation,
because in the real number system, we cannot take square roots of negative numbers. But
what if we agree that there exists a number x such that x=
√−1 ?
Such a number does indeed exist, although it is not a real number. It is known as an "imagi-
nary number". We denote it by i (although some branches of engineering use j instead) and
we’ll assume that the usual rules for algebraic manipulation apply.
Imaginary unit
The number denoted by i that satisfies the condition i2 =−1 is called the
imaginary unit. It follows that
i=
√−1.
The equation x2+ 1 = 0 now has two imaginary solutions, namely i and −i. To check that
x=± i are solutions, substitute into the equation
x2+1= (±i)2+1=−1+1= 0.
12 MATH 1021 Calculus of One Variable
What about the equation x2+9= 0? In this case
x2+9= 0 =⇒ x=±√−9=±√−1×9=±√−1
√
9=±3 i.
It is easy to show by substitution into x2+9= 0 that x=±3 i are both solutions.
Properties of i – The imaginary unit satisfies the following useful relations:
i2 = (−i)2 =−1
i3 = (i2. i) = (−1. i) =−i
i4 = (i2. i2) = (−1).(−1) = 1
i8 = (i4. i4) = (1.1) = 1, and so on.
We are now in a position to introduce a new number set:
Imaginary numbers
Any non–zero real multiple of i is called a purely imaginary number or just
imaginary number. The square of an imaginary number is a negative real
number.
For example
3i, −20i, −i/5 and pii
are all imaginary numbers, and their squares
(3i)2 =−9, (−20i)2 =−400, (−i/5)2 =−1/25, (pi i)2 =−pi2,
are all negative real numbers.
Complex numbers
Suppose now that you are given this equation to solve:
x2−4x+5= 0.
Completing the square and rearranging gives (x−2)2 =−1; that is, x−2=±i or x = 2± i.
These solutions can also be obtained by applying the familiar quadratic formula:
x=
4±√16−20
2
=
4±√−4
2
=
4±√−1√4
2
=
4±2 i
2
= 2± i.
Chapter 1: Real and Complex Numbers 13
These solutions are not purely imaginary, although they do involve an imaginary number.
The solutions 2+ i and 2− i are called complex numbers.
Cartesian form of a complex number
A complex number expressed in the form a+ ib is said to be in Cartesian
form.
• Real numbers are a special case of complex numbers when b= 0.
• Imaginary numbers are a special case of complex numbers when
a= 0.
The complex number 2+ i in the above example is written in Cartesian form with a= 2 and
b= 1.
Real and Imaginary parts
Given a complex number in Cartesian form z= a+ ib:
• The real number a is called the real part of z and we write Re(z) = a.
• The real number b is called the imaginary part of z and we write Im(z) = b.
Examples 1.4a
i) If z= 2+ i then Re(z) = 2 and Im(z) = 1.
ii) If w= 3+8i then Re(w) = 3 and Im(w) = 8.
iii) If z= 1
2
−5i then Re(z) = 1
2
and Im(z) =−5.
iv) The purely imaginary number z=−7i can be written as z= 0−7i. Therefore
Re(z) = 0 and Im(z) =−7.
v) For the real number z= 4= 4+0i we have Re(4) = 4 and Im(4) = 0. ♦
Quadratic equations
If we allow complex numbers as solutions to quadratic equations with real coefficients then
every such quadratic equation will always have two solutions, and they will be either both
real or both complex.
14 MATH 1021 Calculus of One Variable
We can see this in general if we look at the quadratic formula. The solution to the quadratic
equation ax2+bx+ c= 0, where a, b and c are reals, is given by
x=
−b±
√
b2−4ac
2a
.
Whether ax2+bx+ c= 0 has (purely) real or complex roots depends on the sign expression
b2−4ac which is known as the discriminant of the quadratic.
x is
{
real if b2−4ac≥ 0
complex if b2−4ac< 0.
Example 1.4b The solutions of x2+6x+25= 0 must be complex since b2−4ac=−64< 0.
Using the quadratic formula, the solutions are found to be −3+4i and −3−4i. These com-
plex numbers are related; they are complex conjugates of each other. This will be examined
further in the next section. ♦
1.5 Arithmetic in Cartesian form
Complex numbers can be added or multiplied together, subtracted from one another or di-
vided by one another.
Consider two complex numbers z= a+bi and w= c+di. Here the real part of z is a and the
imaginary part of z is b; the real part of w is c and the imaginary part of w is d.
Addition
z+w = (a+bi)+(c+di)
= (a+ c)+(b+d)i
Rule: Add real parts to real parts and imaginary parts to imaginary parts.
Example 1.5a
(3−4i)+(1+2i) = 3+1+(−4+2)i
= 4−2i
♦
Chapter 1: Real and Complex Numbers 15
Subtraction
z−w = (a+bi)− (c+di)
= (a− c)+(b−d)i
Rule: Subtract real parts from real parts and imaginary parts from imaginary parts.
Example 1.5b
(3−4i)− (1+2i) = 3−1+(−4−2)i= 2−6i
♦
Multiplication
zw = (a+bi)(c+di)
= ac+adi+bci+(bd)i2
= (ac−bd)+(ad+ cb)i
Rule: Expand the brackets in the normal way, remembering that i2 can be simplified to −1,
and collect terms into real and imaginary parts.
Example 1.5c
(3−4i)(1+2i) = 3−4i+6i−8i2 = 3+2i+8= 11+2i
♦
Complex conjugate and division
To divide one complex number by another we have to introduce the complex conjugate of a
complex number.
Complex conjugate
The complex conjugate of the number z = a+ ib is the complex number
defined by z= a− ib.
16 MATH 1021 Calculus of One Variable
The geometric interpretation of the complex conjugate z is the reflection of z about the
real axis. The following properties of the complex conjugate can be easily proved from the
definition,
Properties of conjugates
z+w= z+w zw= zw zn = zn
If z= a+ ib then zz¯= (a+ ib)(a− ib) = a2+b2.
Examples 1.5d
i) 3+5i= 3−5i 2−7i= 2+7i
ii) Verify the first property of conjugates in the box above when z= 1+2i and w= 3+ i
z+w= (1+2i)+(3+ i) = 4+3i= 4−3i
z+w= (1+2i)+(3+ i) = (1−2i)+(3− i) = 4−3i
iii) If z is a real number then z= z. For example, if z=
√
2 then z=
√
2.
iv) If z is a purely imaginary number then z=−z. For example 3i=−3i. ♦
Division
If w 6= 0 then to find z
w
we multiply both top and bottom by the complex conjugate of w.
z
w
=
z
w
w
w
=
(a+bi)
(c+di)
(c−di)
(c−di)
=
ac−adi+ cbi− (bd)i2
c2− cdi+ cdi−d2i2
=
(ac+bd)+(cb−ad)i
c2+d2
=
ac+bd
c2+d2
+
cb−ad
c2+d2
i
This process is similar to rationalising the denominator of a quotient of surds. Multiplying by
the complex conjugate of the divisor produces a real number in the denominator and allows
the number to be written in the Cartesian form a+bi.
Chapter 1: Real and Complex Numbers 17
Example 1.5e
5−10i
1+2i
=
(5−10i)(1−2i)
(1+2i)(1−2i)
=
5−20i+20i2
1−2i+2i−4i2
=
−15−20i
5
= −3−4i
♦
Equality of complex numbers
Two complex numbers are equal to each other if and only if both their real and imaginary
parts are equal. In other words, if z = a+ bi and w = c+ di, then z = w if and only if a = c
and b= d.
1.6 The set of complex numbers
We have seen that a real number is a particular type of complex number, one with zero
imaginary part. The complex numbers include real numbers and form a set which contains
the set of real numbers and hence all of the other number sets we have mentioned.
Set of complex numbers
The "set of complex numbers" C is the set of all numbers of the form a+ bi
where a and b are real numbers and i2 =−1.
Using set notation we can write:
C= {a+bi | a,b ∈ R, i2 =−1}.
As we have shown in Figure 1.1, we also have
N⊂ Z⊂Q⊂ R⊂ C .
The set of complex numbers, like the set of real numbers, is closed under addition, subtrac-
tion, multiplication and division. This means that the sum of two complex numbers is another
complex number, and so on.
18 MATH 1021 Calculus of One Variable
The set of complex numbers is not ordered
Complex numbers lack an important property of the real numbers.
The set of real numbers is "ordered"; that is, if we have any two real numbers x and y we can
say that either
x> y or x< y or x= y.
Because of this property we are able to represent real numbers on the real number line.
The set of complex numbers is not ordered. Consider the two complex numbers 2− 3i and
−1+5i. It does not make sense to write
2−3i > −1+5i or −3i < −1+5i.
However, we may write Re(2−3i) > Re(−1+5i) and Im(2−3i) < Im(−1+5i) because
the real part and the imaginary part of a complex numbers are both real numbers.
Because the set of complex numbers is not ordered, complex numbers cannot be represented
as points on a line. Instead, complex numbers are represented as points on a plane.
The complex plane
The "complex plane" or "Argand diagram" allows complex numbers to be represented graph-
ically. The horizontal axis in the complex plane is called the "real axis". All real numbers
lie on the horizontal axis in the complex plane; positive numbers to the right of the origin,
negative numbers to its left.
The vertical axis is known as the "imaginary axis". All purely imaginary numbers lie on the
vertical axis. Each point in the complex plane corresponds to a single complex number. For
example:
-1 2
3i
−2i
2+3i
2−2i
−1+3i
Imaginary axis
Real axis
Chapter 1: Real and Complex Numbers 19
The modulus of a complex number
In Section 1.3 we defined the "modulus" or "absolute value" |x| of a real number x as the
distance on the real number line from x to zero, given by
|x|=
{
x if x≥ 0,
−x if x< 0.
In the case of complex numbers the modulus is also defined to give the distance from the
origin:
Modulus
For a complex number z= a+bi the modulus is defined by |z|=
√
a2+b2.
By Pythagoras’ Theorem, this formula gives the distance from z to the origin of the complex
plane.
b
a+ ib
|z|=
√ a2 +
b
2
a
b
We can express |z| in terms of z and its complex conjugate as follows,
z z= (a+bi)(a−bi) = a2+b2 = |z|2 =⇒ |z|=√z z.
Properties of the modulus
For all complex numbers z= a+ ib and w= c+ id, we have
a) |zw|= |z| |w|,
b) |z/w|= |z|/|w|,
c) |z+w| ≤ |z|+ |w|. This is called the triangle inequality,
d) |z−w| ≥ |z|− |w|.
20 MATH 1021 Calculus of One Variable
To show that (a) is true, we calculate |zw| and |z| |w| separately and show they are equal.
|zw|= |(a+ ib)(c+ id)|
= |(ac−bd)+ i(ad+bc)|
=
√
(ac−bd)2+(ad+bc)2
=
√
a2c2+b2d2−2abcd+a2d2+b2c2+2abcd
=
√
a2c2+b2d2+a2d2+b2c2,
while
|z| |w|=
√
a2+b2
√
c2+d2
=
√
a2c2+b2d2+b2c2+a2d2
= |zw|.
We will show (c) algebraically (a geometric argument can also be used, and this is left as
an exercise). Note that since all quantities are non-negative, proving |z+w| ≤ |z|+ |w| is
equivalent to proving |z+w|2 ≤ (|z|+ |w|)2. Now
|z+w|2 = |(a+ c)+ i(b+d)|2 = (a+ c)2+(b+d)2 = a2+b2+ c2+d2+2(ac+bd).
Similarly,
(|z|+ |w|)2 = |z|2+ |w|2+2|z| |w|= a2+b2+ c2+d2+2|z| |w|.
So we must prove that ac+bd ≤ |z| |w|. Now
|z| |w|=
√
a2+b2
√
c2+d2
=
√
(a2+b2)(c2+d2)
=
√
a2c2+b2d2+a2d2+b2c2
=
√
(ac+bd)2−2abcd+a2d2+b2c2
=
√
(ac+bd)2+(ad−bc)2
≥
√
(ac+bd)2
= |ac+bd|
Now every real number k is less than or equal to its own absolute value |k|. Hence
ac+bd ≤ |ac+bd| ≤ |z| |w|,
and the proof is complete. The proofs of (b) and (d) are left as an exercise.
Chapter 1: Real and Complex Numbers 21
Subsets of the complex plane
The modulus can be used to specify subsets of the set of complex numbers which can be
graphed in the complex plane.
Examples 1.6a
i) {z ∈ C | |z| > 2} is the set of complex numbers z such that z is more than 2 units
distant from the origin. The set is represented by the shaded area below, which extends
indefinitely.
2-2
2i
-2i
ii) {z ∈ C | 1 ≤ |z| ≤ 3} is the set of complex numbers which are between one and three
units distant from the origin.
1-1
i
-i
3-3
3i
-3i
iii) {z ∈ C | |z− 1| < 2}. As with real numbers, |z− 1| is exactly the distance from z
to 1. Hence, this is the set of all complex numbers whose distance from 1 is less than 2.
Geometrically, these are all points in the complex plane that are inside the circle, centre
1, radius 2.
22 MATH 1021 Calculus of One Variable
1-1 3
2
iv) {z ∈ C | |z+ 2− i| > 1}. Here |z+ 2− i| = |z− (−2+ i)| is the distance from the
complex number z to −2+ i. So this set is the set of all complex numbers whose
distance from −2+ i is greater than one unit. In other words, this is the set of points in
the complex plane with are strictly outside the circle of radius 1 and centre −2+ i.
-2
ib
-2+i
v) Here is a different type of subset of the complex numbers: {z ∈C | Imz≤ 0} is the set
of all complex number whose imaginary part is less or equal to zero.
♦
Chapter 1: Real and Complex Numbers 23
Summary of Chapter 1
• Set theory is the natural language used to describe and manipulate
numbers.
• Complex numbers in Cartesian form a+ ib were introduced to find
solutions of quadratic equations with no real roots.
• Arithmetic operations of addition, subtraction, multiplication and di-
vision were introduced in Cartesian form.
• The complex conjugate of a complex number z = a+ ib was defined
as z¯= a− ib.
• The modulus of z= a+ ib was defined as |z|=
√
a2+b2.
• The complex plane, also known as the Argand diagram, gives us a
geometric representation of a complex number z = a+ bi as a point
(a,b) in the Cartesian plane.
Exercises
1.1 In each of the following exercises, perform the indicated operations and give the final
answer in the form x+ yi.
a) (5−2i)+(2+3i)
b) (2− i)− (6−3i)
c) (2+3i)(−2−3i)
d) −i(5+ i)
e) 1/i
f) (a+ ib)(a− ib)
g) 6i/(6−5i)
h) (a+ ib)/(a− ib)
i) 1/(3+2i)
j) i2, i3, i4, . . . , i10
k) (1+ i)/(1− i)
l) [i/(1− i)]+ [(1− i)/i]
m) (1/i)−3i/(1− i)
n) i123−4i9−4i
1.2 If z= 5+12i and w= 3+4i, express
w+ z, z−w, zw and z/w
in the form a+bi. Use these results to verify that
24 MATH 1021 Calculus of One Variable
a) |zw|= |z||w|
b) |z/w|= |z|/|w|
c) |z+w| ≤ |z|+ |w|
d) |z−w| ≥ |z|− |w|
1.3 If z= x+ yi, express each of the following explicitly in terms of x and y.
a) Re(z/z)
b) |(z/z)|
c) Imz3
d) Re z4
e) |z6|
f) |(z+1)/(z−1)|
g) Re(1/z2)
1.4 Simplify the following expressions.
a) Im
1
1+ i
b) Re
(1− i)2
1+2i
c) |cosθ + isinθ |, where θ is any angle
d)
∣∣∣∣1+3i3+ i
∣∣∣∣
e)
∣∣∣∣ (1+ i)6i3(1+4i)2
∣∣∣∣
1.5 Solve the following equations using the quadratic formula.
a) y2+2y+5= 0
b) z2+3z+8= 0
c) t2+ t−1= 0
d) 7a2+8a+4= 0
1.6 If z= 3−2i, plot z, −z, z and −z as points in the complex plane.
1.7 Show that for any complex number z, |z|= |z|.
1.8 If z= z, what can you say about z?
1.9 Prove properties (ii) and (iv) of the modulus, given at the end of the chapter.
[Hint for (iv): write |z|= |(z+w)−w| and use property (iii).]
1.10 Give a geometric justification of the triangle inequality:
|z1+ z2| ≤ |z1|+ |z2| ,
where z1 and z2 are any two complex numbers.
1.11 In each of the following cases, find the set of all points in the complex plane satisfy-
ing the given condition (describe the set, sketch it, and give its cartesian equation, if
appropriate).
Chapter 1: Real and Complex Numbers 25
a) Imz≥ 0
b) 0< Im(z+1)≤ 2pi
c) −1≤ Re z< 1
d) Re(iz) = 3
e) Re(z+2) =−1
f) |z−5|= 6
g) |z+2i| ≥ 1
h) |z+ i|= |z− i|
i) |z+3|+ |z+1|= 4
j) |z+3|− |z+1|=±1
1.12 If z is a variable complex number, mark clearly on an Argand diagram (i.e., on the
complex plane) the regions described by
a) Rez≥−2 and 0≤ Imz≤ 3
b) Rez≥−2 or 0≤ Imz≤ 3
c) 2< |z|< 3 and Rez< 2.5
d) |z−2+ i|> 1 and Rez> 2
e) 1< |z−2+ i|< 3 and Imz≥ 0.
C H A P T E R 2
Polar Forms of Complex Numbers
In the last chapter we introduced the set of complex numbers to solve quadratic equations
and showed how they can be represented as points on the complex plane using the Cartesian
form a+ ib. The polar forms introduced in this chapter simplify arithmetic calculations of
multiplication and division as well as the calculation of integer powers and roots of complex
numbers.
2.1 Standard Polar form
To position the complex number z= a+bi in the complex plane we used the real part a and
the imaginary part b as Cartesian coordinates on the plane. It is also possible to plot the same
number using polar coordinates, that is,
• the distance r = |z|=
√
a2+b2 of the point from the origin, called the modulus and
• the angle θ of the line from the point to the origin, measured anti-clockwise from the
positive real axis. This angle is called the argument of the complex number and denoted
arg(z) as shown in the figure below:
b
z= a+ ib
θ = arg(z)
r =
|z|
a
ib
r cosθ
r sinθ
Elementary trigonometry shows that
tan θ =
b
a
and therefore arg(z) = θ = tan−1
(b
a
)
,
26
Chapter 2: Polar Forms of Complex Numbers 27
where tan−1 is the inverse function of tan. It also shows that
a= r cosθ and b= r sinθ ,
and therefore,
z= a+ ib
= r cosθ + ir sinθ
= r(cosθ + isinθ).
Standard polar form
The "standard polar form" of a complex number z is given by
z= r(cosθ + isinθ)
where r is the modulus and θ its argument.
Recall that the form z = a+ ib introduced in the last chapter is called the "Cartesian form",
in which z is specified by its real part a and its imaginary part b.
Calculating the argument θ
Always plot the complex number to find θ .
Because tanθ has the same values in the first and third quadrants and in the
second and fourth quadrants it is essential that you plot the complex num-
ber on the complex plane when you are finding its argument. Otherwise
you may get the wrong value of θ .
For example, z1 = 2+ 2i is in the first quadrant and z2 = −2− 2i in the third quadrant.
However, tanθ1 =
2
2
= 1 and tanθ2 =
−2
−2 = 1 have the same value. The only way
to determine the correct quadrant is to plot z1 and z2 in the complex plane.
Examples 2.1a
i) Write −3+3i in polar form.
Here r = |−3+3i|=
√
(−3)2+32 =√18= 3√2.
Also, we find that tanθ =
3
−3 = −1. Using the calculator in radian mode, it will tell
you that tan−1(−1)≈−0.7854 which is −pi
4
.
This is not the correct angle. Plotting−3+3i on the complex plane gives the following
picture:
28 MATH 1021 Calculus of One Variable
b
θ
r
−3
3i
which shows that θ = arg(−3+ 3i) is an angle in the second quadrant. By inspection
we can see that arg(−3+3i) = 3pi/4. The diagram easily distinguishes between right
and wrong answers. So the standard polar form is
−3+3i= 3
√
2(cos3pi/4+ isin3pi/4) .
ii) Write −1−√3i in polar form.
The modulus is given by r=
√
(−1)2+(√3)2 =√1+3= 2. We plot−1−√3i in the
complex plane.
b
θ
r
−1
−√3i
We see that arg(−1−√3i) lies in the third quadrant. Since tanθ =√3, the value of θ
is 4pi/3. (We could also write θ = −2pi/3 equally correctly.) Therefore the standard
polar form is
−1−
√
3i= 2(cos4pi/3+ isin4pi/3).
iii) Find the modulus and argument of 3+7i.
The modulus is r =
√
32+72 =
√
58. In the complex plane 3+ 7i lies in the first
quadrant.
Chapter 2: Polar Forms of Complex Numbers 29
b
θ
3
7i
We find that tanθ =
7
3
and so θ = tan−1
7
3
≈ 1.17, therefore the polar form is
3+7i=
√
58
(
cos(tan−1
7
3
)+ isin(tan−1
7
3
)
)
≈
√
58
(
cos1.17+ isin1.17
)
.
iv) Write −29 in polar form.
Although−29 is a real number it can still be written in polar form. Clearly |−29|= 29
and from the complex plane we see arg(−29) = pi .
b
θ
−29
Hence −29= 29
(
cospi + isinpi
)
in polar form.
v) Convert 8
(
cos(−pi/6)+ isin(−pi/6)
)
to Cartesian form.
It is usually much simpler to convert a complex number from polar form to Cartesian
form than to convert a complex number from Cartesian to polar form. All that needs to
be done is to evaluate the cosine and sine and simplify the resulting expression. So
8
(
cos(−pi/6)+ isin(−pi/6)
)
= 8
(√
3
2
− 1
2
i
)
= 4
√
3−4i.
30 MATH 1021 Calculus of One Variable
b
pi
6
4
√
3
−4i
vi) Convert 5
(
cos(pi/2)+ isin(pi/2)
)
into Cartesian form.
Since cos(pi/2) = 0 and sin(pi/2) = 1, we obtain
5
(
cos(pi/2)+ isin(pi/2)
)
= 5
(
0+ i
)
= 5i
b
pi/2
5i
♦
Principal Argument Arg(z)
If we add 2pi to the argument arg(z) of a complex number z, we come back to the same point
on the complex plane. In fact, adding or subtracting any integer multiple of 2pi gives the same
complex number again. Therefore a complex number has an infinite number of arguments
which differ by integer multiples of 2pi .
The most general form of the argument may be expressed in the form
arg(z) = θ +2kpi where k ∈ Z,
where θ is any argument of z. This fact will become important when we take roots of complex
numbers later in this chapter.
Example 2.1b Find an argument θ of the complex number z = −1+ i and then write the
most general form of the argument.
First plot z in the complex plane as shown in the figure below.
Chapter 2: Polar Forms of Complex Numbers 31
b−1+ i
3pi
4
−5pi
4
11pi
4
We can choose θ = 3pi/4, 11pi/4, −5pi/4, and so on. The most general argument may be
expressed in any of the forms
arg(z) = 3pi/4+2kpi where k ∈ Z, or
arg(z) = 11pi/4+2kpi where k ∈ Z, or
arg(z) =−5pi/4+2kpi where k ∈ Z .
♦
To eliminate the problem of having an infinite number of arguments we make the following
definition.
Principal argument
The "principal argument" of z, denoted Arg (z), is the unique argument that
satisfies the condition
−pi < Arg(z)≤ pi.
Referring to Example 2.1b we see that only the argument θ = 3pi/4 satisfies the condition
−pi < θ ≤ pi and therefore the principal argument Arg (z) = 3pi/4.
32 MATH 1021 Calculus of One Variable
A note on special angles
In the examples above you will see that most of the polar angles that we used were angles with
exact sines or cosines, sometimes known as special angles. For example, any angle which is
a multiple of pi/2 has either sine or cosine equal to zero. So cos(pi/2) = 0, sin(pi/2) = 1 and
cospi =−1, sinpi = 0.
The angles pi/6 and pi/3, which correspond to 30 and 60 degrees respectively (and any angles
that are multiples of these) have special values for sine and cosine, as does pi/4 (45 degrees)
and its multiples.
You will have learnt about these special cases at high school. As they are used extensively
in this chapter, it is important that you revise them as soon as possible if you have forgotten
about them. You may find it helpful to look at the right-angle triangles with angle pi/4 or pi/3
and pi/6.
√
2
1
1
1
√
3 22
pi
4
pi
3
pi
6
It is easy to find sines and cosines from these triangles using the basic definitions
sinθ =
opposite
hypotenuse
cosθ =
adjacent
hypotenuse
tanθ =
opposite
adjacent
.
In summary„
• cos(pi/2) = 0 sin(pi/2) = 1,
• cos(pi) =−1 sin(pi) = 0,
• cos(pi/4) = 1/
√
2, sin(pi/4) = 1/
√
2
• cos(pi/3) = 1/2, sin(pi/3) =
√
3/2
• cos(pi/6) =
√
3/2, sin(pi/6) = 1/2.
Chapter 2: Polar Forms of Complex Numbers 33
2.2 Polar exponential form - Euler’s formula
There is a very useful expression known as Euler’s formula that we will prove in Chapter 8 on
Taylor series. It is named after Leonhard Euler, a Swiss mathematician, physicist, astronomer,
geographer, logician and engineer (1707 – 1783):
Euler’s formula
e iθ = cosθ + isinθ .
Using this formula we can write a complex number in polar exponential form as follows:
z= r(cosθ + isinθ) = r e iθ .
Polar exponential form
The polar exponential form of a complex number z= a+ ib is
z= r e iθ ,
where r =
√
a2+b2 is the modulus and θ = arg(z) is an argument.
Examples 2.2a The Cartesian, standard polar and polar exponential forms of the complex
numbers in Examples 2.1a are, respectively,
i) −3+3i = 3√2(cos3pi/4+ isin3pi/4) = 3√2e i(3pi/4)
ii) −1−√3i = 2(cos4pi/3+ isin4pi/3) = 2e i(4pi/3)
iii) 3+7i=
√
58
(
cos(tan−1 7
3
)+ isin(tan−1 7
3
)
)
=
√
58e i(tan
−1 7
3 )
iv) −29= 29(cospi + isinpi) = 29e ipi
v) 5
(
cos(pi/2)+ isin(pi/2)
)
= 5e i(pi/2) ♦
Of course, once we have the polar exponential form, the standard polar and the Cartesian
forms can be written down immediatelly using Euler’s formula.
34 MATH 1021 Calculus of One Variable
2.3 Arithmetic in polar form
The polar forms are particularly useful when multiplying or dividing complex numbers, or
raising a complex number to a power. However, using polar forms for addition and subtrac-
tion is more complicated and gives no extra insight into the problem. Therefore, we will not
consider addition or subtraction in polar form.
Multiplication and division
Let z= reiθ and w= seiφ be any two non-zero complex numbers. Then
Multiplication
zw= reiθ × seiφ = (rs)ei(θ+φ)
Multiply the moduli and add the arguments
Division
z
w
=
reiθ
seiφ
=
(r
s
)
e i(θ−φ)
Divide the moduli and subtract the arguments
Example 2.3a Let z= 6e i(pi/3) and w= 2e i(pi/6). Calculate zw and z/w.
zw= (6e i(pi/3)) (2e i(pi/6)) = (6×2)e i(pi/3+pi/6) = 12e i(pi/2)
z/w= 6e i(pi/3) /2e i(pi/6) = (6/2)e i(pi/3−pi/6) = 3e i(pi/6)
♦
In standard polar notation the above rules become
Multiplication
zw= r
(
cos(θ)+ i sin(θ)
)
s
(
cos(φ)+ i sin(φ)
)
= (rs)
(
cos(θ +φ)+ i sin(θ +φ)
)
Division
z
w
=
cos(θ)+ i sin(θ)
cos(φ)+ i sin(φ)
=
(r
s
)(
cos(θ −φ)+ i sin(θ −φ)
)
.
Notice that the polar exponential form is more intuitive and concise than the standard polar
form.
Chapter 2: Polar Forms of Complex Numbers 35
Raising to an integer power
For every positive integer n, we have
zn = (reiθ )n = (reiθ )× (reiθ )× . . . . . .× (reiθ ) = rneinθ .
In fact, this holds for all integers n, whether positive, negative or zero.
Integer power
zn = (reiθ )n = rneinθ .
To raise a complex number to any integer, raise the modulus to the integer and
multiply the argument by the integer.
In the special case when a complex number of modulus 1 is raised to an integer power, we
have "De Moivre’s theorem", named after the French mathematician Abraham De Moivre
(1667–1754).
De Moivre’s theorem
If a complex number has modulus 1 then
(
eiθ
)n
= einθ . Using Euler’s
formula on both sides of the equation gives
(cosθ + isinθ)n = cos(nθ)+ i sin(nθ), for any n ∈ Z .
This last expression is called De Moivre’s theorem.
Example 2.3b Use the polar exponential form to find z8 when z = 1+
√
3i. Write the final
answer in Cartesian form.
We find that |z|= 2 and arg(z) = pi/3, so z= 2 e i pi/3. Hence
z8 =
(
2 eipi/3
)8
=28 e i 8pi/3
=256 e i 8pi/3
=256 e i 2pi/3 since
(
2pi/3= 8pi/3−2pi)
=256
(
cos2pi/3+ isin2pi/3
)
=−128+128
√
3 i.
♦
Notice that to write the final answer in Cartesian form, it is easier to transform the complex
exponential first to the standard polar form.
36 MATH 1021 Calculus of One Variable
Example 2.3c Use the polar exponential form to find z2w3 and z2/w3
when z= 2 e i pi/4 and w= 3 e i 3pi/2.
We first calculate z2 and w3, to obtain
z2 = 4 e i pi/2 and w3 = 27 e i 9pi/2.
Therefore
z2w3 = 4 e i pi/2×27 e i 9pi/2 = 108 e i 5pi = 108
(
cos5pi + i sin5pi
)
=−108.
Similarly,
z2
w3
=
4 e i pi/2
27 e i 9pi/2
=
4
27
e−i 4pi =
4
27
.
♦
Example 2.3d Simplify e i 15pi/7.
We have
e i 15pi/7 = e i
(
pi/7+2pi
)
= e i pi/7× e i 2pi
= e i pi/7
(
cos2pi + isin2pi
)
= e i pi/7 (1+ i 0)
= e i pi/7.
♦
The last example demonstrates once again that arguments are only determined up to integer
multiples of 2pi; that is, e i θ is the same as e i φ when θ and φ differ by an integer multiple of
2pi . It is useful to remember that e2pii = 1 and that for every integer n, e2npii = 1.
Equality of complex numbers
If r e i θ = s e i φ , then r = s and θ = φ +2kpi , for some integer k.
Examples 2.3e
i) Find (2+2i)(1−√3i) in polar exponential and standard polar forms.
First, let us put both numbers into polar form. This simplifies the multiplication and
we will also need these numbers in polar form for the next example. It is essential to
draw a diagram:
Chapter 2: Polar Forms of Complex Numbers 37
b
b
θ
φ 21
2i
−√3i
Here |2+2i|=√4+4=√8= 2√2 and |1−√3i|=√1+3= 2. From the diagram,
θ = arg(2+2i) is in the first quadrant and φ = arg(1−√3i) is in the fourth quadrant.
Since tanθ = 1, θ = pi/4 and since tanφ =
√
3, φ =−pi/3. So we have
(2+2i)(1−
√
3i) = 2
√
2 e ipi/4 2e−ipi/3
= ei
(
pi/4+(−pi/3)
)
= 4
√
2e−i(pi/12)
= 4
√
2
(
cos(−pi/12)+ i sin(−pi/12)).
ii) Find (2+2i)/(1−√3i).
The numbers (2+2i) and (1−√3i) are already in polar form from the previous exam-
ple.
(2+2i)
(1−√3i) =
2
√
2e ipi/4
2e−ipi/3
=
√
2e
i
(
pi/4−(−pi/3)
)
=
√
2e i(7pi/12).
iii) Find
(
(2+2i)/(1−√3i))6.
The quotient has already been calculated in polar form in the previous example.
(
(2+2i)
(1−√3i)
)6
=
(√
2e i(7pi/12)
)6
=
(
2
1
2
)6
e i(7pi/2)
= 23 e i(7pi/2).
♦
38 MATH 1021 Calculus of One Variable
2.4 Roots of complex numbers
Recall the process of finding the nth root of a positive real number: we say that x is a nth root
of a if xn = a and we write x= a1/n. For example, 3= 91/2 because 9= 32.
By analogy with roots of real numbers, the nth root of a complex number w is a complex
number z such that zn = w and we write z= w1/n.
Every non–zero complex number (which includes every real number) has two complex square
roots, three complex cube roots, four complex fourth roots and so on. In general:
Roots of complex numbers
Every non–zero complex number has exactly n distinct complex nth roots.
Therefore if we seek to find all cube roots of a complex number, for example, we know that
there will be three of them. Knowing how many roots to look for is useful in deciding how
many different values of k to use in finding the roots explicitly.
IMPORTANT –When finding roots of complex numbers, express it in polar
form z = eiφ first, where φ is any value of the argument and then add an
integer multiple of 2pi ,
z= eφ+2kpi , k ∈ Z.
Then calculate
z1/n = e(φ+2kpi)/n.
The n complex roots are obtained by taking n consecutive values of k.
Example 2.4a Find all fifth roots of −√3− i, that is, all z such that z5 =−√3− i.
First, we put −√3− i into polar form. The modulus of −√3− i is |−√3− i|= 2.
Plotting −√3− i on the complex plane we see that the principal argument
of −√3− i is −5pi
6
.
b
−√3
−i
−5pi
6
Chapter 2: Polar Forms of Complex Numbers 39
Therefore
−
√
3− i= 2e−i
(
5pi/6
)
= 2e−i
(
5pi/6+2kpi
)
.
Taking the fifth root, gives
z= 21/5 e−i
(
5pi/6+2kpi
)
/5 = 21/5 e−i
(
pi/6+2kpi/5
)
.
For k = 0,1,2,3,4. we obtain the five different values of z:
z= 21/5 e−i
(
pi/6
)
,
z= 21/5 e−i
(
17pi/30
)
,
z= 21/5 e−i
(
29pi/30
)
,
z= 21/5 e−i
(
41pi/30
)
,
z= 21/5 e−i
(
53pi/30
)
.
These are the five fifth roots of −√3− i.
♦
Example 2.4b Find the cubic roots of w=−2+2i.
Write the number in polar form: The modulus is
|w|= |−2+2i|=
√
8.
The diagram shows that the principal argument is θ = 3pi/4. However, remember that when
finding roots the most general form of the argument must be used by adding integer multiples
of 2pi , that is,
θ = 3pi/4+2kpi, k ∈ Z .
b
3pi
4
-2
2i
Therefore,
z=
[√
8 ei(3pi/4+2kpi)
]1/3
=
(√
8
)1/3
e
i
(
3pi/4+2kpi
)
/3
=
√
2 e
i
(
pi/4+2kpi/3
)
=
√
2 eiφ where φ = pi/4+2kpi/3.
40 MATH 1021 Calculus of One Variable
Since k can be any integer it appears at first sight that there are infinitely many complex num-
bers z whose cube is −2+ 2i. However, close inspection shows that after three consecutive
values of k we come back to the same point in the complex plane. To see this, let k = 0,1,2.
When k = 0, we obtain
z =
√
2 ei pi/4
=
√
2(cospi/4+ isinpi/4)
=
√
2
( 1√
2
+ i
1√
2
)
= 1+ i.
Plotting this answer on the complex plane we get:
b
pi
4
1
i 1+ i
When k = 1, we obtain
z =
√
2 e
i
(
pi/4+2pi/3
)
=
√
2 e
i
(
11pi/12
)
=
√
2
(
cos
11pi
12
+ isin
11pi
12
)
When k = 2, we obtain
z =
√
2 e
i
(
pi/4+4pi/3
)
=
√
2 e
i
(
19pi/12
)
=
√
2 e
−i
(
5pi/12
)
(after subtracting 2pi),
=
√
2
(
cos
(−5pi
12
)
+ isin
(−5pi
12
))
If we choose other values of k it turns out that we simply replicate one of the three values of
z that we’ve already calculated. For example, if k =−1, then
z =
√
2 e
i
(
pi/4+(−2pi/3)
)
=
√
2 e
−i
(
5pi/12
)
,
Chapter 2: Polar Forms of Complex Numbers 41
which is one of the values already found.
If all three distinct solutions are plotted on the complex plane we see that all lie on the circle
of radius
√
2 centred on the origin, and each is separated from the others by an angle of 2pi/3.
b
b
b pi
4
−5pi
12
11pi
12
2pi
3
2pi
3
2pi
3
√
2
√
2i
−√2
−√2i
♦
Complex roots of real numbers
How many complex roots does a real number have? Let us look at the fourth roots of 16. You
already know that 24 = (−2)4 = 16. Hence 2 and −2 are fourth roots of 16. Are there other
fourth roots?
Write the number 16 in polar form: The modulus is 16 and the principal argument is 0. The
most general form of the argument is θ = 0+2kpi, k ∈ Z, therefore
16= 16 ei
(
0+2kpi
)
=⇒ z= 161/4 ei
(
0/4+2kpi/4
)
= 2 ekpi/2
where k is any integer. We shall now choose various values of k to find explicit values of z.
When k = 0 we obtain
z= 2ei0 = 2(cos0+ isin0) = 2.
When k = 1 we obtain
z= 2eipi/2 = 2
(
cos
pi
2
+ isin
pi
2
)
= 2i.
When k = 2 we obtain
z= 2eipi = 2(cospi + isinpi) =−2,
42 MATH 1021 Calculus of One Variable
and when k = 3 we obtain
z= 2ei3pi/2 =
(
cos
3pi
2
+ isin
3pi
2
)
=−2i.
All other values of k give one of the four answers already found, namely ±2,±2i. For
example, if k = 7, we obtain z= 2
(
cos7pi/2+ isin7pi/2
)
=−2i.
Therefore, 16 has two real fourth roots
(± 2) but it has four complex fourth
roots
(±2,±2i). When these roots are plotted on the complex plane, they all
lie on the circle of radius 2 centred at 0, spaced pi/2 apart.
b
b
b
b
pi
2
pi
2
pi
2
pi
2
2
2i
−2
−2i
2.5 Roots of polynomial equations
In Chapter 1 we looked at the solutions of quadratic equations to motivate the introduction of
complex numbers. Quadratic equations are a special case of more general equations called
polynomial equations.
Polynomial expressions
A polynomial in z is an expression of the form
anz
n+an−1zn−1+an−2zn−2+ · · ·+a1z+a0
where z is the "variable" and the numbers an,an−1, . . . ,a0 are the coefficients.
Polynomial equations
If we set the expression equal to zero, we obtain a polynomial equation
anz
n+an−1zn−1+an−2zn−2+ · · ·+a1z+a0 = 0.
Chapter 2: Polar Forms of Complex Numbers 43
If an 6= 0 then the polynomial is said to have "degree" n. The term anzn is known as the
"leading term".
Roots of polynomial equations
The "roots of a polynomial equation" are the numbers z which satisfy
anz
n+an−1zn−1+an−2zn−2+ · · ·+a1z+a0 = 0.
A polynomial equation of degree n has at most n complex roots. All, some or
none of these roots may be real.
For example, the problem “solve z4−16= 0 over the real numbers”, or equivalently,
find the real roots of z4−16= 0
has the answer z = 2 or − 2. If the problem is changed slightly to read “solve z4− 16 = 0
over the complex numbers”, or
find the complex roots of z4−16= 0,
then the correct answer is z = 2, 2i,−2 or − 2i. Clearly this polynomial equation has two
real roots but has four complex roots.
Complex roots of real polynomials
Polynomial equations of degree n have at most n distinct complex roots, as some roots might
be repeated. For example, the polynomial
z2−2z+1= (z−1)2
has a double root at z= 1, also called a root of multiplicity 2.
If n is a positive integer then the equation zn = 0 has a root of multiplicity n at z = 0. It is
true that every polynomial equation of degree n has exactly n complex roots, counted with
multiplicity.
We have already seen in Chapter 1 that when a quadratic equation with real coefficients has
non–real complex roots, then these roots come in complex conjugate pairs. So for example,
the equation z2+4z+5= 0 has roots z=−2+ i and z=−2− i.
44 MATH 1021 Calculus of One Variable
In fact:
Complex roots of real polynomials
If the coefficients in a polynomial equation are all real then all of the non–real
complex roots occur in complex conjugate pairs.
For example, the polynomial equation z4− 16 = 0 that was discussed earlier in this chapter
has two real roots (2 and −2) and two imaginary roots (2i and −2i), and these imaginary
roots are complex conjugates of each other. The coefficients of this polynomial, 1 and −16
are both real and so we expect complex roots will occur in complex conjugate pairs.
By contrast, the polynomial z3− i= 0, does not have all real coefficients; the coefficients are
1, which is real, and −i, which is not real. The roots of z3− i= 0 are
√
3
2
+
1
2
i,
−√3
2
+
1
2
i, −i.
Although they are all non–real complex numbers, they do not occur in complex conjugate
pairs.
Complex conjugate roots
If one complex root of a polynomial equation with real coefficients is known
then its complex conjugate can immediately be written down to give another
root.
Example 2.5a Find all roots of z4−18z2+192z−175= 0, given that 3−4i is a root.
Since the coefficients of the polynomial are real, if 3−4i is a root, then its complex conjugate
3+4i is also a root. We construct the quadratic expression with these two roots,[
z− (3−4i)][z− (3+4i)]= z2−6z+25.
Next, we use polynomial long division to divide this quadratic into the original polynomial
as follows,
z2 +6z −7
z2−6z+25 )z4 −18z2 +192z −175
z4 −6z3 +25z2
6z3 −43z2 +192z
6z3 −36z2 +150z
−7z2 +42z −175
−7z2 +42z −175
0
Chapter 2: Polar Forms of Complex Numbers 45
Therefore
z4−18z2+192z−175
z2−6z+25 = z
2+6z−7. The quotient is another quadratic which
can be easily factorized z2+6z−7= (z−1)(z+7) and therefore
z4−18z2+192z−175 = (z2+6z−7)(z2−6z+25)
= (z−1)(z+7)[z− (3−4i)][z− (3+4i)].
So the four roots of the original degree 4 polynomial are: 1, −7, 3−4i, 3+4i.
♦

Technical aside It is not difficult to prove rigorously that if a polynomial equation with
real coefficients has complex roots then these roots occur in complex conjugate pairs.
First we need to show that z+w= z+w and that zw= zw. Try doing this by writing z= a+ ib
and w= c+ id and calculating z+w and zw.
Then, let us consider a polynomial
p(z) = anz
n+an−1zn−1+an−2zn−2+ · · ·+a1z+a0
where an, an−1, an−2, . . . , a0 are all real.
Suppose there is some complex number v which is a root of the polynomial, so that p(v) = 0.
If we take complex conjugates of both sides of the equation we have p(v) = 0= 0 and hence
0 = p(v)
= anvn+an−1vn−1+an−2vn−2+ · · ·+a1v+a0
= anvn+an−1vn−1+an−2vn−2+ · · ·+a1v+a0
= an vn+an−1 vn−1+an−2 vn−2+ · · ·+a1 v+a0
= anvn+an−1vn−1+an−2vn−2+ · · ·+a1v+a0
= an(v)
n+an−1(v)n−1+an−2(v)n−2+ · · ·+a1(v)+a0
= p(v)
Therefore since p(v) = 0, v is a root. That is, both v and its conjugate v are roots.
As you read through the above proof try to work out why each line follows from the previous
line. ⊳
2.6 Sine and cosine in terms of exponentials
Using Euler s formula, we have for all real θ ,
(2.6a) eiθ = cosθ + isinθ .
46 MATH 1021 Calculus of One Variable
Replacing i with −i, gives
(2.6b) e−iθ = ei(−θ) = cos(−θ)+ isin(−θ) = cosθ − isinθ .
Adding equations (2.6a) and (2.6b) we obtain
eiθ + e−iθ = cosθ + isinθ + cosθ − isinθ = 2cosθ ,
which rearranges to give cosθ =
eiθ + e−iθ
2
.
Subtracting equations (2.6a) and (2.6b) gives
eiθ − e−iθ = cosθ + isinθ − cosθ + isinθ = 2isinθ ,
which rearranges to give sinθ =
eiθ − e−iθ
2i
.
Sine and cosine in terms of exponentials
For all real θ , cosθ =
eiθ + e−iθ
2
and sinθ =
eiθ − e−iθ
2i
.
These expressions for cosθ and sinθ in terms of eiθ will be used in Chapter 11 to derive
trigonometric identities which can be used to solve integration problems involving powers of
cosθ and sinθ .
2.7 Complex exponential function
We have defined eiθ as cosθ + isinθ for any θ ∈ R. This has given us a way of calculating
complex exponentials where the exponent is a purely imaginary number. So, for example,
e3i = cos3+ isin3≈−0.989+0.141i.
Note that we are using radian measure, not degrees. The next step is to extend this so we can
define exponentials of any complex number.
Complex exponential function
Let z= x+ iy be a complex number with x and y real, the complex exponential
function is defined as
ez = ex+iy = exeiy = ex
(
cosy+ isiny
)
.
Chapter 2: Polar Forms of Complex Numbers 47
Study this definition carefully, especially the last expression:
ez = ex(cosy+ isiny).
Notice that since x and y are real and ex is real and positive, this displays ez as a complex
number in polar form. We can therefore read off the modulus and argument of ez.
Modulus and argument of ez
When z= x+ iy with x and y real, |ez|= ex and arg(ez) = y.
The usual rules for multiplying, dividing and taking integer powers still apply. If z = x+ iy
and w = u+ iv we know that z+w = (x+ u)+ i(y+ v), z−w = (x− u)+ i(y− v) and nz =
nx+ iny. Hence
ez× ew = (ex eiy)(eu eiv) = ex+u ei(y+v) = ez+w,
ez
ew
=
ex eiy
eu eiv
= ex−uei(y−v) = ez−w,
(ez)n = (ex eiy)n = (ex)n (eiy)n = enx einy = enz, for any integer n.
We can now calculate the value of ez for any complex number z.
Example 2.7a Express the complex exponentials e0, e2+4i, e−1+ipi/4, e−1+i17pi/4 and ex
where x is real, as complex numbers in Cartesian form.
Using the definition of ez, we obtain
e0 = e0+0i = e0(cos0+ isin0) = 1(1+ i0) = 1,
e2+4i = e2(cos4+ isin4) = e2 cos4+ ie2 sin4≈−4.83−5.59i,
e−1+ipi/4 = e−1(cos
pi
4
+ isin
pi
4
) = e−1 cos
pi
4
+ ie−1 sin
pi
4
≈ 0.26+0.26i,
e−1+i17pi/4 = e−1(cos
17pi
4
+ isin
17pi
4
) = e−1 cos
17pi
4
+ ie−1 sin
17pi
4
≈ 0.26+0.26i,
ex = ex+i0 = ex(cos0+ isin0) = ex(1+ i0) = ex.
That the third and fourth results are equal should be no surprise, since pi/4 and 17pi/4 differ
by an integer multiple of 2pi . The last result shows that when z equals the real number x, the
complex expression ez agrees with the usual real exponential ex. ♦
48 MATH 1021 Calculus of One Variable
Example 2.7b Find a complex number z such that ez = 1+ i.
First write 1+ i in polar exponential form. We have 1+ i =
√
2eipi/4, and so we require
z = x+ iy such that ez = exeiy =
√
2eipi/4. This gives us ex =
√
2, from which we find that
x= 1
2
ln2, and y= pi/4+2kpi , where k is any integer.
We have in fact found infinitely many appropriate values of y to put together with a uniquely
determined value of x. That is, there are infinitely many values of z satisfying ez = 1+ i. Here
are some of them:(1
2
ln2
)
+ i
(
pi/4
)
,
(1
2
ln2
)
+ i
(
9pi/4
)
,
(1
2
ln2
)
+ i
(
17pi/4
)
,
(1
2
ln2
)− i (7pi/4), (1
2
ln2
)− i (15pi/4)
and so on. Some are plotted in the complex plane in the figure below. (Note that the scales
on the horizontal and vertical axes are different).
1
2
ln2+ ipi
4
1
2
ln2− i7pi
4
1
2
ln2+ i9pi
4
1
2
ln2+ i−15pi
4
4pii
2pii
−2pii
−4pii
0
b
b
b
b
♦
The previous example illustrates a most interesting property of the complex exponential func-
tion, namely its periodicity.
All points whose imaginary parts differ by integer multiples of 2pi are mapped to the same
point by the complex exponential function, because e2kpii = 1 for every integer k, and hence
for all z ∈ C, ez = ez+2kpii for any integer k. In particular, ez = ez+2pii for all z ∈ C, indicating
that ez is periodic with period 2pii.
The complex exponential function is a periodic function, with period 2pii.
Chapter 2: Polar Forms of Complex Numbers 49
An amazing formula
If we substitute θ = pi into Euler’s formula eiθ = cosθ + i sinθ , it gives
eipi = cospi + isinpi =−1.
Rearranging we obtain
A remarkable formula
eipi +1= 0.
This is a most remarkable expression. One equation of great simplicity contains five of the
most important numbers in mathematics: e, pi, i, 1 and 0. What is also remarkable is the fact
that these five numbers come from very different areas of mathematics, like pi from geometry,
e from calculus and i from solving quadratic equations.
Summary of Chapter 2
• This chapter introduced three ways of expressing a complex number:
a) Cartesian form z= a+ ib
b) standard polar form z= r(cosθ + isinθ)
c) polar exponential form z= reiθ
• The Cartesian form is convenient when adding or subtracting complex
numbers.
• The polar forms greatly simplify multiplication, division and the cal-
culation of powers and roots of complex numbers.
• Euler’s formula eiθ = cosθ + isinθ allows trigonometric functions to
be written in complex form and motivates the definition of the complex
exponential function.
• If one complex root of a polynomial equation with real coefficients is
known then its complex conjugate can immediately be written down to
give another root.
• The expressions for cosθ and sinθ in terms of e± iθ can be used to
derive trigonometric identities which are helpful in solving certain types
of integration problems.
50 MATH 1021 Calculus of One Variable
Exercises
2.1 For each of the following numbers, give the numerical value of the real part x, the
imaginary part y, the modulus r and the principal value of the argument θ . Plot the
number as a point in the complex plane.
a) 1− i√3
b) 1/(1− i)
c) (i+
√
3)2
d) 2(cos(pi/6)+ isin(pi/6))
e)
(
1+ i
1− i
)2
f)
3+ i
2+ i
2.2 Write each of the following complex numbers in polar form.
−4i, −2+2i, 1− i.
Use your results to perform the following operations in polar form.
a) (−2+2i)(1− i)
b) −4i/(−2+2i)
c) (1− i)6
d) (−2+2i)15
2.3 Use de Moivre’s theorem to simplify
a)
(
cos(2pi/3)+ isin(2pi/3)
)9
b)
(
cos(pi/3)+ isin(pi/3)
)4 c)
(
cos(2pi/3)− isin(2pi/3)
)6
d)
(
sin(2pi/3)+ icos(2pi/3)
)9
.
2.4 Write the following complex numbers in the form reiθ :
a) 1+ i
b) −√3− i
c) −2i
d) −3
2.5 Calculate (−√3− i)12.
2.6 Find
∫ pi/4
0
sin4θ dθ using the expression for sin4θ in terms of cos4θ and cos2θ in
Example 12.1c.
2.7 Find a formula for cos6θ in terms of powers of sinθ and cosθ .
2.8 Find all complex numbers z such that ez = i.
2.9 Recall that if p(z) is a polynomial with real coefficients and if w ∈ C is a root of p(z)
then so is w.
Find the roots of the quadratic equation
q(z) = z2−3(1+ i)z−2+6i= 0.
Verify that ifw is a root of q(z) thenw is not a root. Explain why this does not contradict
the statement at the start of this question.
Chapter 2: Polar Forms of Complex Numbers 51
2.10 Find all the roots of f (z) = z4−3z3+7z2+21z−26, given that 2−3i is a root.
2.11 Find all the roots of z4−5z3+4z2+2z−8, given that 1− i is a root.
C H A P T E R 3
Functions
The concept of function is fundamental to the study of all branches of mathematics. In this
chapter we review the definition of a function and discuss some of their important properties.
3.1 Functions – definitions and examples
You would have often used the phrase “y is a function of x”, or the equation y = f (x), to
indicate that y is a variable which depends on the variable x. If we denote the values over
which x can vary as the set A, and if we observe that the values of y belong to a set B, then
we say that f is a function from A to B, written f : A→ B.
Definition of function
If A and B are sets then a function from A to B (written f : A→ B) is a rule
f which assigns to each element x in A exactly one element in B, denoted by
f (x).
We call f (x) the image of x under f and we say f maps x to f (x) or x is
mapped to f (x) by f .
The rule f is sometimes a simple equation giving f (x) in terms of x, such as f (x) = x2.
However, the rule may take other forms. In the case where A is a finite set, for example, the
rule may take the form of a list of the values of f (x), as x runs through A.
Domain, Codomain and Range of a function
Given a function f from A to B, f : A→ B, we call A the domain and B the
codomain of the function.
The range of f is the set of values { f (x) | x ∈ A}. Thus the range is always
the set B or a proper subset of B.
It is sometimes useful to think of a function f as a sort of machine which accepts an element
from the domain as input, processes it, and produces an output. The set of all possible outputs
forms the range.
52
Chapter 3: Functions 53

Figure 3.1: In the diagram A is the Domain and B the Codomain of f (x). In this example the
Range is a proper subset of the Codomain B.
Example 3.1a Let A = {a,b,c,d}, B = {1,2,3,4,5} and consider the rule f given by the
list of values
f (a) = 2, f (b) = 1, f (c) = 4, f (d) = 4.
Then f is a function from A to B with domain A = {a,b,c,d}, codomain B = {1,2,3,4,5}
and range {1,2,4}.
A B
a
2b
1
c 4
d
3
5 ♦
Example 3.1b Consider the function f : R→ R given by the parabola f (x) = x2. In this
case the domain of f is R, the codomain is also R and the range consists of the positive real
numbers f (x)≥ 0.
♦
54 MATH 1021 Calculus of One Variable
Functions from R to R
Many of the functions studied in elementary calculus have domain the set of real numbers R
(or subsets of R), and outputs which are also real numbers. These are the functions we study
in this subject.
Natural domain
In circumstances when the domain is not specified explicitly, it should be
assumed that the domain of f is the set of all real numbers x for which the
defining formula gives a well defined, unique real number as the value for
f (x). Such a set of numbers is called the "natural domain" of f .
Examples 3.1c
i) Assuming that the domain is a subset of R, the natural domain of the function given by
the formula f (x) =
1
x
is the set {x∈R |x 6= 0}. Zero cannot be included in the domain,
since there is no real number
1
0
.
ii) In order to determine the natural domain of the function given by the formula
f (x) =
√
x2−4, where x is real, note that we must have x2− 4 ≥ 0 (since we cannot
take the square root of a negative number). That is, we must have x≥ 2 or x≤−2. The
natural domain is therefore {x ∈ R | |x| ≥ 2}= (−∞,−2] ⋃ [2, ∞). ♦
Other types of functions
Not all functions have inputs and outputs in the set of real numbers R. We have already seen
some examples of functions where the domain is the set of complex numbers. Consider the
following examples of functions, with their natural domains and their ranges.
Examples 3.1d
i) The function f : N→ N given by f (x) = x2 has domain the set of natural numbers. Its
range is the set of all perfect squares, {0,1,4,9, . . .}.
ii) The function f : C→ C with f (z) = z2 has domain the set of complex numbers. The
outputs are also complex numbers. What is the range? We’re interested to find out if
the set of outputs occupies the whole of C or if it is confined to a proper subset of C. It
turns out that the range is C.
iii) The function f : R→ C defined by f (x) = cosx+ isinx is a function from the real
numbers to the complex numbers. Note that the modulus of cosx+ isinx is always
equal to 1, whatever the value of x, and so the range of f is the set of complex numbers
with modulus 1.
Chapter 3: Functions 55
iv) The function r : R → V defined by r(t) = cos t i+ sin t j+ t k is a vector function,
with domain the set of real numbers and outputs a subset of the set of vectors in 3-
dimensional space. ♦
Modulus function
Considering the function f : C→ R given by the formula f (z) = |z| for all z in C. It is easy
to see that every output is a non-negative real number. The modulus of 0 is 0 and every other
complex number has positive modulus, since the modulus of z measures the distance from
the origin to z in the complex plane. For this function the range of f is equal to the set
{x ∈ R | x≥ 0}= [0,∞).
Natural domain and range of ez
Let’s now return to the complex exponential function which we defined earlier in the chapter.
It is a function f : C→ C given by the formula f (z) = ez, or more usefully,
f (z) = f (x+ iy) = ex(cosy+ isiny),
where z= x+ iy in Cartesian form and ex is just the ordinary real exponential function of the
real variable x.
Recall that |ez|= ex. Therefore no matter which complex number z we select as an input, the
complex number ez has modulus which is strictly positive, never zero. This tells us that the
complex number 0 is never an output for the complex eponential function, so the range is not
equal to the whole of C but must be a proper subset of C.
We now make use of another fact previously mentioned, namely that argez = y. Since y can
be any real number, the outputs ez can have any argument we please simply by selecting that
value as the value of y.
Putting all this information together shows that ez can have any positive modulus and, inde-
pendently, any argument. That is, the complex exponential function has natural domain C
and range {w ∈ C | w 6= 0}.
Pictures of functions
Recall that the "graph" of a function f : R→ R is the set of points {(x, f (x)) | x ∈ R} in the
ordinary Cartesian plane; we say that this is the graph y = f (x). Graphical representations
are possible because both the inputs and outputs of functions from R to R can be represented
as points on a Cartesian plane.
56 MATH 1021 Calculus of One Variable
Vertical line test
One special property of the graph of a functions is that every line x= a parallel
to the y-axis intersects the graph in exactly one point. The uniqueness of the
point of intersection is a consequence of the requirement in the definition of a
function that, given a in the domain of f , f (a) must be unique.
A curve which is such that some vertical line intersects it more than once is
not the graph of a function.
Graph of a function.
Vertical line
intersects the
graph twice.
Not a function.
If A and B are finite sets, it is sometimes useful to represent a function f : A→ B by an "arrow
diagram". For example, the arrow diagram of the function f in Example 3.1a is:
A B
a
2b
1
c 4
d
3
5
Here, if x is in A and y is in B then we draw an arrow from the dot labelled x to the dot
labelled y when f (x) = y. Note that there is exactly one arrow emanating from each of the
dots representing the elements of A, and each arrow lands on a dot representing an element
of B.
3.2 Combining functions
Suppose f and g are two functions whose ranges are contained in a set in which it is possible
to add, subtract, multiply and divide. The functions f and g can then be combined in vari-
ous ways to form new functions. In particular, they can be added, subtracted, multiplied or
divided to form the functions f +g, f −g, f g or f/g respectively.
Chapter 3: Functions 57
The functions f +g, f −g, f g have formulas ( f +g)(x) = f (x)+g(x), ( f −g)(x) = f (x)−
g(x), and ( f g)(x) = f (x)×g(x). These functions are defined for all values of x which lie in
the domain of f and also in the domain of g. That is, if the domain of f is A and the domain
of g is B, then the domain of each of f +g, f −g, f g is the intersection of A and B, namely
A∩B.
The function f/g is given by the formula ( f/g)(x) =
f (x)
g(x)
and is defined for all values of
x which lie in the domain of f and also in the domain of g, such that g(x) 6= 0; that is, the
domain of f/g is {x ∈ A∩B | g(x) 6= 0}.
Composite functions
Another way to combine two functions is to allow one function to operate on the output of
the other. The result is known as a composite function.
Composite functions
Given two functions f and g, the "composite function" g◦ f is defined by the
expression (g◦ f )(x) = g( f (x)) for all x in the domain of f such that f (x) is
in the domain of g.
Note that the output u = f (x) of the function f is used as the input of the function g. If the
range of f is a subset of the domain of g, then (g◦ f )(x)will be defined for all x in the domain
of f .
Examples 3.2a
i) Suppose f (x) =
√
x, and g(x) = 1−x2. The natural domain of f is {x ∈R | x≥ 0}, and
the natural domain of g is R. The range of f is a subset of R, and so the domain of the
composite g◦ f is equal to the domain of f .
(g◦ f )(x) = g( f (x)) = g(√x) = 1− (√x)2 = 1− x for x≥ 0.
Note that the natural domain of the function 1− x is R. Here, however, we are consid-
ering it as the composite function g◦ f where f (x) =√x, and so the domain is [0, ∞).
On the other hand, the range of g is {x∈R | x≤ 1}, which is not a subset of the domain
of f . The only values of x for which f ◦g is defined are those such that g(x) = 1−x2 ≥
0; that is, for −1≤ x≤ 1.
( f ◦g)(x) = f (g(x)) = f (1− x2) =
√
1− x2 for |x| ≤ 1.
Note that in general f ◦g 6= g◦ f . Note also that g◦ f is quite different from the product
g f of g and f . In this example, (g f )(x) = g(x)× f (x) =√x(1− x2).
ii) If f (x) = ex and g(x) = cosx, then (g ◦ f )(x) = cos(ex), ( f ◦ g)(x) = ecosx, and
( f g)(x) = (g f )(x) = ex cosx. ♦
58 MATH 1021 Calculus of One Variable
3.3 Injective and inverse functions
Here we revise the important concept of injective or ono-to-one functions. This is the crucial
property of a function that guarantees the existence of an inverse.
Injective functions
A function f : A→ B is said to be injective or one-to-one on the domain A if
distinct elements in A are mapped to distinct elements in B; that is, if x1 6= x2
implies that f (x1) 6= f (x2), for all x1, x2 ∈ A.
Equivalently, f is injective if f (x1) = f (x2) implies that x1 = x2, for all
x1, x2 ∈ A.
Examples 3.3a
i) The function f with domain R defined by f (x) = 3x+2 is injective, since if 3x1+2=
3x2+2 then x1 = x2, for all pairs x1, x2.
ii) The function f with domain R defined by f (x) = x2 is not injective since, for example,
f (−2) = f (2). This also shows that the function with domain C and formula f (z) = z2
is not injective, since R⊂ C.
iii) The sine function with domain R is not injective. If sinx1 = sinx2, then x1 is not
necessarily equal to x2. However the sine function with domain [−pi2 , pi2 ] is indeed
injective. (Draw the graph!) This example illustrates the fact that the property of
injectivity depends on the domain we choose.
iv) The complex exponential function with domain C given by f (z) = ez is not injective
because ez = ez+2pii.
v) The arrow diagram on the left represents an injective function, but that on the right does
not.
♦
Chapter 3: Functions 59
Given the graph of a function from R to R, the "horizontal line test" is a simple way to
determine whether or not the function is injective.
The horizontal line test
A function f :R→R is injective if no horizontal line intersects its graph more
than once.
Examples 3.3b
i) Any horizontal line intersects y= x exactly once so this function is injective on R.
ii) There are horizontal lines which intersect y = x2 twice, and so the parabola is not an
injective function on R. ♦
The last example shows that in order to find an inverse of y= x2 the domain must be restricted
to some subset of R so that on the new domain the function is injective.
Inverse functions
Suppose that the function f : A→ B has domain A and range B, and that on this domain, f
is an injective function. Then we may define an inverse function f−1 : B→ A with domain B
and range A. The inverse function f−1 has the following property:
Property of the inverse function
f−1( f (x)) = x for all x ∈ A and f ( f−1(x)) = x for all x ∈ B. This means that
f−1(y) = x if and only if y= f (x).
The notation f−1 always means the inverse function. It does not denote the
reciprocal 1/ f (x).
The inverse function f−1 “undoes” what the function f does, and viceversa. That is, if x ∈ A
and y ∈ B, then y = f (x) if and only if x = f−1(y). Thus to obtain a formula for the inverse
function we must solve y= f (x) for x, that is, make x the subject of the equation.
If f is a real valued invertible function of one real variable, the point (x,y) is on the graph of
f if and only if (y,x) is on the graph of f−1. This is because
(x,y) is on the graph of f ⇐⇒ y= f (x)
⇐⇒ x= f−1(y)
⇐⇒ (y,x) is on the graph of f−1.
60 MATH 1021 Calculus of One Variable
Hence the graph of such a function and the graph of its inverse are always reflections of one
another in the line y= x.
The following is a summary of the procedure for finding an inverse function formula for f (x).
Calculating the inverse function of f (x)
• If f is not injective on its natural domain, we must restrict the domain to
some subset A on which f is injective. This involves choosing a section
of the graph of f where the function is strictly increasing or strictly
decreasing.
• The formula for the inverse function f−1 : B→ A is found by rearrang-
ing the equation y= f (x) to make x the subject, x= f−1(y).
• Finally, we swap x and y to get the inverse in standard notation with x
and y the independent and dependent variables, respectively.
Example 3.3c Possibly the most well-known pair of inverse functions are the exponential
and logarithm functions.
The function f :R→ (0,∞), given by f (x) = ex, is clearly injective. Notice that with domain
R, the range of f is (0,∞). Its inverse, f−1 has domain (0,∞) and range R and is given by
the rule f−1(y) = x if and only if y = ex. That is, f−1(y) = lny. The graph of the logarithm
function is a reflection in the line y= x of the graph of the exponential function. ♦
y= xy= ex
y= lnx
Chapter 3: Functions 61
Example 3.3d Find the inverse of the function f (x) = 4x−1.
First we note that f is injective on R. For any x1, x2, if f (x1) = f (x2) then 4x1−1= 4x2−1
and this implies that x1 = x2. The range of the function is clearly R.
Now, make x the subject x=
y+1
4
and therefore, f−1(y) =
y+1
4
.
Now swap x and y, and the formula for f−1 is then written in the more customary notation
using x to stand for the independent variable and y for the dependent variable,
y= f−1(x) =
x+1
4
. ♦
Example 3.3e The natural domain of the function f given by f (x) =
√
ex+1 is R, and
the range is (1,∞). It is an increasing function (and hence injective). To see why, either
note that it’s the composite of two increasing functions, or apply the derivative test. Since
f ′(x) =
ex
2
√
ex+1
> 0 for all x∈R, it is clear that f is increasing. So f has an inverse function
g : (1,∞)→ R. To find the formula for g, we let y =√ex+1 and rearrange the equation to
make x the subject, as follows:
y=
√
ex+1 ⇒ y2 = ex+1
⇒ ex = y2−1
⇒ x= ln(y2−1).
So y= g(x) = ln
(
x2−1) for x ∈ (1,∞). ♦
Example 3.3f The function f : R→ R given by f (x) = x2 is not injective on its natural
domain.
However, if we restrict the domain to some subset of R on which the function is injective,
then we can find an inverse. For example, if A= {x∈R | x≥ 0}, then the range of f is also A.
In this case the function f : A→ A has an inverse f−1, also with domain and range A, defined
by f−1(x) =
√
x.
Note that it is possible to choose any domain over which the function is injective in order to
be able to define an inverse. For example, we could take the set S = {x ∈ R | x ≤ 0} as the
new domain of f . The range of f is still the set A, and now the inverse function f−1 : A→ S
is defined by f−1(x) =−√x.
62 MATH 1021 Calculus of One Variable
Or we could take, say, the interval (2,4) as the new domain of f . This time the range is the
interval (4,16) and once again we would have an invertible function, namely the function
f−1 : (4,16)→ (2,4) given by f−1(x) =√x.
f :A→A, f (x)=x2 f :S→A, f (x)=x2
2 4
f :(2,4)→(4,16), f (x)=x2
The corresponding inverse functions are shown below:
f−1:A→A, f−1(x)=√x f−1:A→S, f−1(x)=−√x
4 16
f−1:(4,16)→(2,4), f−1(x)=√x
♦
3.4 Inverse trigonometric functions
Example 3.4a The sine function has an inverse if we restrict its domain to the interval
[−pi/2,pi/2] over which the function is strictly increasing, the corresponding range being
[−1,1]. Then the inverse function sin−1 has domain [−1,1] and range [−pi/2,pi/2]. This is
called the principal value range.
pi
2
−pi
2
1
−1
y= sinx
b
pi
2
b−pi
2
b
1
b
−1
y= sin−1 x
♦
The table and figures below show the graphs of sinx, cosx and tanx with restricted domains
and their corresponding inverse functions.
Chapter 3: Functions 63

3.5 Hyperbolic functions and their inverses
Certain combinations of the exponential functions ex and e−x occur so frequently in mathe-
matical applications that they have been given special names. The two basic combinations
are the hyperbolic sine, or "sinh" (pronounced “shine”), and the hyperbolic cosine, or "cosh"
64 MATH 1021 Calculus of One Variable
(pronounced “cosh”). They are defined on the domain R by the following formulas:
coshx=
ex+ e−x
2
sinhx=
ex− e−x
2
Although these functions are defined in terms of exponentials, they are called cosh and sinh
because they have many properties which are similar to those of the trigonometric sine and
cosine functions.
For example, note that cosh(−x) = coshx and therefore coshx is an even function whose
graph is symmetric about the y-axis as shown in the figure below, just as cosx is.
Similarly, sinh(−x) =−sinhx and therefore sinhx is an odd function whose graph is skew-
symmetric about the y-axis (symmetric about the origin), just as sinx is.
1
y= coshx y= sinhx
The graph of coshx is called a catenary because it gives the shape of a chain hanging under
gravity. The word catenary is derived from the Latin word for "chain." The German mathe-
matician Joachim Jungius (1587-1657) showed that the shape of the hanging chain is coshx,
disproving Galileo’s claim that the curve was a parabola.
Some identities
The following identities are easily verified from the definitions.
cosh2 x− sinh2 x= 1
sinh(x+ y) = sinhx coshy+ coshx sinhy
cosh(x+ y) = coshx coshy+ sinhx sinhy
For example,
cosh2 x− sinh2 x =
(
ex+ e−x
2
)2
−
(
ex− e−x
2
)2
=
(
e2x+2+ e−2x
4
)
−
(
e2x−2+ e−2x
4
)
=
4
4
= 1
Chapter 3: Functions 65
The above identity explains the term “hyperbolic” functions and shows another similarity
with sine and cosine (the “circular” functions). Squaring and adding the parametric equations
of a circle x= cos t and y= sin t, gives the Cartesian equation of the unit circle x2+ y2 = 1.
In contrast, squaring and adding the parametric equations x= cosh t and y= sinh t, gives the
Cartesian equation of the hyperbola x2− y2 = 1.
Derivatives
Since we know that
d
dx
(ex) = ex, the derivatives of the hyperbolic functions are easily found.
Again note the similarity with the trigonometric functions (note carefully the differences as
well!).
d
dx
(coshx) = sinhx
d
dx
(sinhx) = coshx

Technical aside Analogously with the trigonometric functions, we may also define
tanhx=
sinhx
coshx
cothx=
coshx
sinhx
sechx=
1
coshx
cosechx=
1
sinhx
Note that cothx and cosechx are not defined when x= 0 since sinh0= 0. ⊳
Inverse sinh function From the graph it is clear that the function f : R→ R, f (x) = sinhx,
is injective on R. The function therefore has an inverse, called the "inverse hyperbolic sine
function" usually written as sinh−1 and has domain and range equal to R.
Note that sinh−1 y= x if and only if y= sinhx. The graph of sinh−1 x is obtained by reflecting
the graph of sinhx about the line y= x.
y= sinhx y= sinh
−1 x
66 MATH 1021 Calculus of One Variable
To find the formula for sinh−1(x), we note that if y= sinhx, then
y=
ex− e−x
2
⇒ 2y= ex− e−x
⇒ 2y= e
2x−1
ex
⇒ e2x−2yex−1= 0.
This is a quadratic in ex with solution
ex =
2y±
√
4y2+4
2
= y±
√
y2+1.
Since ex is always positive, we accept the + sign and reject the − sign, obtaining
ex = y+
√
y2+1. Taking natural log on both sides gives x = ln(y+
√
y2+1). Finally,
swapping x↔ y gives the expression
y= ln(x+
√
x2+1),
which is the formula for the inverse hyperbolic sine function
sinh−1(x) = ln(x+
√
x2+1).
Inverse cosh function
Inspecting the graph of coshx given earlier, we observe that the function f : R→ R,
f (x) = coshx is clearly not injective on R. However, choosing just the right hand half of the
curve (domain [0,∞)) will give us an injective function whose range is [1,∞).
The inverse f−1 : [1,∞)→ [0,∞) is called the "inverse hyperbolic cosine function", usually
written as cosh−1. For x≥ 0, cosh−1 y= x if and only if y= coshx .
1
y= coshx y= cosh−1 x
1
As an exercise, show that with domain [0,∞) for the cosh function, the formula for cosh−1 is
given by
cosh−1(x) = ln(x+
√
x2−1)
for all x ∈ [1,∞). What would the formula be if we choose the domain of cosh to be (−∞,0]?
Chapter 3: Functions 67
Summary of Chapter 3
• Functions are defined in this chapter and their domain, codomain and
range are introduced.
• Composite functions (also known as functions of a function) are also
studied and examples given to illustrate their application.
• The natural domain of a function f is the set of all real numbers x for
which the defining formula gives a well defined, unique real number as
the value for f (x).
• Injective or one-to-one functions are defined and the horizontal line
test is introduced to identify them.
• The inverse trigonometric functions sin−1(x), cos−1(x) and tan−1(x)
are defined by restricting the domain of sinx, cosx and tanx, respec-
tively.
• The hyperbolic functions sinhx and coshx are defined in terms of the
exponential function and their inverses sinh−1(x) and cosh−1(x) are
identified.
Exercises
3.1 Which of the following curves are the graphs of functions f : A→ R?
a) A= {x ∈ R | −1≤ x≤ 1}
x2+ y2 = 1
b) A= R
68 MATH 1021 Calculus of One Variable
y= x2
c) A= {x ∈ R |x 6= 0}
xy= 1
d) A= {x ∈ R | |x| ≥ 1}
x2− y2 = 1
3.2 Which of the arrow diagrams represent functions from A to B?
a) A B
a
b
c
d
1
2
3
4
5
b) A B
a
b
c
1
2
3
c) A B
a
b
c
d
1
2
3.3 Draw the graphs of the following functions, assuming that their domains are as large as
possible.
a) f (x) = x−2
b) g(x) =
x2−4
x+2
.
3.4 Find the largest subsets of R which are suitable domains of the following functions:
a) f (x) = sin(x+pi)
b) f (x) = x3+ x
1
3
c) f (x) = ex
d) f (x) =−e−x
e) f (x) =
√
x−2
f) f (x) =
√
x2+4
g) f (x) = ln(x2+1)
h) f (x) = ln(x+1)
Chapter 3: Functions 69
3.5 Find the ranges of the following functions f : A→ R, where f and A are given below.
a) f (x) =
√
x+8, A= [−8,∞).
b) f (x) =
√
x+8, A= (0,∞).
c) f (x) = ln
(
x2+2
)
, A= R.
d) f (x) = |3sinx|, A= R.
e) f (x) = cos2 x, A=
[
0, pi
2
]
.
f) f (x) = x+ cos2 x, A= R.
g) f (x) = cos(x2), A= [−1,1].
h) f (x) =
√
ex+3, A= R.
3.6 For what values of x is it possible to form the composite functions f ◦ g and g ◦ f , in
the following cases?
a) f (x) =
√
x, g(x) = ex b) f (x) = lnx, g(x) =−x2.
3.7 Check that f : [−1,∞) → [0,∞) with f (x) = √x+1 and g : [0,∞) → [−1,∞) with
g(x) = x2−1 are inverse to each other.
3.8 For each function f (x) below, find a domain A and range B such that f : A→ B has an
inverse. In each case, find a formula for the inverse.
a) f (x) = 5x+1
b) f (x) =
1
x
c) f (x) = ln
(
ex−1)
d) f (x) =
√
lnx−1
e) f (x) =
√(
lnx
)2
+5
f) f (x) = e2x+1
C H A P T E R 4
Limits and Continuity
Limits are essential to the definition of derivative of a function which allows the calculation
of the tangent to a curve or the instantaneous speed of a moving car. The concept of limit also
appears in a natural way when attempting to calculate the area under a curve, which leads to
the definition of the definite integral.
In this chapter we look at the concept of limit of a function from an intuitive point of view,
what is called the informal definition of limit, and then study some of the basic laws that
limits satisfy. The concept of limit is used to introduce the important definition of continuous
functions.
An introduction to the formal definition of limit (not examinable here) is given in Appendix
A for those interested in this topic.
4.1 Informal definition of limit
Suppose that f is a real-valued function defined at all points in an open interval I ⊆ R con-
taining the point a, except possibly at a itself. In many cases we will observe that if we take
values of x that are close to a but not equal to a then the values of f (x) will become as close
as we like to a point L. Informally, we say that f (x) has limit L as x approaches a.
Informal definition of limit
We say “the limit of a function f (x), as x approaches a, is equal to L” if we
can make the values of f (x) as close to L as we like by taking x sufficiently
close to a, from both the left and right sides of a but not equal to a.
In this case we write
lim
x→a f (x) = L.
Referring to Figure 4.1, the notation x→ a− means that x approaches a from the left (x< a),
and x→ a+ means that x approaches a from the right (x> a).
For the limit to exist the function f (x) must approach the same value L when x approaches a
from the left and from the right. That is limx→a− f (x) = limx→a+ f (x) = L.
70
Chapter 4: Limits and Continuity 71
Figure 4.1: Informal definition of limit – The function need not be defined at x= a.
Example 4.1a Use this informal idea of limit to study the behaviour of the function f de-
fined by the parabola f (x) = x2+1 for values of x near 2.
We start by constructing two tables with values of f (x) for x close to 2 from the left x→ 2−
and close to 2 from the right x→ 2+, but not equal to 2.
x< 2 1.5 1.8 1.95 1.99 1.995 1.999 1.9999
f (x) 3.3500 4.2400 4.8025 4.9601 4.9800 4.9960 4.9996
x> 2 2.5 2.2 2.05 2.01 2.005 2.001 2.0001
f (x) 7.2500 5.8400 5.2025 5.0401 5.0200 5.0040 5.0004
From the tables we see that when x is getting closer to 2 (on either side of 2) f (x) is getting
closer to 5. It is apparent that we can make the values of f (x) as close as we like to 5 by
taking x sufficiently close to 2. We express this behaviour of f (x) by saying:
“The limit of the function f (x) = x2+1 as x approaches 2 is equal to 5 and write”
lim
x→2
(
x2+1
)
= 5.
72 MATH 1021 Calculus of One Variable
♦
Example 4.1b Using a calculator in radian mode construct a table of values to six decimal
places to guess the value of lim
x→0
sinx
x
.
The function
sinx
x
is not defined at x = 0. However, we only need values of f (x) near x = 0
and not at x= 0 itself.
Since the values of f (x) do not change by replacing x with −x, only one table is necessary to
tabulate vales of f (x) for x on either side of zero.
x ±0.5 ±0.3 ±0.1 ±0.01
sinx
x
0.958851 0.985067 0.998334 0.999983
From the values on the table we guess that lim
x→0
sinx
x
= 1, which is the correct answer as
confirmed by the graph of f (x) in Figure 4.2 below.
y=
sinx
x
Figure 4.2:
In Appendix B we prove that lim
x→0
sinx
x
= 1 using a geometric argument. ♦
Example 4.1c Investigate lim
x→0
sin
pi
x
. Calculating the function for some small values of x, we
get f (1) = sinpi = 0, f (0.1) = sin10pi = 0, f (0.01) = sin100pi = 0, and so on. On the basis
of these results we might be tempted to guess that
lim
x→0
sin
pi
x
= 0.
However, this limit is wrong because the values of sin
pi
x
oscillate between 1 and -1 infinitely
many times as x approaches 0 as shown in the graph in Figure 4.3. Since the values of the
function do not tend to a fixed number, the limit does not exist. ♦
Chapter 4: Limits and Continuity 73
y= sin
pi
x
Figure 4.3:
4.2 One-sided limits
Consider the function f (x) which has the following graph:
K limx→a+ f (x) = K
L limx→a− f (x) = L
a
b
bc
and suppose that we wanted to calculate limx→a f (x). As we mentioned earlier we use a−
to indicate that x approaches a from the left, that is we approach a through values of x that
are less than a. Similarly, a+ is used to indicate that x approaches a from the right, that is,
through values of x that are greater than a.
It is easy to see that if x approaches a along the x–axis from the left then limx→a− f (x) = L;
whereas, if x approaches a along the x–axis from the right then limx→a+ f (x) = K.
Since L 6= K it follows that the ordinary limit, limx→a f (x), does not exist. However, we can
still define one-sided limits as follows,
74 MATH 1021 Calculus of One Variable
Left-hand limit
We say that L is the limit of f (x) as x approaches a from the left if we can
make the values of f (x) as close to L as we like by taking x sufficiently close
to a and such that x is less than a. We write
lim
x→a−
f (x) = L.
Right-hand limit
We say that K is the limit of f (x) as x approaches a from the right if we can
make the values of f (x) as close to K as we like by taking x sufficiently close
to a and such that x is greater than a. We write
lim
x→a+
f (x) = K.
Example 4.2a Consider the function
f (x) =
{
1, if x≥ 0,
−1, if x< 0,
which has the following graph:
b
bc
1
-1
x
f (x)
At the point x= 0 the limit of f (x) as x appproaches 0 depends on the direction of approach.
Chapter 4: Limits and Continuity 75
As x→ 0+ (through values greater than 0), the limit is 1. However, as x→ 0− (through values
less than 0), the limit is −1.
Conclusion
Since the value of the limit cannot be both +1 and −1 we say that the limit
does not exist. However, the one sided limits do exist and
lim
x→0+
f (x) = 1 and lim
x→0−
f (x) =−1.
♦
4.3 The basic limit laws
The limit laws that we introduce in this section consist of a set of rules for calculating the
limit of one function in terms of the limits of simpler functions.
The sum law
Suppose that limx→a f (x) and limx→a g(x) both exist. Then
lim
x→a
(
f (x)+g(x)
)
= lim
x→a f (x)+ limx→ag(x).
The difference law
Suppose that limx→a f (x) and limx→a g(x) both exist. Then
lim
x→a
(
f (x)−g(x)
)
= lim
x→a f (x)− limx→ag(x).
The product law
Suppose that limx→a f (x) and limx→a g(x) both exist. Then
lim
x→a
(
f (x)×g(x)
)
=
(
lim
x→a f (x)
)
×
(
lim
x→ag(x)
)
.
The quotient law
Suppose that limx→a f (x) and limx→a g(x) both exist and that limx→a g(x) 6= 0.
Then
lim
x→a
f (x)
g(x)
=
limx→a f (x)
limx→a g(x)
.
76 MATH 1021 Calculus of One Variable
The power law
Suppose that limx→a f (x) exists. Then
lim
x→a
[
f (x)
]n
=
[
lim
x→a f (x)
]n
where n is a positive integer.
The root law
Suppose that limx→a f (x) exists. Then
lim
x→a
n
√
f (x) = n
√
lim
x→a f (x) where n is a positive integer.
If n is even, we assume that limx→a f (x)> 0.
4.4 Limits at infinity – Horizontal asymptotes
The definition of limx→a f (x) given in Section 4.1 tells us about the expected behaviour of
f (x) as x approaches the finite number a. We can also ask how f (x) behaves as x becomes
arbitrarily large and positive (x→ ∞) and arbitrarily large and negative (x→−∞).
Looking at the graph of the function f (x) =
1
x
below, it is intuitively clear that
lim
x→∞
1
x
= lim
x→−∞
1
x
= 0
.
y= 1
x
We observe that the curve gets closer to the x-axis (the line y= 0) when x→±∞. In this case
we say that the line y = 0 is a horizontal asymptote of the curve y =
1
x
. In general, we have
Chapter 4: Limits and Continuity 77
the following definition:
Horizontal asymptotes
The line y= L is called a horizontal asymptote of the curve y= f (x) if either
lim
x→∞ f (x) = L or limx→−∞ f (x) = L
Important results
We will always assume the following basic results when calculating limits as x→±∞:
lim
x→−∞e
x = 0, lim
x→∞e
−x = 0, lim
x→∞
1
x
= 0, lim
x→−∞
1
x
= 0.
Example 4.4a Note that limx→±∞ sinx and limx→±∞ cosx do not exist. These functions os-
cillate between ±1 and never settle down to a finite limit as x→±∞. ♦
Limits at infinity of rational functions
Rational functions are expressions of the form f (x) =
p(x)
q(x)
, where p(x) and q(x) are both
polynomials. When calculating limx→±∞
p(x)
q(x)
, the numerator and denominator both tend to
infinity and we obtain what is known as an “indeterminate form” of type
∞
∞
. In this case we
may use the following technique:
To find the limits of rational functions as x→±∞,
divide top and bottom by the largest power of x appearing in the denominator.
The following examples illustrate this idea.
Example 4.4b Find limx→−∞
3x3−2
x4+4x2−1 . The numerator approaches −∞ and the denom-
inator approaches ∞. Since the largest power of x appearing in the denominator is x4, we
divide top and bottom by x4,
lim
x→−∞
3x3−2
x4+4x2−1 = limx→−∞
(3x3−2)/x4
(x4+4x2−1)/x4 = limx→−∞
3
x
− 2
x4
1+
4
x2
− 1
x4
=
0−0
1+0−0 = 0.
♦
78 MATH 1021 Calculus of One Variable
Example 4.4c Now consider limx→∞
x2+1
2x2−1. The numerator and denominator both ap-
proach ∞. Dividing top and bottom by x2 gives
lim
x→∞
x2+1
2x2−1 = limx→∞
1+
1
x2
2− 1
x2
=
1+0
2−0 =
1
2
.
♦
Example 4.4d Next, consider limx→−∞
x4+4x2−1
3x3−2 . The numerator approaches ∞ and the
denominator approaches −∞. We divide top and bottom by the largest power of x in the
denominator, which is x3.
lim
x→−∞
x4+4x2−1
3x3−2 = limx→−∞
x+
4
x
− 1
x3
3− 2
x3
=
−∞+0+0
3+0
=−∞.
♦
4.5 Infinite limits – Vertical asymptotes
The term “infinite limit” is used to describe the situation where function values become arbi-
trarily large in magnitude (either positive or negative) as x approaches either a fixed point a
or as x→±∞. Loosely speaking, we say the function “approaches infinity” or “approaches
minus infinity”.
Polynomial functions such as f (x) = x4 or g(x) = x3 approach infinity or minus infinity as
x→±∞. Strictly speaking, limits such as
lim
x→0+
1
x
, lim
x→0−
1
x
, lim
x→∞x
3
and so on do not exist, since they are not finite numbers. Nevertheless, to know whether
such functions are becoming arbitrarily large in either a positive or negative direction is often
extremely useful information and so we adopt an “abuse of notation” and use the symbol ∞
as if it were a number, as follows.
Infinite limits
We write limx→a f (x) = ∞ if f (x) can be made to exceed any positive number
we please by choosing x sufficiently close to a .
We write limx→∞ f (x) = ∞ if f (x) can be made to exceed any positive number
we please by choosing x sufficiently large and positive.
We write limx→−∞ f (x) = ∞ if f (x) can be made to exceed any positive num-
ber we please by choosing x sufficiently large and negative.
Chapter 4: Limits and Continuity 79
Similar statements hold when the infinite limit is −∞.
A word of warning: the limit laws requires that limx→c f (x) and limx→±∞ f (x) exist, in other
words, they must be finite. Consequently, the limit laws do not apply to infinite limits. This
means that when infinite limits are involved we must exercise some care.
Example 4.5a Consider limx→∞
(
x3−x2+3x+6). If the sum and product rules did actually
apply here then this limit would be ∞3−∞2+ 3∞+ 6. What is the meaning of sums and
differences of infinity? Expressions such as this need careful scrutiny. The main point here
is that if x is very large and positive then x3 is much bigger than −x2+3x+6. Therefore for
very large x, x3− x2+3x+6≈ x3; consequently,
lim
x→∞
(
x3− x2+3x+6)= lim
x→∞x
3 = ∞.
This calculation agrees with the graph of y= x3− x2+3x+6.
0 1 2 3 4 5−1−2−3−4−5
0
−50
−100
−150
50
100
y= x3− x2+3x+6
Similarly, limx→−∞
(
x3− x2+3x+6)= −∞. The term x3 dominates all other terms and the
limit as x→−∞ depends on its behaviour. ♦
80 MATH 1021 Calculus of One Variable
Vertical asymptotes
Going back to the function f (x) =
1
x
and its graph (reproduced below again for convenience),
we see that the curve approaches +∞ as x approaches 0 through positive values and −∞ as x
approaches 0 through negative values.
y= 1
x
This behaviour motivates the following definition,
Vertical asymptotes
The line x = a is called a vertical asymptote of the curve y = f (x) if at least
one of the following statements is true:
lim
x→a f (x) = ∞ limx→a−
f (x) = ∞ lim
x→a+
f (x) = ∞
lim
x→a f (x) =−∞ limx→a− f (x) =−∞ limx→a+ f (x) =−∞
Example 4.5b Consider the function f (x) =
x
x−2. Intutively we can see that
lim
x→2+
= ∞ and lim
x→2−
=−∞,
and therefore the line x= 2 is a vertical asymptote. ♦
4.6 The squeeze law
The squeeze law
Suppose that g(x) ≤ f (x) ≤ h(x), for all x near a, and that limx→a g(x) = L
and limx→a h(x) = L. Then limx→a f (x) exists and, moreover,
lim
x→a f (x) = L.
The next examples show just how useful the squeeze law is.
Chapter 4: Limits and Continuity 81

Figure 4.4: The function f (x) is squeezed or sandwiched between h(x) and g(x). and so
limx→a f (x) = limx→a g(x) = limx→a h(x)
Example 4.6a As our first application of the squeeze law, we calculate lim
x→0
xsin
1
x
. Notice
that xsin
1
x
is not defined at x= 0 and therefore 0 does not belong to the domain.
Before we calculate the limit, look at the graph of y= xsin
1
x
below and guess what the value
of lim
x→0
xsin
1
x
should be. In order to apply the squeeze law notice that for all x 6= 0,
y= xsin
1
x
Figure 4.5:
∣∣∣sin 1
x
∣∣∣≤ 1.
This inequality is our starting point. Multiplying both sides by |x| gives
∣∣∣xsin 1
x
∣∣ ≤ |x| or
82 MATH 1021 Calculus of One Variable
alternatively
−|x| ≤ xsin 1
x
≤ |x|.
Referring to Figure 4.4 we can see that in this example f (x) = xsin
1
x
, g(x) = −|x| and
h(x) = |x|. Since limx→0 |x|= 0, by the squeeze law
lim
x→0
xsin
1
x
= 0,
which is consistent with the graph in Figure 4.5. ♦
Example 4.6b Show that lim
x→∞sin(e
−x2) = 0.
First we use the squeeze law to show that limx→∞ e−x
2
= 0. Since x < x2 for all x > 1, it
follows that ex < ex
2
because the exponential is an increasing function.
Therefore, taking reciprocals gives e−x > e−x2 > 0, which is valid for x> 1. Taking limits on
this expression gives
lim
x→∞e
−x > lim
x→∞e
−x2 > 0
and since lim
x→∞e
−x = 0 by our previous assumptions, it follows by the squeeze law that
lim
x→∞e
−x2 = 0 also.
Finally, by the composition law,
lim
x→∞sin(e
−x2) = sin( lim
x→∞e
−x2) = sin0= 0.
♦
4.7 Continuous and discontinuous functions
We have seen that the intuitive notion of limit of a function at a point a is the unique number
L such that the function values can be made as close as we please to L by selecting domain
elements sufficiently close to a, in both directions, from the right and from the left.
It is important to stress that when taking limits of a function at a point, the point in question
need not be in the domain of the function. We are interested only in the values of the function
near the point.
However, when the point in question is in the domain of the function, then the obvious ques-
tion arises: if limx→a f (x) exists, does limx→a f (x) = f (a)? In other words, is the limit equal
to the value of the function at that point?
If the answer to this question is yes, we say the function is "continuous at the point".
Chapter 4: Limits and Continuity 83
Continuous functions
A real valued function f of one real variable is said to be "continuous at the
point" a in its domain if limx→a f (x) exists and equals f (a).
Functions which are continuous at every point of their domains are said to be
"continuous functions".
If f is continuous at a point a this definition implies that the next three conditions are auto-
matically satisfied:
a) f (a) is defined, that is, the point a is in the domain of f .
b) limx→a f (x) = L exists, that is, L is finite.
c) limx→a f (x) = f (a), that is, L= f (a).
If any of these conditions is not satisfied we say that f is discontinuous at a, or that f has a
discontinuity at x= a.
Continuity of Elementary Functions
We will always assume that the elementary functions ex, lnx, sinx, cosx, tanx,xa and poly-
nomials are continuous on their natural domains. Similarly, any functions formed by adding,
multiplying, dividing and composing such functions are continuous, provided we exclude
division by zero.
One-sided continuity
It is also possible to define
One-sided continuity
A function f is continuous from the right at a point a if
lim
x→a+
f (x) = f (a),
and f is continuous from the left at a if
lim
x→a−
f (x) = f (a).
Example 4.7d below shows a function that is continuous from the right and discontinuous
from the left.
84 MATH 1021 Calculus of One Variable
Continuity on a closed interval
Continuity on a closed interval
A function f is continuous on the closed interval [a,b] if it is continuous
at every point inside the interval and is continuous from the right at a and
continuous from the left at b.
Limits of continuous functions
To calculate the limit of a continuous function we use the following property:
Direct substitution rule
Suppose we know that a function f (x) is continuous at a point a and we wish
to calculate limx→a f (x). In this case all we have to do is substitute x= a into
the function to obtain the value f (a). Then
lim
x→a f (x) = f (a).
Example 4.7a Calculate limx→2 f (x), where f (x) =
(
x3+2x2−4x+1).
Since we know all polynomials are continuous functions for x ∈ R, using the direct substitu-
tion rule, gives
lim
x→2
f (x) = f (2)
=
(
23+2×22−4×2+1)
= 9.
♦
To calculate the limit of the composition of continuous functions we use the following law:
The composition law
Suppose that f (x) is a continuous function and that limx→a g(x) = L. Then
lim
x→a f (g(x)) = f
(
lim
x→ag(x)
)
= f (L).
Consequently, if f (x) and g(x) are both continuous functions then f (g(x)) is
also continuous.
Chapter 4: Limits and Continuity 85
Example 4.7b Calculate limx→pi/2 esin(x).
Since ex and sinx are continuous functions in R, by the composition law, the composite
function esin(x) is also continuous in R. Therefore limx→pi/2 esin(x) = esin(pi/2) = e1 = e ≈
2.718. ♦
Example 4.7c Consider the function f given by f (x) =
2− sin(lnx)
x2+2
. What is its limit as
x→ 4?
We notice that the function is continuous at x= 4. Therefore, we may use the direct substitu-
tion rule:
lim
x→4
f (x) = f (4) =
2− sin(ln4)
18
.
♦
Discontinuous functions
Most of the functions we will come across in this course are continuous. To see some discon-
tinuous functions we generally have to consider functions which have gaps or jumps in their
graphs. Here are a few examples to illustrate functions of one variable which are discontinu-
ous at a point in their domain.
Example 4.7d Jump discontinuity
Consider the function
f (x) =
{
1, if x≥ 1,
−1, if x< 1,
which has the following graph: The function has domain R and is continuous at each x ∈ R
1
1
Figure 4.6:
except (intuitively) at the point x = 1. At this point, the limit of f (x) as x appproaches 1
depends on the direction of approach. As x→ 1+ (through values greater than 1), the limit
is 1. As x→ 1− (through values less than 1), the limit is −1. However, the value of the limit
86 MATH 1021 Calculus of One Variable
cannot be both +1 and −1. Hence this limit cannot exist and so f (x) is not continuous at
x= 1. It has a jump discontinuity at x= 1.
It should be clear that there is no way in which we could re-define f (1) in order to make this
function continuous on R. However, it can be said that f is continuous at x = 1 from the
right. ♦
Example 4.7e Infinite discontinuity
Next let f (x) be the function with domain R defined by
f (x) =


1
x
, if x 6= 0,
0, if x= 0.
Then y= f (x) has the following graph (Note the dot drawn at the origin indicating the value
of f (x) when x= 0)
b
y=
1
x
As x approaches 0 through positive values, 1/x becomes increasingly large and positive and
approaches infinity. As x approaches 0 through negative values, 1/x become increasingly
large and negative and approaches minus infinity. Thus limx→0
1
x
does not exist and f (x) is
said to have an infinite discontinuity at x= 0.
Setting f (0) = 0 is quite arbitrary; if we define f (0) = a for any a ∈ R, this function would
still fail to be continuous at x= 0 because limx→0
1
x
does not exist. ♦
Example 4.7f Removable discontinuity
Consider once again the function f (x) =
sinx
x
that is undefined at x = 0 and therefore is
discontinuous at x= 0. As a matter of fact, it is an indeterminate form of type
0
0
. However, in
Example 4.1b we showed that limx→0
sinx
x
= 1. Here again is the graph of y=
sinx
x
showing
the gap at x = 0 where f (x) is undefined. The fact that the limit exists and is finite, means
that it is possible to define a new function
g(x) =


sinx
x
, if x 6= 0,
1, if x= 0,
which is continuous for all real values of x. Because we were able to eliminate the disconti-
nuity at x= 0, f (x) is said to have a removable discontinuity at that point. ♦
Chapter 4: Limits and Continuity 87
y=
sinx
x
Figure 4.7:
We now turn to an example of a function with domain R which is discontinuous at every
point x ∈ R.
Example 4.7g Discontinuous at every point x ∈ R
Recall that Q is the set of rational numbers. Let
f (x) =
{
1, if x ∈Q,
−1, if x /∈Q .
The graph of f (x) is a bit hard to draw because the function jumps too quickly between ±1;
however, it looks something like the following:
1
−1
♦
88 MATH 1021 Calculus of One Variable
Summary of Chapter 4
• Limits were introduced in a heuristic, informal manner.
When we say that f (x) has a limit L as x approaches a we are excluding
the point a itself and considering only the behaviour of f (x) near a,
irrespective of whether the function is defined at a or not.
• A function is continuous at x = a if limx→a f (x) exists and is equal to
the value of the function f (a).
• The direct substitution rule says that if we want limx→a f (x) and we
know that f (x) is continuous at x = a, then we simply substitute x = a
and find f (a).
• Discontinuous functions – We looked at examples of the following
types of discontinuities:
– Jump discontinuities,
– Infinite discontinuities,
– Removable discontinuities, and
– Functions that are discontinuous everywhere.
Exercises
4.1 Use continuity to calculate:
a) lim
x→−1
(6x4−2x3+ x2+ x−1)
b) lim
x→−3
x4+ x2−6
x+9
c) lim
x→2
x2−4
x+2
d) lim
x→0
x3−2x+1
2x4−5x+2.
4.2 Calculate the following limits by cancelling out the common factor:
a) lim
x→1
x3−1
x−1
b) lim
x→3
x3−27
x−3
c) lim
x→−1
3x+
4x2− x
x+2
d) lim
x→5
(2x2−10x
x−5
)2
4.3 Calculate the limit of the following rational functions:
Chapter 4: Limits and Continuity 89
a) lim
x→∞
3x3−2x2+4x−6
4x3+17x−24 .
b) lim
x→−∞
17x2−18x+1
4x4−15x3+2x2+1.
c) lim
x→−∞
1+ x2
1− x2 .
d) lim
x→∞ −7x
3+4x2+2x− 1
x
.
e) lim
x→4
1
x−4.
4.4 Use the squeeze law to calculate
a) lim
x→0
x2 sin
(1
x
)
b) lim
x→1
|x−1|sin
( 1
x−1
) c) limx→0 xcos
(2
x
)
4.5 Compute the following limits. Hint: let x=
1
t
.
a) lim
t→∞
sin t
t
. b) lim
t→∞ t sin
(1
t
)
.
4.6 Sketch the graphs of the functions specified by the formulas. For each function, state
whether limx→1 f (x) exists, and if it does, find its value.
a) f (x) = |x−1|
b) f (x) =
{
0, when x= 1,
x, when x 6= 1.
4.7 Use continuity to calculate the limits:
a) lim
x→1
(x2+3)cos(pix)
b) lim
x→2
esin(pix)
c) lim
x→0
(
ex+3−
√
x2+1
)
.
C H A P T E R 5
Differentiation
In this chapter we use the concept of limit developed in Chapter 4 to define the derivative of a
function. The basic limit laws studied in that chapter can be used to derive the basic rules of
differentiation including the chain rule for differentiating composite functions. We also look
at “implicit differentiation“ which allows us to calculate the tangent line to a curve given by
an implicit function.
5.1 The derivative at a point
Recall that the derivative of a function f at a point x = a in its domain can be interpreted as
the slope of the tangent line to the graph y= f (x) at the point (a, f (a)). In the Cartesian plane
the slope is often said to be given by “rise over run”, where the run is the distance measured
in the positive x direction and the rise is the distance measured in the positive y direction.
This is illustrated in the diagram below.
x
y
b
a
f (a)
y= f (x)
run
rise
Consider points P and Q on the graph below with coordinates (a, f (a) and (a+h, f (a+h)),
respectively. Note that h can be positive or negative – in Figure 5.1 below, we show h positive.
Intuitively, the tangent line to the curve at P can be considered as the limiting position of the
secant line PQ as h approaches 0. We can write down the slope of the secant PQ using the
“rise over run” idea, obtaining
Slope of PQ=
f (a+h)− f (a)
h
.
90
Chapter 5: Differentiation 91
Figure 5.1:
Then the derivative of f at the point x= a is defined as follows:
The derivative of a function f at a point a is
f ′(a) = lim
h→0
f (a+h)− f (a)
h
,
provided this limit exists. Note that in this definition f ′(a) is just a number.
An alternative notation for f ′(a) is
d f
dx
∣∣∣
x=a
.
It is important to emphasize that the limit must exist. The reason is that there are functions
which do not have a derivative at one or more points of their domains.
Example 5.1a The classic example of a function that does not have a derivative at a point,
is the absolute value function f (x) = |x| defined by the expression
f (x) = |x|=
{
x if x≥ 0,
−x if x< 0,
whose graph has a V-shape with its point at the origin as shown in Figure 5.2.
This function does not have a derivative at x= 0, for
f ′(0) = lim
h→0
f (0+h)− f (0)
h
= lim
h→0
|h|
h
,
and this limit does not exist as the next example shows. ♦
92 MATH 1021 Calculus of One Variable
0 1 2 3 4−1−2−3−4−5
0
−1
1
2
3
4
x
y
Figure 5.2:
Example 5.1b Explain informally why limh→0
|h|
h
does not exist, and hence why the abso-
lute value function f (x) = |x| does not have a derivative at x= 0.
Since
|h|=
{
h if h≥ 0,
−h if h< 0,
it follows that
|h|
h
=
{
1 if h≥ 0,
−1 if h< 0,
and of course y =
|h|
h
is not defined when h = 0. Its graph is shown in the diagram below,
and demonstrates that there is no tendency for the function values to become close to a single
number which could be regarded as the (unique) limit as h approaches 0 through both positive
and negative values.
bc
bc
1
-1
h
y
Observe that if we get closer to x= 0 from the right, the limit is 1, and if we approach x= 0
from the left, the limit is -1. Since limx→0− 6= limx→0+ , the derivative of f (x) = |x| does not
exist at x= 0. ♦
Chapter 5: Differentiation 93
5.2 The derivative as a function
In the previous section we studied the derivative of a function f at a fixed point x = a. The
answer is a real number that represents the slope of the tangent at that point. We now change
our emphasis and replace the number a by the variable x in the definition of derivative at a
point:
The derivative of f as a function is
f ′(x) = lim
h→0
f (x+h)− f (x)
h
,
provided the limit exists. Note that in this definition f ′(x) is an actual func-
tion.
The new function f ′ is called the derivative of the function f . Its value at x can be interpreted
geometrically as the slope of the tangent at the point (x, f (x)).
Differentiable function
If f : I→ R is a real-valued function of one variable and its derivative f ′(x)
is defined for all x in the open interval I, we say f is differentiable on I.
Alternative notation
It is common practice to use ∆x instead of h and define ∆y= f (x+∆x)− f (x). With this new
notation the definition of the derivative of f becomes
f ′(x) = lim
∆x→0
∆y
∆x
,
provided the limit exists. The numbers ∆x and ∆y are called the increments of the variables
x and y, respectively.
Example 5.2a Use the definition of derivative as a function to calculate the derivative f ′(x)
of the function f (x) = x2− x.
94 MATH 1021 Calculus of One Variable
Applying the formula we obtain
f ′(x) = lim
h→0
f (x+h)− f (x)
h
= lim
h→0
[
(x+h)2− (x+h)]− [x2− x]
h
= lim
h→0
x2+2hx+h2− x−h− x2+ x
h
= lim
h→0
h(2x−1)+h2
h
= lim
h→0
[
(2x−1)+h]
= (2x−1).
This is the same answer we would get if we used the basic rules of differentiation that we are
revising below. ♦
The following important result shows how the properties of continuity and differentiability
are related:
Theorem
If a function f is differentiable at a point a, then f is also continuous at a.
The converse of this theorem is not true; there are functions that are continuous at a point but
not differentiable at that point.
Example 5.2b For example, the function f (x) = |x| is continuous at x= 0 because
lim
x→0
f (x) = lim
x→0
|x|= 0= f (0).
However, we showed in Example 5.1b that f is not differentiable at x = 0, showing that the
converse theorem does not hold. This behaviour is shown in the plot of f (x) = |x| (reproduced
0 1 2 3 4−1−2−3−4−5
0
−1
1
2
3
4
x
y
here again) where we can see the cusp or corner of the curve at x= 0. ♦
Chapter 5: Differentiation 95
Second and higher order derivatives
If a function f is differentiable its derivative f ′ is also a function which may have its own
derivative, denoted by
(
f ′
)′
= f ′′. This new function f ′′ is called the second derivative of
the original function f . An alternative notation for the second derivative of y= f (x) is
d
dx
(dy
dx
)
=
d2y
d x2
.
Similarly, the third derivative f ′′′ is the derivative of the second derivative. In general, the
nth derivative of f is denoted by f (n) and is obtained from f by differentiating n times. If
y= f (x) then we write the nth derivative as
y(n) = f (n)(x) =
dny
d xn
.
Example 5.2c Find the third derivatives of the functions a) f (x) = 3x4+ x2
and b) g(x) = sinx.
a) f (x) = 3x4+ x2, f ′(x) = 12x3+2x, f ′′(x) = 36x2+2, f ′′′(x) = 72x.
b) g(x) = sinx, g′(x) = cosx, g′′(x) =−sinx, g′′′(x) =−cosx. ♦
5.3 Basic rules of differentiation
Suppose that f and g are differentiable functions of one variable on the interval I. The
following well–known formulas give the derivatives of k f , f +g, f −g, f g and f/g.
The constant multiple rule
(k f )′ = k f ′, k constant
The sum/difference rule
( f ±g)′ = f ′±g′
The product rule
( f g)′ = f ′g+ f g′
The quotient rule
( f
g
)′
=
f ′g− f g′
g2
96 MATH 1021 Calculus of One Variable
The power rule
(xa)′ = axa−1 (a constant)
Even though we are not proving these results here, we should emphasize that the rules are a
direct consequence of the limit laws introduced in Chapter 4.
Examples 5.3a Use the basic rules of differentiation to check the following derivatives:
i)
d
dx
(
4x3−2x+4cosx− 1
ex+ x
)
= 12x2−2−4sinx+ e
x+1
(ex+ x)2
.
ii)
d
dx
(
4xex
2
)
= 4ex
2
+8x2ex
2
.
iii)
d
dx
(
3sin(2x+1)
)
= 6cos(2x+1).
iv)
d
dx
( 7x+1
3x2−1
)
=−21x
2+6x+7
(3x2−1)2 . ♦
5.4 The chain rule
The chain rule is used to find the derivative of a “composite function”, also known as a
“function of a function”. Recall the definition of composite function given in Section 3.2.
Composite function
Given two functions f and g, the "composite function" f ◦g is defined by the
expression ( f ◦ g)(x) = f (g(x)) for all x in the domain of g such that g(x) is
in the domain of f .
When dealing with a composite function f ◦ g it is often useful to think of g as the “inside”
function, and of f as the “outside” function. Using an intermediate variable u we can write a
function of a function in the form
( f ◦g)(x) = f (g(x)) =⇒ y= f (u) where u= g(x).
For example, the function y= (cosx)3 may be expressed in the form y= u3 where u= cosx.
Using this notation the Chain Rule for differentiation may be written as follows:
Chapter 5: Differentiation 97
Chain Rule
If y= f (u) where u= g(x) then
dy
dx
=
dy
du
du
dx
Example 5.4a Use the chain rule to find the derivatives of the functions:
a) y= sin(ex). In this case we let y= sinu where u= ex. Therefore
d
dx
sin(ex) =
dy
du
du
dx
= cos(u)× ex = cos(ex)× ex = ex cos(ex).
b) y= ecosx. In this case we let y= eu where u= cosx. Therefore
d
dx
ecosx =
dy
du
du
dx
= eu× (−sinx) = ecosx× (−sinx) =−sinxecosx.
c) y= (x3−1)100. In this case we let y= u100 where u= x3−1. Therefore
d
dx
(x3−1)100 = dy
du
du
dx
= 100u99× (3x2) = 300x2(x3−1)99
. ♦
Basic table of derivatives
f (x) f ′(x)
sinx cosx
cosx −sinx
tanx sec2 x
ex ex
lnx 1/x (x> 0)
xa axa−1 (a constant)
98 MATH 1021 Calculus of One Variable
5.5 Implicit differentiation
Implicit differentiation allows us to calculate the derivative of functions that are defined im-
plicitly by a relation of the form F(x,y) = 0. Sometimes it is possible to solve for y and
obtain an explicit function of x in the usual form y= f (x) but this is not always the case. For
example, consider the relations given by
x2+ y2−9= 0 or x2y2+ xsiny−4= 0.
The first equation represents a circle and it is possible to make y the subject to obtain the two
explicit functions
y=
√
9− x2 and y=−
√
9− x2,
corresponding to the upper and lower semicircles, respectively. It is now possible to find the
slope of the tangent
dy
dx
using ordinary differentiation.
However, the functions defined by the second relation cannot be expressed in explicit form
y= f (x). You may convince yourself by trying to make y the subject as an exercise.
Nevertheless, it is still possible to calculate the derivative using the method of implicit differ-
entiation:
Implicit differentiation consists of differentiating both sides of the relation
F(x,y) = 0 with respect to x and then solving the resulting equation for
dy
dx
.
Important
Wemust remember to use the chain rule everytime we differentiate a function
of y with respect to x.
Example 5.5a Use implicit differentiation to find the derivative
dy
dx
if F(x,y)= 0 is the circle
x2+y2−9= 0. Then find the equation of the tangent to the circle at the point (x,y) = (√5, 2).
Differentiate both sides of x2+ y2−9= 0 with respect to x to give
d
dx
(
x2+ y2−9)= d
dx
(
0
)
,
d
dx
(
x2
)
+
d
dx
(
y2
)− d
dx
(
9
)
=
d
dx
(
0
)
,
2x+2y
dy
dx
−0= 0,
x+ y
dy
dx
= 0.
Making
dy
dx
the subject in the last equation we obtain
dy
dx
=−x
y
. Note that the derivative
is both, function of x and function of y also.
Chapter 5: Differentiation 99
The equation of the tangent line is given by y− y0 = m(x− x0), where (x0,y0) = (
√
5,2) and
the slope m=
dy
dx
=−x
y
=−
√
5
2
. Therefore the equation of the tangent is
y−2=−
√
5
2
(x−
√
5).
♦
In the next example we simplify the notation and use y′ instead of
dy
dx
for the derivative.
Example 5.5b Calculate y′ if sin
(
x+ y
)
= y2 cosx.
Differentiating implicitly both sides with respect to x and remembering to use the product
and chain rules, we have
cos
(
x+ y
)× (x+ y)′ = 2yy′ cosx− y2 sinx,
cos
(
x+ y
)× (1+ y′)= 2yy′ cosx− y2 sinx.
Expanding and solving for y′, gives
y′ =
y2 sinx+ cos
(
x+ y
)
2ycosx− cos(x+ y) .
♦
Logarithmic differentiation
The calculation of derivatives of complicated functions involving products, quotients or pow-
ers can sometimes be simplified by taking the natural logarithm of both sides of the equation
and then using implicit differentiation.
Example 5.5c Differentiate the function y=
(
x3+1
x7/9
)1/4
.
First take the natural logarithm of both sides:
lny= ln
(
x3+1
x7/9
)1/4
=
1
4
ln
(
x3+1
x7/9
)
=
1
4
ln(x3+1)− 7
36
lnx.
We now differentiate implicitly:
1
y
dy
dx
=
3x2
4(x3+1)
− 7
36x
or
dy
dx
= y
[
3x2
4(x3+1)
− 7
36x
]
=
(
x3+1
x7/9
)1/4[
3x2
4(x3+1)
− 7
36x
]
.
♦
100 MATH 1021 Calculus of One Variable
Example 5.5d Differentiate the function y= xx.
Taking natural logarithm of both sides gives
lny= x lnx
and differentiating implicitly:
1
y
dy
dx
= x× 1
x
+ lnx= 1+ lnx =⇒ dy
dx
= y(1+ lnx) = xx(1+ lnx).
♦
5.6 The Mean Value Theorem
The Mean Value Theorem is not examinable but we will use it to prove the Lagrange form
of the remainder of Taylor series in Chapter 7 and to prove the Fundamental Theorem of
Calculus in Chapter 10.
Theorem 5.6a (The Mean Value Theorem) Suppose that f is differentiable on some inter-
val I containing the points a and b. Then there exists a number c in the open interval (a,b)
such that
f ′(c) =
f (b)− f (a)
b−a .
In other words, the Mean Value Theorem guarantees that there is at least one point c ∈ (a,b)
for which the tangent line to f at x = c has the same slope as the slope of the secant line
joining points (a, f (a)) and (b, f (b)). The following picture illustrates this idea.
b
b
f (a)
f (b)
a b
The “dashed” line is the secant line joining points (a, f (a)) and (b, f (b)). For the particular
function drawn above, there are two points c between a and b where the tangent line has the
same slope as the secant – these tangent lines are also drawn. If you think about it a bit you’ll
understand why we need the function to be differentiable and hence continuous.
Chapter 5: Differentiation 101
Summary of Chapter 5
• The derivative of a function is defined in terms of limits and an exam-
ple of calculation from first principles is given for a simple function to
illustrate the definition.
• The basic rules of differentiation were introduced along with exam-
ples that show that the calculation of derivatives is greately simplified
by these rules.
• The chain rule is used to calculate derivatives of composite functions
(function of a function).
• Implicit differentiationmakes use of the chain rule to calculate deriva-
tives of implicit functions.
• Logarithmic differentiation is an application of implicit differentia-
tion to functions involving products, quotients and powers.
• TheMean Value Theorem introduced in this chapter will be used later
on in the proof of the Fundamental Theorem of Calculus.
Exercises
5.1 Differentiate the following functions:
a) f (x) = ex+5
b) f (x) = (ln4)ex
c) f (x) = xex
d) f (x) =
x2+5x+2
x+3
e) f (x) = (x+1)99
f) f (x) = xe−x2
g) f (t) = ecos t
h) f (t) = et cos3t
i) f (t) = ln
(
cos(1− t2))
j) f (x) = (x+ sin5 x)6
102 MATH 1021 Calculus of One Variable
k) f (x) = sin(sin(sinx))
l) f (x) = sin(6cos(6sinx))
5.2 For each of the following functions f , find f ( f ′(x)) and f ′( f (x)).
a) f (x) =
1
x
,
b) f (x) = x2,
c) f (x) = 2,
d) f (x) = 2x.
5.3 Use implicit differentiation to find
dy
dx
if:
a) x3+ y3 = 1
b) x2+ xy− y2 = 4
c) 4cosx siny= 1
d)
√
xy= 1+ x2y
e) 2
√
x+
√
y= 3
5.4 Use implicit differentiation to find the equation of the tangent line to the following
curves at the given point:
a) x2+ xy+ y2 = 3 at the point (1,1) – This is an ellipse.
b) x2+2xy− y2+ x= 2 at the point (1,2) – This is a hyperbola.
C H A P T E R 6
Applications of Differentiation
Optimizing functions of one variable is an extremely useful application of calculus. In this
chapter we develop various procedures such as the first and second derivative tests to identify
local and global extrema of such functions. We also study methods to find points of inflection
and identify concavity that will allow us to sketch functions of one variable. Other important
applications of differentiation are investigated including L’Hopital’s rule for finding limits of
indeterminate forms.
6.1 Optimizing functions of one variable
Problem optimization means finding the optimal or best way of solving a given problem.
Examples are available in the physical, biological, behavioral sciences as well as in finance
and economics. We might be interested in minimizing the cost of manufacturing a certain
item or in maximizing the profit in performing certain transactions, and so on.
Mathematically, these problems can be reduced to finding the extrema of a function f (x),
that is, identifying the maximum or minimum values of the function in some part of its
domain. There are two types of extrema, local and global:
Local extrema
We say that f has a local maximum (or relative maximum) at a point a if
f (a)≥ f (x) when x is near the point a.
Similarly, we say that f has a local minimum (or relative minimum) at a
point a if f (a)≤ f (x) when x is near the point a.
Global extrema
A function f has a global maximum (or absolute maximum) at a point a if
f (a)≥ f (x) for all x in its domain.
Similarly, f has a global minimum (or absolute minimum) at a point a if
f (a)≤ f (x) for all x in its domain.
103
104 MATH 1021 Calculus of One Variable
The following is an important result that guarantees the existence of global extrema for
functions defined on closed intervals:
The Extreme Value Theorem
If f is continuous on a closed interval [a,b], then f attains a global maximum
value f (c) and a global minimum value f (d), where c and d are some real
numbers on [a,b].
If a function is defined on an open interval of the form (a,b) where the end-points are
excluded, it need not have global maxima and minima.
Example 6.1a Functions without local or global maxima or minima
Consider the function f : (0,∞)→ (0,∞) given by f (x) = 1
x
. There is no global maximum
because the value of
1
x
can be made as large as we like simply by choosing the positive
number x sufficiently close to (but not equal to) 0, and there is no global minimum since
1
x
can be made as small as we like by choosing x sufficiently large and positive.
However if we change the domain from (0,∞) to [1,2], we see that the global maximum is 1
and the global minimum is
1
2
. There are no local maxima or minima in either case.
y= 1
x
b
b
1 2
In the next example the function (whose domain is R) has a local minimum, a global mini-
mum and no local or global maxima. ♦
Example 6.1b Suppose that f (x) = x2+ x.
b
−1
2
Chapter 6: Applications of Differentiation 105
Completing the square in the form f (x) = (x+ 1
2
)2− 1
4
shows that the smallest value of f
occurs when x=−1
2
, that is, f (−1
2
) =−1
4
. Therefore f (x)≥ f (−1
2
) =−1
4
for every x in any
open interval about −1
2
and hence f has a local minimum at x=−1
2
.
As the domain of f is R, it is clear that f has no global maximum since f (x) may be made as
large as we please by choosing x sufficiently large. The global minimum is the same as the
local minimum, namely −1
4
at the point x=−1
2
. ♦
Critical points
Critical points are important because they are potential candidates to become local extrema
and therefore also potential global extrema.
A critical point of f (x) is a number c in the domain of f , where either f ′(c) =
0 or where f ′(c) does not exist.
The following result allows us to identify where local maxima and local minima are located:
Property of critical points
If f is differentiable at the point c and has a local maximum or minimum at c,
then f ′(c) = 0.
This result will not be proved formally here, but it is easy to convince yourself that it is true
by looking at the following diagram, which illustrates the case of a maximum point.
x
y
b
c
y= f (x)
b
b
P
Q
R
The plot shows that the point P at x= c is a maximum point. Therefore, there is an interval I
about c such that f (x)≤ f (c) for all x ∈ I. Select two values of x in I, one to the left of c and
106 MATH 1021 Calculus of One Variable
the other to the right. Let R and Q be the corresponding points on the graph, and let PR, PQ
be the secants as shown.
The slope of secant PQ is negative and the slope of secant PR is positive. The derivative
f ′(c) is the limiting value of each of these slopes, as the points Q and Rmove along the curve
towards P. Since the secants PR have positive slope their limit cannot be negative. Similarly,
the secants PQ have negative slope and so their limit cannot be positive.
As the derivative of f exists at c, these two slopes must ultimately tend towards a common
value, which must be zero. Thus f ′(c) = 0. A similar situation occurs for the case of a local
minimum.
Unfortunately the converse of this property is not always true: if f ′(c) = 0 it does not always
follow that c is a maximum or minimum point. The standard counter example is the cubic
function f (x) = x3 whose graph is shown below.
x
y= x3
The point x= 0 is a critical point because f ′(0) = 0 and yet the origin is neither a minimum
nor a maximum. It is what we call a point of inflection to be introduced in the next sections.
Nevertheless, finding the critical points gives us a list of possible places where maxima or
minima may exist.
Calculation of global extrema
If the function f is continuous in a closed interval [a,b], the global maximum and global
minimum values occur either at the critical points or at the endpoints of the interval.
Calculation of global extrema
a) Find the values of f at the critical points on [a,b].
b) Calculate the values of f at the endpoints of the interval. That is, calcu-
late f (a) and f (b).
c) Compare all the numbers obtained in the first two steps. The largest of
all the values is the global maximum value and the smallest of these
values is the global minimum value.
Chapter 6: Applications of Differentiation 107
Example 6.1c Find the global maximum and minimum values of the function
f (x) = x3−3x2−9x−5, 1) on the interval [0,4] and 2) on the interval [−2,4].
The derivative is f ′(x) = 3x2−6x−9= 3(x+1)(x−3), therefore setting f ′(x) = 0 gives the
critical points x=−1 or x= 3.
1) In this case we do not have to consider x = −1 because it is outside the interval [0,4].
Using the three step process defined above to calculate global extrema, we have
a) The value of f at the critical point x= 3 inside [0,4] is f (3) =−32.
b) The values of f at the endpoints of the interval are f (0) =−5 and f (4) =−25.
c) Comparing these three numbers, shows that the global maximum is f (0) =−5 and the
global minimum f (3) =−32.
0 1 2 3 4 5−1−2
0
−10
−20
−30
−40
−50
10
20
30
40
y= x3−3x2−9x−5
2) Now consider f on the interval [−2,4]. Again, using the same method,
a) Both critical points are now inside the interval [−2,4] so we get f (−1) = 0
and f (3) =−32.
b) The values of f at the endpoints of the interval are f (−2) =−7 and f (4) =−25.
c) Comparing these four numbers shows that in this case the global maximum is
f (−1) = 0 and the global minimum is still f (3) =−32.
Note that when the domain is R, f has no global maximum or minimum because
lim
x→−∞ f (x) =−∞ and limx→∞ f (x) = ∞. ♦
6.2 Increasing and decreasing functions
Recall that f ′(x) represents the slope of the curve y= f (x) at the point x, which by definition
is the slope of the tangent. Therefore, f ′(x) tells us the direction of the curve at the point x.
108 MATH 1021 Calculus of One Variable
In particular, it can tell us whether it increases or decreases near the point.
Increasing/Decreasing test
a) If f ′(x)> 0 on an interval, then f is increasing on that interval.
b) If f ′(x)< 0 on an interval, then f is decreasing on the interval.
These tests also help us identify the nature of a critical point at x= c. For example, if f has
a local maximum at a point, it must first increase to that point and then decrease. Similarly,
if f has a local minimum at a point, it must decrease to the point and then increase. The
diagram below illustrates maximum and minimum critical points with f ′(c) = 0.
max
f ′ > 0 f ′ < 0
min
f ′ < 0 f ′ > 0
These observations form the basis of the first derivative test to identify the nature of the
critical points:
First Derivative Test
Suppose that c is a critical point of a continuous function f .
a) If f ′ changes from positive to negative at c, then f has a local maximum
at c.
b) If f ′ changes from negative to positive at c, then f has a local minimum
at c.
c) If f ′ does not change sign at c, then f has no local maximum or mini-
mum at c.
Note that the First Derivative Test is a direct consequence of the Increasing/Decreasing Test.
Example 6.2a Consider the function f (x) = x2 ex. Find the critical points and draw a sign
diagram to indicate the regions where f (x) increases or decreases, identifying the critical
points as local maxima or local minima. Use this information to draw a sketck of f (x)
showing the most salient features.
The natural domain of f is the whole of the real line−∞ < x< ∞ so there are no singularities.
Differentiating f (x) = x2ex using the product rule, gives
f ′(x) = x2ex+ ex 2x
= x(x+2)ex,
Chapter 6: Applications of Differentiation 109
and hence there are two critical points at x = 0 and x = −2 such that f ′(0) = f ′(−2) = 0.
The values of the function at those two points are f (0) = 0 and f (−2) = 4e−2 ≈ 0.5.
Now to determine the nature of the critical points (0,0) and (−2,4e−2) we look at the sign of
f ′(x) on either side of x= 0 and x=−2 and use the first derivative test.
It is helpful to draw a sign diagram for f ′(x) as follows:
x<−2 x=−2 −2< x< 0 x= 0 x> 0
f ′(x) +ve 0 −ve 0 +ve
slope ր −→ ց −→ ր
Table 1
Thus the graph slopes up to a local maximum at (−2,4e−2), then down to a local minimum
at x = 0, then up again after that. We note also that f (x) = x2ex ≥ 0 for all x and that
limx→−∞ = 0 and limx→∞ = ∞.
We now have enough information at our disposal to be able to sketch the graph. It is imme-
diately clear that for large
0 1 2 3 4−1−2−3−4−5
1
2
3
4
x
y
positive x the function takes on values much bigger than 4e−2; so (−2,4e−2) is not a global
maximum, but only a local maximum. The function value at x=−2 is bigger than the values
immediately around, but by no means the biggest value the function attains. On the other
hand, the point (0,0) is a global minimum, since x2ex ≥ 0 for all x. This function does not
have a global maximum since it goes all the way up to +∞. ♦
Example 6.2b In the previous example, we saw that f (x) = x2ex does not have a global
maximum over its natural domain −∞ < x < ∞. In this example we restrict the domain to
the closed interval [−3,1] and so the Extreme Value Theorem asserts that there will be both,
global maximum and global minimum. Now use the three step process to calculate global
extrema.
a) Both critical points x= 0 and x=−2 are inside the interval [−3,1]. The function values
are f (0) = 0 and f (−2) = 4e−2 ≈ 0.5.
110 MATH 1021 Calculus of One Variable
b) The values of f at the endpoints of the interval are f (−3) = 9e−3 ≈ 0.45 and f (1) =
e≈ 2.72.
c) Comparing these four values, shows that the global maximum is f (1) = e ≈ 2.72 and
the global minimum f (0) = 0.
0 1−1−2−3
1
2
3
e
x
y
♦
Maxima and minima when the derivative is undefined
The maximum and minimum points we have encountered so far have all occurred at points
where the first derivative is zero (that is, at points where the curve has a horizontal tangent).
We now turn to cases in which the first derivative is undefined.
We have already shown in Example 5.1b that the absolute value function
f (x) = |x|=
{
x if x≥ 0,
−x if x< 0,
does not have a derivative at x= 0. In fact, its graph has a V-shape with a cusp at the origin.
Because there is no tangent at 0, we say that the graph is not smooth at the origin, even though
it is continuous there. Indeed, it is certainly defined at x = 0 and |0| = 0. As this example
shows, very frequently, cusps will be local maximum or minimum.
As for critical numbers with f ′(x) = 0, the determining feature of a maximum at a cusp is
that f ′(x) changes from positive to negative as x increases through the value x0. Similarly, a
minimum at a cusp is characterized by f ′(x) changing from negative to positive at the critical
number.
f ′ undefined
f ′ > 0 f ′ < 0
f ′ undefined
f ′ < 0 f ′ > 0
f ′ undefined
f ′ > 0
f ′ > 0
The diagrams illustrate a cusp that is a local maximum, a cusp that is a local minimum and a
cusp that is neither a local maximum nor a local minimum.
Chapter 6: Applications of Differentiation 111
Example 6.2c The first derivative of f (x) = (x−1)2/5 is
f ′(x) =
2
5
(x−1)−(3/5)
=
2
5(x−1)3/5 ,
which is undefined at x= 1. There are no values of x that make f ′(x) zero; so x= 1 is the only
critical number. The corresponding critical point is (1,0). The sign diagram for the slope is
shown in the table below.
x< 1 x= 1 x> 1
f ′(x) −ve undefined +ve
slope ց undefined ր
Table 2
We can now draw the graph of f (x) clearly showing that the cusp at (1,0) is also a global
minimum.
0 1 2 3 4 5−1−2−3−4
1
x
y
♦
6.3 Concavity and points of inflection
Let x = c be a point in the domain of a function f at which the derivative f ′(c) is defined.
The tangent to the graph at the point (c, f (c)) can then be drawn.
Concavity
a) If the graph of f lies above the tangent line for all points close to
(c, f (c)) then we say that the graph is concave upward at this point.
b) If the graph lies below the tangent line at all points close to (c, f (c))
then the graph is said to be concave downward at this point.
Note from the diagram that if the concavity is upwards, going from left to right, the slope of
the tangent increases. This means that the derivative f ′ is an increasing function and therefore
its derivative f ′′ is positive.
112 MATH 1021 Calculus of One Variable
concave upward
c
f (c)
x
f (x)
c
f (c)
concave downward
x
f (x)
Similarly, if the concavity is downwards, going from left to right, the slope of the tangent de-
creases. This means that the derivative f ′ is a decreasing function and therefore its derivative
f ′′ is negative.
These observations lead to the following test, which may be proved using the Mean Value
Theorem:
Concavity Test
a) If f ′′(x) > 0 for all x in an interval I, then the graph of f is concave
upwards on I.
b) if f ′′(x) < 0 for all x in an interval I, then the graph of f is concave
downwards on I.
A point at which the concavity changes from upward to downward (or vice versa) is called
a point of inflection. Since the sign of the second derivative changes as we pass through a
point of inflection, at the point of inflection itself the second derivative must either be zero or
undefined.
0 1 2
1
c
f (c)
f(x) = x3−3x2+2x+0.5
{
f ′′ = 0
point of inflection
f ′′ < 0
f ′′ < 0concave downward
concave upward
1
0
1
f ′′ undefined
point of inflection
} (c, f (c))=(0,0.5)
f(x) = 0.5− 3
√
x5
concave downward
concave upward
horizontal tangent
Observe that although the graph and its tangent line at the point (c, f (c)) have the same slope,
as they must by the definition of “tangent”, the fact that (c, f (c)) is a point of inflection means
that the graph actually crosses the tangent at (c, f (c)) as shown in the diagrams.
It is helpful to locate points of inflection when sketching graphs, since changes in concavity
are significant aspects of a curve’s shape.
Chapter 6: Applications of Differentiation 113
Second derivative test
Since we are discussing applications of second derivatives, we should mention the Second
Derivative test for identifying the nature of local extrema. The method assumes that the
second derivative of a function f exists.
Second Derivative Test
If f and f ′ are differentiable functions and f ′(c) = 0 (that is, x= c is a critical
point of f ), then
a) If f ′′(c)> 0, there is a local minimum of f at x= c,
b) if f ′′(c)< 0, there is a local maximum of f at x= c, and
c) if f ′′(c) = 0, we cannot draw any conclusions without further work.
To understand why this works, consider the following diagram which illustrates case (a). If
f ′′(c)> 0 then we also have f ′′(x)> 0 for all x sufficiently close to c, and hence f ′(x) must
be an increasing function in some interval containing c.
Since f ′(c) = 0, this means that a little to the left of c we must have f ′(x) negative and a little
to the right we must have f ′(x) positive. In other words, f has a local minimum at x= c.
c
f ′(x)< 0 f ′(x)> 0
Similar reasoning holds for case b). The examples below show why we can make no conclu-
sion when f ′′(c) = 0, in case c).
Critical points and concavity
Note that if f ′(c) = 0 and f is concave upward at (c, f (c)) then the critical point (c, f (c)) is
a local minimum of f . Similarly, if f is concave downward at a critical point then the point
is a local maximum. However, if f ′(c) = 0 and (c, f (c)) is a point of inflection (as well as
being a critical point) then it is neither a local maximum nor a local minimum.
This is because the tangent at (c, f (c)) must be horizontal (since f ′(c) = 0) and the graph
must cross the tangent at (c, f (c)) (since we have a point of inflection); so f (x) > f (c) on
one side of x= c and f (x)< f (c) on the other side.
The simplest example of this is the graph of y = x3 at the origin where both f ′(0) = 0 and
f ′′(0) = 0. Hence it must be a point of inflection as well as a critical point.
114 MATH 1021 Calculus of One Variable
By contrast, y= x4 also has the property that f ′(0) = f ′′(0) = 0, however, in this case (0,0)
is a local minimum. This example shows that the condition f ′′(x) = 0 is no guarantee that we
have a point of inflection.
Examples 6.3a
i) Let f (x) = x3−3x2−9x−5. Then f ′(x) = 3x2−6x−9= 3(x−3)(x+1); so x=−1
and x = 3 are the only critical points of f . The second derivative is f ′′(x) = 6(x− 1),
so f ′′(−1) =−12< 0 and f ′′(3) = 12> 0; therefore, x =−1 is a local maximum for
f and x= 3 is a local minimum.
ii) Next, consider the function g(x) = x3. The first and second derivatives are g′(x) = 3x2
and g′′(x) = 6x so x= 0 is the only critical point of y= x3. For this function the second
derivative test is of no help in deciding the nature of the critical point at x= 0. However,
as g′(x) = 3x2 we see that g′(x) ≥ 0 for all x; hence, there is no change in the sign of
g′(x) around x= 0 and we have a point of (horizontal) inflection.
iii) Finally, consider h(x)= x4. This time, h′(x)= 4x3 and h′′(x)= 12x2, so x= 0 is the only
critical point and h′′(0) = 0. Unlike the last example, the sign of h′(x) changes around
x= 0: more precisely, h′(x)< 0 if x< 0, and h′(x)> 0 if x> 0. Hence, h is decreasing
on (−∞,0) and increasing on (0,∞), and therefore, x= 0 is a local minimum. ♦
x
y= x3
x
y= x4
The graphs of y = x3 and y = x4 both satisfy f ′(0) = f ′′(0) = 0. In both cases the x-axis is
the tangent to the graph at the origin. For y = x3 the graph crosses its tangent at the origin,
indicating a point of inflection. The graph of y = x4 is concave upward at the origin, indicating
a local minimum.
Chapter 6: Applications of Differentiation 115
6.4 Curve sketching
The following list provides the steps necessary to sketch a curve and is intended as a guide
only. Not every item is relevant to every function.
CURVE SKETCHING SUMMARY
a) Domain – Find the set of values of x for which the function f (x) is
defined. This could be the natural domain or some specified subset of
the natural domain (Section 3.1).
b) Intercepts – Find the y-intercept from y= f (0) and the x-intercepts by
solving the equation f (x) = 0.
c) Horizontal asymptotes – Calculate limx→∞ f (x) and limx→−∞ f (x). If
either limit is finite equal to L, then the line y= L is a horizontal asymp-
tote (Section 4.4)
d) Vertical asymptotes – If there is a value x= a such that
limx→a+ f (x) = ±∞ or limx→a− f (x) = ±∞ then there is a vertical
asymptote at x= a (Section 4.5).
e) Critical points – Find the critical points c of f where f ′(c) = 0 or f ′(c)
does not exist. (Section 6.1).
f) Intervals of increase or decrease – Using the critical points found
above, draw a sign diagram to find the intervals on which f ′(x)> 0
( f increasing) and the intervals on which f ′(x)< 0 ( f decreasing) (Sec-
tion 6.2). This process will also identify the nature of the critical points
(maximum, minimum or neither) using the First Derivative Test.
g) Concavity and points of inflection – Calculate f ′′(x) and use the con-
cavity test. Find the points of inflection by solving f ′′(x) = 0
(Section 6.3).
You may draw another sign diagram (or combine with previous one)
to find the intervals on which f ′′(x) > 0 ( f concave upwards) and the
intervals on which f ′′(x)< 0 ( f concave downwards)
h) Sketch the curve – Put together the information obtained above to draw
the graph of f (See the example below).
116 MATH 1021 Calculus of One Variable
Example 6.4a Sketch the graph of f (x) = x2/3(x−5) indicating the most important features
according to the list of steps above.
a) Domain – The natural domain of the function f (x) = x2/3(x−5) consists of all x such
that x ∈ R.
b) Intercepts – The y-intercept is y= f (0) = 0, and since f (0) = f (5) = 0 the x-intercepts
occur when x= 0 and x= 5.
c) Horizontal asymptotes – Since x2/3 is the cube root of x2, then it is zero at x= 0 and
positive elsewhere. Thus x2/3(x−5) has the same sign as x−5 for all nonzero values
of x. This means that f (x) < 0 for x < 5 and f (x) > 0 for x > 5. This also implies
that limx→∞ x2/3(x− 5) = ∞ and limx→−∞ x2/3(x− 5) = −∞. Therefore there are no
horizontal asymptotes.
d) Vertical asymptotes – Because f (x) = x2/3(x−5) is defined for all values of x, there
are no points x= a such that limx→a+ f (x) =±∞ or limx→a− f (x) =±∞. That is, there
are no vertical asymptotes.
e) Critical points – To find the critical points we calculate the derivative
f ′(x) = x2/3+(x−5) 2
3
x−1/3
= x2/3+
2(x−5)
3x1/3
=
3x+2(x−5)
3x1/3
=
5x−10
3x1/3
=
5
3
(x−2)
x1/3
So f ′(2) = 0 and is undefined when x= 0. Therefore, there are two critical points, one
at (0,0) and the other at (2,−3 3√4)≈ (2,−4.8).
Regarding the critical point (0,0), we had already found that f (0) = 0 and that f (x)< 0
for nearby values of x on either side of x= 0. This shows that (0,0) is a local maximum,
and it might have led us to expect that the derivative would be zero at x= 0.
In fact, there is a cusp at x= 0, as can be seen by inspecting the values of the derivative
for points close to zero. If x is small and positive then x1/3 (the cube root of x) is also
small and positive, whereas x− 2 is close to −2. So (x−2)
x1/3
is a negative number of
large magnitude.
On the other hand, when x is negative and close to zero, x1/3 is also negative and close
to zero, while x− 2 is still close to −2; so in this case (x−2)
x1/3
is positive and of large
magnitude. So we have an extremely sharp cusp at the origin, the graph being nearly
vertical on either side of the cusp.
Since f (0) = f (5) = 0 and f (x) < 0 for 0 < x < 5, it is clear that the critical point
(2,−3 3√4) must be a local minimum.
Chapter 6: Applications of Differentiation 117
f) Intervals of increase or decrease – Look at the Table below that shows the sign dia-
gram for f ′(x).
g) Concavity and points of inflection – Calculate f ′′(x):
f ′′(x) =
5
3
(
x1/3− (x−2) 1
3
x−(2/3)
x2/3
)
=
5
3
(3x1/3− (x−2)x−2/3)
3x2/3
=
5
3
(3x− (x−2))
3x4/3
=
5
9
(2x+2)
3x4/3
=
10
9
(x+1)
x4/3
.
In the first line of the above equation we applied the quotient rule for differentiation
and in the third line we multiplied both the numerator and the denominator by x2/3.
So f ′′(−1) = 0 and at that point f (−1) = (−1)2/3(−1−5) =−6. Since x4/3 is never
negative we see that f ′′ < 0 when x<−1, and f ′′ > 0 for x>−1 (excluding x= 0).
In summary, the concavity is downward for x < −1, the point (−1,−6) is a point of
inflection, and the concavity is upward for −1< x< 0 and for x> 0.
h) Sketch the curve – We now put together all the information obtained above to draw
the following sign diagram:
x <−1 x=−1 −1 2
dy/dx +ve +ve +ve undef. −ve 0 +ve
slope ր ր ր ↑↓ ց −→ ր
y y <−6 y =−6 −6y>−3 3√4 y =−3 3√4 −3 3√4< y
d2y/dx2 −ve 0 +ve undef. +ve +ve +ve
concavity downward inflection upward undef. upward upward upward
Table 3
118 MATH 1021 Calculus of One Variable
Finally, after calculating that f (−4)≈−22.6 and f (6)≈ 3.3, we are able to draw the graph
quite accurately. In order to exaggerate the change in concavity at (−1,−6) we have used
different scales for x and y.
0−1−2−3−4 1 2 3 4 5
0
−3
−6
−9
−12
−15
−18
−21
−24
y = 5x−1
(dotted line)
Point of inflection
x
y
The graph of f(x) = x2/3(x−5). The line y= 5x−1 is the tangent at (−1,−6); it crosses
the graph at the point of inflection (−1,−6).
♦
6.5 L’Hôpital’s rule
In Chapter 4 we mentioned indeterminate forms in the context of the quotient law for calcu-
lating limits. The quotient law states that
lim
x→c
f (x)
g(x)
=
limx→c f (x)
limx→c g(x)
.
If both limits are 0 or both are ∞, then the limit of the ratio f/g can be 0, ∞ or any real
number, in other words, the limit of the ratio is indeterminate. These two cases lead to two
types of indeterminations:
Case 1 – When both, the numerator and denominator tend to infinity, we obtain what is
known as an “indeterminate form” of type
∞
∞
.
Case 2 – Similarly, if both, the numerator and denominator tend to zero, we obtain what is
called an “indeterminate form” of type
0
0
.
l’Hôpital’s rule provides a powerful method for calculating the limits of these two types of
Chapter 6: Applications of Differentiation 119
indeterminate forms.
L’Hôpital’s rule
(
0
0
form
)
Suppose that limx→c f (x) = 0, limx→c g(x) = 0 and that limx→c
f ′(x)
g′(x)
exists.
Then
lim
x→c
f (x)
g(x)
= lim
x→c
f ′(x)
g′(x)
.
L’Hôpital’s rule
(±∞
±∞ form
)
Suppose that limx→c f (x) =±∞, limx→c g(x) =±∞ and that limx→c f
′(x)
g′(x)
exists. Then
lim
x→c
f (x)
g(x)
= lim
x→c
f ′(x)
g′(x)
.
In both the
0
0
form and
±∞
±∞ form of L’Hôpital’s rule, we also allow c to be ±∞.
Important
When applying the rule more than once to the same expression, is important to ensure that
the conditions are satisfied at each step or we will get the wrong answer.
Examples 6.5a
i) We calculate limx→∞
ex
x
.
As x→ ∞ this expression has the form ∞
∞
and so we may apply l’Hôpital’s rule. This
gives
lim
x→∞
ex
x
= lim
x→∞
ex
1
= ∞.
What this is saying is that the function y= ex grows much more quickly than the func-
tion y= x; that is, if x is large enough then ex dominates x.
ii) Find limx→∞
ex
x2
.
This time we must use l’Hôpital’s rule twice; after the first application the limit still
cannot be evaluated, but the expression remains in the form appropriate for l’Hôpital’s
rule to be applied again,
lim
x→∞
ex
x2
= lim
x→∞
ex
2x
= lim
x→∞
ex
2
= ∞.
120 MATH 1021 Calculus of One Variable
Therefore ex dominates x2 as x becomes very large. More generally, it can be shown
that ex dominates xn, for any n > 0. (Convince yourself that limx→∞
ex
xn
= ∞ for any
integer n> 0.)
iii) Consider limx→∞
6x2+2x−6
2x2−7x+4. Applying l’Hôpital’s rule twice we find that
lim
x→∞
6x2+2x−6
2x2−7x+4 = limx→∞
12x+2
4x−7 = limx→∞
12
4
= 3.
Check that the two applications of l’Hôpital’s rule are allowed. Of course, we could
also evaluate this limit by dividing both the numerator and the denominator by x2.
iv) Now for a slightly harder example. Consider limx→0+ x lnx. Recall that x→ 0+ in-
dicates that we consider the limit only as x approaches 0 from the right. Happily,
l’Hôpital’s rule can also be used to evaluate such one-sided limits. However on the
face of it, l’Hôpital’s rule does not seem to apply in this particular problem because
x lnx is not in the form of a fraction f (x)/g(x).
However, it is possible to rewrite the expression as follows
x lnx= (lnx)/(
1
x
).
Since lim
x→0+
lnx=−∞ and lim
x→0+
1
x
= ∞ we see that the new expression is an indetermi-
nate form and so l’Hôpital’s rule can be used to compute this limit. We have
lim
x→0+
x lnx= lim
x→0+
lnx
1
x
= lim
x→0+
1
x
− 1
x2
= lim
x→0+
−x2
x
= lim
x→0+
−x= 0.
v) Finally, the following calculation contains a mistake; find it.
lim
x→1
x3− x2+2x−1
x2−2x−2 = limx→1
3x2−2x+2
2x−2 = limx→1
6x−2
2
= 2.
What is the actual value of this limit? ♦
Example 6.5b To conclude this section we give the harder example
lim
x→∞
(
1+
1
x
)x
.
Having x appear as an exponent creates some problems; we can get rid of the exponent by
using the formula
f (x) = exp
[
ln f (x)
]
which is a direct consequence of fact that the exponential and the natural log are inverse
functions. Therefore,(
1+
1
x
)x
= exp
[
ln
(
1+
1
x
)x]
= exp
[
x ln
(
1+
1
x
)]
= exp
[ ln(1+ 1
x
)
1/x
]
.
Chapter 6: Applications of Differentiation 121
Hence,
lim
x→∞
(
1+
1
x
)x
= exp
[
lim
x→∞
ln
(
1+ 1
x
)
1/x
]
= exp
[
lim
x→∞
1
1+ 1x
× (− 1
x2
)
− 1
x2
]
= exp
[
lim
x→∞
1
1+ 1
x
]
= exp
[ 1
1+0
]
= exp[1] = e.
This limit is both unexpected and amazing. Our working shows that if x is very large then
e≈
(
1+
1
x
)x
.
For example, taking x= 1000 we find that
(
1+
1
1000
)1000
= 2.7169 . . .
and e= 2.7182818 . . .. You might like to experiment with your calculator to see what happens
with larger values of x. ♦
Summary of Chapter 6
• Optimization of functions is related to the problem of identifying their
global extrema. The Extreme Value Theorem guarantees the existance
of global maxima and minima for functions that are continuous on a
closed interval [a,b].
• Curve sketching is another important application of differentiation. A
summary of the most important tools needed for sketching is provided
in Section 6.4.
• L’Hôpital’s rule is concerned with the application of differentiation to
the calculation of limits of indeterminate forms.
Exercises
6.1 Find the derivatives of the following functions:
122 MATH 1021 Calculus of One Variable
a) f (x) = 7x3−2x+ 1
x
,
b) f (x) = (x+ cosx)3,
c) f (x) = sin
(
cos(x2)
)
,
d) f (x) = x3 sinx+ excosx,
e) f (x) = 7x8+
1
1+ cosx2
.
6.2 Find the equation of the tangent line to the function f (x) = 4xex+ cosx at the point
x= 0.
6.3 Find the global minimum and maximum values of the following functions on the closed
interval [−3,3].
a) f (x) = x2ex
b) f (x) = 3x2−6
c) f (x) = |x−1|+2
d) f (x) = cos
(x
4
)
6.4 Show that the polynomial
f (x) = 1+ x+ x3+ x5+ x7
is strictly increasing over the whole real line.
6.5 Find the value a such that the function
f (x) = x lnx
is strictly decreasing for x ∈ (0,a) and strictly increasing when x> a.
6.6 Use l’Hôpital’s rule to find lim
x→∞
sinhx
coshx
. Now find the limit using the definition of sinh
and cosh in terms of exponentials and compare the results.
6.7 Use l’Hôpital’s rule to find the following limits:
a) lim
x→0
xcoshx
sinx
b) lim
r→0
sinr2
r2
c) lim
x→1
lnx
(x−1)
d) lim
x→0+
x lnx
e) lim
x→0
ex−1
sinx
f) lim
x→0
1− cosx
x2
g) lim
x→0
ex− x−1
cosx−1
h) lim
x→1
x2−1
lnx
i) lim
x→0
sinx
1− ex
j) lim
x→0
cosx−1
xsinx
C H A P T E R 7
Taylor Polynomials
As we saw in Chapter 2, a polynomial is a function of the form
f (x) = a0+a1x+a2x
2+ · · ·+anxn,
where the ak are constants and n is a non-negative integer. "Taylor polynomials" are special
polynomials which are used to approximate other types of functions near particular points.
The reason for this is that polynomials are easy to evaluate, requiring only addition and
multiplication.
Other types of functions, for example the trigonometric functions or logarithmic functions,
are very difficult to evaluate exactly. (Have you ever thought about the process used by your
calculator in finding, say, sin2 or ln10?) In this chapter we also learn to determining how
good an approximation it provides.
7.1 An approximation for ex
Let’s start by thinking about how we might construct a cubic (polynomial of degree 3) to
approximate ex near the point x= 0,
p(x) = a0+a1x+a2x
2+a3x
3.
a) Firstly, we would certainly want the value of p(x) to be the same as the value of ex at
x= 0. So let’s take p(0) = a0 = e
0 = 1, and thus p(x) = 1+a1x+a2x
2+a3x
3.
b) Next, perhaps it would be a good idea to make the first derivatives the same, so that
the graphs of p(x) and ex have the same slope at x = 0. The derivative of ex is ex,
while p′(x) = a1+ 2a2x+ 3a3x2. At x = 0 we have p′(0) = a1 = e0 = 1 and hence
p(x) = 1+ x+a2x
2+a3x
3.
c) If we make the second derivatives the same as well, the graphs would then have the
same concavity at x= 0. The second derivative of ex is ex. (Indeed all derivatives of ex
are ex.) Now, p′′(x) = 2a2+6a3x and p′′(0) = 2a2. Since we want this to equal e0 = 1,
we’ll let a2 =
1
2
, and p(x) = 1+ x+
x2
2
+a3x
3.
d) Now let’s match up the third derivatives also. Since p′′′(0) = 6a3, we should take
6a3 = 1, a3 =
1
6
and p(x) = 1+ x+
x2
2
+
x3
6
.
123
124 MATH 1021 Calculus of One Variable
The graph below shows that p(x) does indeed match ex very closely near x= 0.
0 1 2 3−1−2−3
ex
1+ x+ x
2
2
+ x
3
6
10
Suppose we use the same idea to construct a polynomial of degree n (with n > 3). That is,
we look for values of a0, a1, a2, . . . , an such that the polynomial
Pn(x) = a0+a1x+a2x
2+ · · ·+anxn
is equal to ex at x= 0 and has each of its first n derivatives at x= 0 equal to the corresponding
derivative of ex at 0. (Note that for a polynomial of degree n all derivatives of order higher
than n are 0.) Since every derivative of ex is equal to 1 at x = 0, we want to choose the
coefficients of Pn(x) so that the first, second, third, . . ., n
th derivatives of Pn(x) are all equal to
1.
Notation The abbreviations that we commonly use for first, second or third derivatives of a
function f , namely f ′ =
d f
dx
, f ′′ =
d2f
dx2
, f ′′′ =
d3f
dx3
are inconvenient for higher derivatives.
For this reason we introduce the notation f (k) to mean the kth derivative of f . That is, f (k) =
dk f
dxk
. Sometimes it is convenient to allow k to equal 0 in this notation, so that we have f (0),
by which we mean simply f .
Using this notation, we want to find Pn(x) = a0+a1x+a2x
2+ · · ·+anxn such that P(k)n (0) = 1
for 0≤ k ≤ n. The first few derivatives of Pn(x) are as follows:
Pn(x) = a0+a1x+a2x
2+ · · ·+anxn
P′n(x) = a1+2a2x+3a3x
2+ · · ·+nanxn−1
P′′n (x) = 2a2+2 ·3a3x+3 ·4a4x2+ · · ·+(n−1)nanxn−2
P′′′n (x) = 2 ·3a3+2 ·3 ·4a4x+3 ·4 ·5a5x2+ · · ·+(n−2)(n−1)nanxn−3
P
(4)
n (x) = 2 ·3 ·4a4+2 ·3 ·4 ·5a5x+ · · ·+(n−3)(n−2)(n−1)nanxn−4
Now think about what happens to Pn(x) once it has been differentiated k times (0 ≤ k ≤
n). All terms in powers of x smaller than k have been differentiated away to zero. The
term in xk will have been multiplied by k on the first differentiation, then by k− 1, then
Chapter 7: Taylor Polynomials 125
by k− 2, and so on. After k differentiations, the term in xk will have been multiplied by
k(k− 1)(k− 2) . . .2.1 = k!. All terms in Pn(x) with powers higher than k will still contain a
power of x after k differentiations. We therefore have the following:
P
(k)
n (x) = k!ak+ terms containing x (k < n)
...
...
P
(n)
n (x) = n!an
Substituting x= 0 into the equations above, we find:
Pn(0) = a0 = 1
P′n(0) = a1 = 1
P′′n (0) = 2a2 = 1, =⇒ a2 =
1
2
P
(3)
n (0) = 3!a3 = 1, =⇒ a3 = 1
3!
...
...
...
P
(k)
n (0) = k!ak = 1, =⇒ ak = 1
k!
...
...
...
P
(n)
n (0) = n!an = 1, =⇒ an = 1
n!
Hence the polynomial
Pn(x) = 1+ x+
x2
2!
+
x3
3!
+ · · ·+ x
n
n!
is equal to ex at x= 0, and each of its first n derivatives is equal to the corresponding derivative
of ex at x = 0. Let’s see how well this polynomial approximates ex. The following diagrams
show the graphs of P6(x) and P7(x) superimposed on the graph of e
x.
0 1 2 3 4 5 6−1−2−3−4−5−6−7
ex
P7(x)
P6(x)
100
126 MATH 1021 Calculus of One Variable
We see that the polynomials of degree 6 and degree 7 look virtually identical with ex for
−2.5≤ x≤ 2.5. Taking higher values of n makes
Pn(x) = 1+ x+
x2
2!
+
x3
3!
+ · · ·+ x
n
n!
a better approximation to ex in the sense that it matches the exponential function over a larger
interval. Here is the graph of P10(x), superimposed on the graph of e
x.
0 1 2 3 4 5 6 7 8−1−2−3−4−5−6−7−8−9
ex
P10(x)
1000
It is somewhat surprising that the polynomial is a good match for ex for −6≤ x≤ 6 since we
set out to find an approximation to ex around x = 0. What is even more surprising is that if
we let n become infinitely big, then the “infinite polynomial”
Pn(x) = 1+ x+
x2
2!
+
x3
3!
+ · · ·
is exactly equal to ex for all values of x! (This “infinite polynomial” is more correctly known
as a "power series". We will deal with series in the next chapter.)
7.2 Taylor polynomials about x= 0
The polynomial we found in the previous section,
Pn(x) = 1+ x+
x2
2!
+
x3
3!
+ · · ·+ x
n
n!
is called the nth order "Taylor polynomial" of ex about x= 0.
The process we used for constructing this polynomial is precisely the one that we use for
constructing Taylor polynomials for other functions. That is, given a function f (x), we find
a polynomial Pn(x) such that Pn(0) = f (0), P
′
n(0) = f
′(0), P′′n (0) = f ′′(0), . . ., P
(n)
n (0) =
f (n)(0). Of course, the function f must be differentiable n times at x= 0.
Chapter 7: Taylor Polynomials 127
As we saw in the previous section, if Pn(x) = a0+a1x+a2x
2+ · · ·+anxn, then the kth deriva-
tive of Pn at x= 0, P
(k)
n (0) = k!ak, for 0≤ k ≤ n.
Since we want P
(k)
n (0) to equal f
(k)(0), we choose ak =
f (k)(0)
k!
for each k, 0≤ k≤ n. (Note
that 0!= 1, and f (0) = f .)
Taylor polynomial about x= 0
The nth order "Taylor polynomial" of a function f about x= 0 is
Pn(x) = f (0)+ f
′(0)x+
f ′′(0)
2!
x2+
f (3)(0)
3!
x3+ · · ·+ f
(n)(0)
n!
xn.
In sigma notation: Pn(x) =
n
∑
k=0
f (k)(0)
k!
xk
Taylor polynomials about x= 0 are known asMaclaurin polynomials
You should memorise this formula. Constructing a Taylor polynomial for a function f is
simply a matter of finding the values of the derivatives of f at zero, and substituting into this
formula. For example, since every derivative of f (x) = ex is ex, and e0 = 1, it is immediately
obvious that the nth order Taylor polynomial for the exponential function ex is
Pn(x) = 1+ x+
x2
2!
+
x3
3!
+ · · ·+ x
n
n!
.
The sine function is also easy to deal with. Let f (x) = sinx, and we have the following:
f (x) = sinx f (0) = 0
f ′(x) = cosx f ′(0) = 1
f ′′(x) =−sinx f ′′(0) = 0
f (3)(x) =−cosx f (3)(0) =−1
f (4)(x) = sinx f (4)(0) = 0
f (5)(x) = cosx f (5)(0) = 1
f (6)(x) =−sinx f (6)(0) = 0
f (7)(x) =−cosx f (7)(0) =−1
Therefore the Taylor polynomial of order 7 for f (x) = sinx, about x= 0, is
P7(x) = x− x
3
3!
+
x5
5!
− x
7
7!
.
Note that P8(x) is the same as P7(x), since f
(8)(0) = sin0= 0.
128 MATH 1021 Calculus of One Variable
It is clear that successive derivatives of sinx, evaluated at 0, will continue to follow the pat-
tern we can see in the table above. (That is, starting with the 0th derivative, the pattern
is 0, 1, 0,−1, 0, 1, 0,−1, . . ..) Hence, successive terms in the Taylor polynomial for sinx
will follow the pattern that is obvious in P7(x), and the general polynomial, of order 2n+ 1,
(n= 0, 1, 2, . . .) for the sine function, is
x− x
3
3!
+
x5
5!
− x
7
7!
+ · · ·+(−1)n x
2n+1
(2n+1)!
.
It is always worthwhile attempting to find a pattern in the values of derivatives when calcu-
lating Taylor polynomials, since finding a pattern allows us to write down an expression for
the general polynomial.
Normally we would look for the general polynomial of order n, but in the case of sinx it is
easier to write one in terms of odd numbers 2n+ 1 only, since all the coefficients of even
powers are zero, and the polynomial consists of terms in odd powers of x only.
The following diagram compares the graph of sinx with those of the Taylor polynomials of
order 7 and order 9.
sinx
x− x3
3!
+ x
5
5!
− x7
7!
+ x
9
9!
x− x3
3!
+ x
5
5!
− x7
7!
Both polynomials are very good approximations over the interval [−pi,pi], although they both
eventually diverge dramatically from sinx. Of course, if we know sinx for −pi ≤ x ≤ pi , we
can calculate it for any value of x using the periodicity of the sine function.
Examples 7.2a
i) If f (x) = cosx, then f (0) = 1, and the values of the first 6 derivatives at x = 0 are
1, 0,−1, 0, 1, 0. The Taylor polynomial of order 6 for cosx is therefore
1− x
2
2!
+
x4
4!
− x
6
6!
.
This time we have an even function, and its Taylor polynomial contains even powers
only. As with the sine function, the pattern continues and the polynomial of order 2n
for cosx is
1− x
2
2!
+
x4
4!
− x
6
6!
+ · · ·+(−1)n x
2n
(2n)!
.
ii) Find the Taylor polynomial of order 3 for the function f (x) = x3− 7x2+ 4x− 2. You
should find that it is precisely x3− 7x2+ 4x− 2. What about the Taylor polynomial
Chapter 7: Taylor Polynomials 129
of order 4? The 4th derivative of f is 0, so the 4th order Taylor polynomial is also
x3−7x2+4x−2 (as, indeed, is any Taylor polynomial of higher degree).
In general, if f is a polynomial of degree n, then its Taylor polynomial of order n or
higher is equal to f . Can you verify this statement?
iii) Let f (x) =
1
1− x . Then:
f (x) =
1
1− x f (0) = 1
f ′(x) =
1
(1− x)2 f
′(0) = 1
f ′′(x) =
2
(1− x)3 f
′′(0) = 2
f (3)(x) =
3 ·2
(1− x)4 f
(3)(0) = 3!
f (4)(x) =
4 ·3 ·2
(1− x)5 f
(4)(0) = 4!
f (5)(x) =
5 ·4 ·3 ·2
(1− x)6 f
(5)(0) = 5!
The pattern is clear; f (n)(x) =
n!
(1− x)n+1 , and f
(n)(0) = n!. The Taylor polynomial of
order n for
1
1− x is therefore
Pn(x) = 1+ x+
2
2!
x2+ 3!
3!
x3+ 4!
4!
x4+ · · ·+ n!
n!
xn
= 1+ x+ x2+ x3+ x4+ · · ·+ xn.
iv) Show that the Taylor polynomials for coshx and sinhx, about x = 0, are
n
∑
k=0
x2k
(2k)!
and
n
∑
k=0
x2k+1
(2k+1)!
respectively. Compare these polynomials with those for cosx and sinx.
♦
7.3 Taylor polynomials about x= a
In order to be able to use the formula given in the previous section for a Taylor polynomial,
we must have a function f (x) which not only exists at x= 0, but is also differentiable n times
at x = 0. Not all functions satisfy this requirement. The function f (x) = lnx is an obvious
example. It is not possible, therefore to approximate lnx with a polynomial using the formula
in the previous section. It is possible, however, to find Taylor polynomials about values other
than zero.
130 MATH 1021 Calculus of One Variable
Taylor polynomial about x= a
The nth order "Taylor polynomial" of a function f about x= a is
f (a)+ f ′(a)(x−a)+ f
′′(a)
2!
(x−a)2+ f
(3)(a)
3!
(x−a)3+ · · ·
· · ·+ f
(n)(a)
n!
(x−a)n
=
n
∑
k=0
f (k)(a)
k!
(x−a)k.
In order to use this formula the function f must exist at a and be differentiable n times at a.
Note that the formula given in the previous section, for a Taylor polynomial about 0, is just a
special case of this formula (with a= 0).
Example 7.3a Let us find a Taylor polynomial for f (x) = lnx. As mentioned above, we
cannot find one about x= 0. We must choose a value for a at which lnx exists, and is differ-
entiable n times. Suppose we choose a= 1, which should make evaluation of the derivatives
relatively simple. (There is no point in making things hard for ourselves!) So we have:
f (x) = lnx f (1) = 0
f ′(x) = 1/x f ′(1) = 1
f ′′(x) =−1/x2 f ′′(1) =−1
f (3)(x) = 2/x3 f (3)(1) = 2
f (4)(x) =−3!/x4 f (4)(1) =−3!
f (5)(x) = 4!/x5 f (5)(1) = 4!
...
...
...
...
f (n)(x) = (−1)n+1(n−1)!/xn f (n)(1) = (−1)n+1(n−1)!
The Taylor polynomial of order n for lnx, about x= 1, is:
Pn(x) = (x−1)+ −1
2
(x−1)2+ 2
3!
(x−1)3+ −3!
4!
(x−1)4+ · · ·+ (−1)
n+1(n−1)!
n!
(x−1)n
= (x−1)− (x−1)
2
2
+
(x−1)3
3
− (x−1)
4
4
+ · · ·+(−1)n+1 (x−1)
n
n
A somewhat neater and more user-friendly polynomial for approximating the logarithm func-
tion is obtained by letting t = x−1. Then x = 1+ t, and we have the following Taylor poly-
nomial for ln(1+ t), about t = 0 :
t− t
2
2
+
t3
3
− t
4
4
+ · · ·+(−1)n+1 t
n
n
.
♦
Chapter 7: Taylor Polynomials 131
7.4 Taylor’s formula – The remainder term
An approximation is not particularly useful if we do not have some idea of how good the
approximation is. If we approximate the value of a function by a value of its Taylor polyno-
mial, then it is important that we are able to say what the error in the approximation might
be. We do this by looking at the "remainder term", which is simply the difference between a
function and its Taylor polynomial. In the next chapter we will see that the remainder term
plays another important role as well.
Definition of remainder term
Given a function f (x) and its Taylor polynomial Pn(x) of order n, the "remain-
der term", Rn(x), is the difference between f (x) and Pn(x).
Rn(x) = f (x)−Pn(x).
In our discussion of the remainder term we will restrict our attention to Taylor polynomials
about x= 0.
Note – The fact that the remainder term is denoted by Rn(x) indicates that the difference
between a function and its Taylor polynomial depends on both n and x. The following ex-
pression for Rn(x) is called the "Lagrange form" of the remainder. (There are other forms of
the remainder which will not concern us here.)
Lagrange form of the remainder
Rn(x) =
f (n+1)(c)
(n+1)!
xn+1 for some c between 0 and x.
Note that the formula for the remainder term Rn(x) is easy to remember, since it is the same
as the last term of Pn+1(x), with 0 replaced by c. You will need to memorise this formula
and be able to use it to estimate errors in various Taylor approximations. If we substitute the
expression for Rn(x) into the equation
f (x) = Pn(x)+Rn(x)
we obtain the formula known as "Taylor’s formula":
132 MATH 1021 Calculus of One Variable
Taylor’s formula
f (x)= f (0)+ f ′(0)x+
f ′′(0)
2!
x2+
f (3)(0)
3!
x3+ · · ·+ f
(n)(0)
n!
xn+
f (n+1)(c)
(n+1)!
xn+1
for some c between 0 and x.
If we let n= 0 in Taylor’s formula, we have
f (x) = f (0)+ f ′(c)x, or f ′(c) =
f (x)− f (0)
x−0 for some c between 0 and x.
This is precisely the Mean Value Theorem (see Section 5.6). In fact, Taylor’s formula can be
considered as a generalised form of the Mean Value Theorem.
7.5 How good is the Taylor polynomial approximation?
Note that the Mean Value Theorem does not tell us how to find the value of c, and it is almost
always impossible to do so. In other words, we cannot expect to find an exact value for the
remainder. Often, however, we can find a bound on the value of the remainder that will tell
us how good (or bad) the Taylor polynomial approximation is.
Example 7.5a Suppose we want to estimate the value of cos(1) using P4(1), where P4(x) is
the Taylor polynomial of order 4 for cos(x) about x = 0. Find an upper bound for the error,
ie., find a bound for the remainder term.
Using Taylor’s formula we obtain cos(x) = P4(x)+R4(x) where P4(x) = 1− x
2
2!
+
x4
4!
and
R4(x) = f
(5)(c)
x5
5!
=−sin(c)x
5
5!
for 0< c< x.
Since we know that |sin(c)| ≤ 1 for any value of c, taking absolute values gives
|R4(x)|=
∣∣∣sin(c)x5
5!
∣∣∣≤ ∣∣∣x5
5!
∣∣∣= ∣∣∣ x5
120
∣∣∣.
Letting x= 1 gives
|R4(1)| ≤ 1
120
≈ 0.0083
and therefore
cos(1) = P4(1)+R4(1) = 1− 1
2!
+
1
4!
+R4(1)≈ 0.54166±0.00833,
or equivalently
0.53333< cos(1)< 0.54999.
The exact value from a calculator is cos(1) = 0.540302. ♦
Chapter 7: Taylor Polynomials 133
Taylor polynomials are also useful in calculating approximate values of definite integrals
in cases where standard methods of finding anti-derivatives fail. Polynomials are easy to
integrate, and we can use the remainder term to estimate the maximum possible error in the
approximation.
Example 7.5b In this example we estimate the value of the definite integral
∫ 1
0
ex
2
dx using
a Taylor polynomial. Note that there is no simple function which is an anti-derivative of ex
2
,
and so it is not possible to evaluate this integral by anti-differentiation.
The Taylor polynomial of degree 5 for ex is 1+ x+ x
2
2!
+ x
3
3!
+ x
4
4!
+ x
5
5!
, so Taylor’s Formula
gives
ex = 1+ x+
x2
2!
+
x3
3!
+
x4
4!
+
x5
5!
+R5(x),
where R5(x) = e
c x
6
6!
for some c between 0 and x.
This is an identity, true for all values of x, and so we can replace x by x2 to obtain
ex
2
= 1+ x2+
x4
2!
+
x6
3!
+
x8
4!
+
x10
5!
+R5(x
2),
and R5(x
2) = ec
x12
6!
for some c between 0 and x2.
Integrating both sides of this equation we have∫ 1
0
ex
2
dx=
∫ 1
0
(
1+ x2+
x4
2!
+
x6
3!
+
x8
4!
+
x10
5!
+R5(x
2)
)
dx
=
[
x+
x3
3
+
x5
5×2! +
x7
7×3! +
x9
9×4! +
x11
11×5!
]1
0
+
∫ 1
0
R5(x
2)dx
= 1+
1
3
+
1
10
+
1
42
+
1
216
+
1
1320
+
∫ 1
0
R5(x
2)dx
= 1.46253 . . .+
∫ 1
0
R5(x
2)dx.
Now, what can be said about the value of
∫ 1
0
R5(x
2)dx? We know that R5(x
2) = ec
x12
6!
for
some c between 0 and x2, and since we are integrating from 0 to 1, x lies between 0 and 1, as
does x2. So c also lies between 0 and 1, and hence 1= e0 < ec < e1 < 3. Therefore∫ 1
0
x12
6!
dx<
∫ 1
0
R5(x
2)dx< 3
∫ 1
0
x12
6!
dx,
and so
0.0001<
[
x13
13×6!
]1
0
<
∫ 1
0
R5(x
2)dx< 3
[
x13
13×6!
]1
0
< 0.0004.
We may therefore conclude that
∫ 1
0
ex
2
dx= 1.463 correct to 3 decimal places. ♦
134 MATH 1021 Calculus of One Variable
7.6 Proof of the remainder formula
The proof of the Lagrange form of the remainder relies on the Mean Value Theorem given in
Chapter 4. It is not part of the examinable material of the course but it is given here because
it is a very important application of the MVT.
We define an unlikely looking function g(t) as follows:
g(t) = f (x)− f (t)− f ′(t)(x− t)− f
′′(t)
2!
(x− t)2−·· · · · ·
· · · · · ·− f
(n)(t)
n!
(x− t)n−Rn(x)(x− t)
n+1
xn+1
.
The introduction of this particular function g in the proof probably seems somewhat myste-
rious. However, it was carefully chosen so that after applying the Mean Value Theorem, we
will obtain the formula we want.
First, we regard x as fixed at some non-zero value, and show that g(x) = 0 and g(0) = 0.
Substituting x for t into the formula for g(t) gives
g(x) = f (x)− f (x)−0−0−·· · · · ·−0= 0.
Substituting 0 for t into the formula for g(t) gives
g(0) = f (x)− f (0)− f ′(0)x− f
′′(0)
2!
x2−·· · · · ·− f
(n)(0)
n!
xn−Rn(x)x
n+1
xn+1
,
that is,
g(0) = f (x)−
(
f (0)+ f ′(0)x+
f ′′(0)
2!
x2+ · · · · · ·+ f
(n)(0)
n!
xn
)
−Rn(x),
in other words,
g(0) = f (x)−Pn(x)−Rn(x) = 0.
Now differentiate g(t) with respect to t (treating x as a constant):
g′(t) =0− f ′(t)− ( f ′′(t)(x− t)− f ′(t))
−
(
f ′′′(t)
2!
(x− t)2− f
′′(t)
2!
2(x− t)
)
−
(
f (4)(t)
3!
(x− t)3− f
′′′(t)
3!
3(x− t)2
)
−·· ·−
(
f (n+1)(t)
n!
(x− t)n− f
(n)(t)
n!
n(x− t)n−1
)
+Rn(x)
(n+1)(x− t)n
xn+1
Chapter 7: Taylor Polynomials 135
Most terms in this expression cancel out in pairs, and we are left with
g′(t) = Rn(x)
(n+1)(x− t)n
xn+1
− f
(n+1)(t)
n!
(x− t)n.
Now, since g(x) = 0 and g(0) = 0 we can apply the Mean Value Theorem. This guarantees
that there exists some number c between 0 and x such that g′(c) = 0. That is,
0= Rn(x)
(n+1)(x− c)n
xn+1
− f
(n+1)(c)
n!
(x− c)n
and solving for Rn(x) gives
Rn(x) =
f (n+1)(c)
(n+1)!
xn+1.
This completes the proof of the formula for the remainder term.
Summary of Chapter 7
• Taylor polynomials are introduced to approximate functions near a
point x = a in terms of the values of the function and its derivatives
evaluated at that point.
• The main reason for approximating functions by polynomials is that
polynomials are easy to calculate, requiring only addition and multipli-
cation.
• The remainder term is derived to be able to give an idea of the error
occured in the approximation. In other words, it will be able to tell us
how good the approximation is.
Exercises
7.1 Suppose f (x) = sinx and we want a straight-line approximation to f (x) accurate to
within 0.05 on an interval [0,h]. Use a calculator to find how big h can be (to
within 0.01 say) for:
a) The constant approximation (that is, the Taylor polynomial of order zero).
b) The approximation by the tangent line at x = 0 (that is, the Taylor polynomial of
order one).
7.2 Calculate the first 5 nonzero terms of the Taylor series for tanx at x= 0.
136 MATH 1021 Calculus of One Variable
7.3 Calculate the Taylor polynomial of order 4 at x= 0 for the function f (x) = e−x2 .
7.4 What happens when you calculate the Taylor polynomial at x = 0 of f (x), where f is
already a polynomial in x? Experiment to see what happens in the following examples:
a) (1+ x)2
b) (1+ x)3
c) (1+ x)n, for n a positive integer.
7.5 Compute the first 4 nonzero terms of the Taylor polynomials for the following functions
a) ln(1+ x)
b) secx
c) sinhx
d) coshx
e)
√
1+ x
7.6 Find the Taylor polynomial of order 3, about x= 1, for x3−3x2+3x−1.
7.7 Show that ex can be approximated to within 0.025 by the polynomial
1+ x+
x2
2!
+
x3
3!
+
x4
4!
for all x in the interval [−1,1]. (You may assume e< 3.)
7.8 Show that cosx= 1− x
2
2!
+
x4
4!
+R(x), where |R(x)| ≤ |x|
6
6!
.
The next four exercises concern the Mean Value Theorem (MVT), which is not ex-
aminable; nevertheless we encourage you to try these problems.
7.9 Use the Mean Value Theorem to show that if f (0) = 0 and f ′(x)≥ 7 for all x≥ 0, then
f (x)≥ 7x for all x≥ 0.
7.10 Suppose that f is a differentiable function with f (0) = 200 and f ′(x)<
1
2
for all x. Use
the Mean Value Theorem to show that f (1000)< 700.
7.11 Use the Mean Value Theorem to show that there is at least one point on the graph of
y= x7+3x4−4x+5
between x= 0 and x= 1 where the tangent to the curve is horizontal.
7.12 Use the Mean Value Theorem to prove the following tests for increasing and decreasing
functions that were referred to in Section 6.2.
a) Suppose f is a differentiable function such that f ′(x) > 0 for all x in some open
interval I. Prove that f is an increasing function on I. (That is, for any two points
a, b ∈ I such that a< b, use the MVT to prove that f (a)< f (b).)
b) Similarly, prove that if f ′(x) < 0 for all x in some open interval I then f is a
decreasing function on I.
C H A P T E R 8
Taylor Series
This chapter begins with an extremely brief introduction to the theory of infinite series. In-
finite series are important in many areas of mathematics, and the associated theory is exten-
sive. Our introduction to the theory is just sufficient to allow us to extend the idea of a Taylor
polynomial to “infinite polynomials”, or Taylor series, that were mentioned in the previous
chapter. Using the remainder term we are able to show that some functions f (x) are equal to
their Taylor series for all values of x. The complex exponential function is defined as a series,
and this leads us to revisit Euler’s formula which we first met in Chapter 2.
8.1 Infinite series
An "infinite sequence" is simply an infinite list of numbers a0, a1, a2, . . . ,an, . . ..
If we add up the terms in such a sequence we obtain an expression of the form
a0+a1+a2+ . . .+an+ . . .=
∞
∑
n=0
an.
We call such an expression an "infinite series" (or simply a "series"). It is not immediately
clear, however, what the sum of infinitely many numbers might mean.
A familiar example is the geometric series
1+
1
2
+
1
4
+
1
8
+
1
16
+ · · · .
You will have learnt in high school that this series is equal to 2, and if you start adding up the
terms 1+ 1
2
+ 1
4
+ · · · you will see that you quite quickly obtain a number close to 2.
However, it is not hard to see that each successive term is exactly half the difference between
2 and the sum of all the preceding terms, so that the sum will never actually reach 2. No
matter how many terms you have managed to add together, there are still infinitely many left!
So what do we mean, exactly, when we say
1+
1
2
+
1
4
+
1
8
+
1
16
+ · · ·= 2?
We need a definition, and for this purpose we introduce the idea of partial sums.
137
138 MATH 1021 Calculus of One Variable
Partial sum of an infinite series
The kth "partial sum", Sk, of the series
∞
∑
n=0
an is the sum of all the terms from
a0 to ak.
Sk = a0+a1+a2+ . . .+ak =
k
∑
n=0
an.
For any series we therefore have a sequence of partial sums,
S0 = a0
S1 = a0+a1
S2 = a0+a1+a2
...
Sk = a0+a1+a2+ . . .+ak,
and we define the sum
∞
∑
n=0
an as the limit of these partial sums Sk as k→ ∞, if it exists.
The sum of an infinite series
The "sum" of the infinite series
∞
∑
n=0
an is the limit, as k→ ∞, of the partial
sums Sk = ∑
k
n=0 an, provided this limit exists,
∞
∑
n=0
an = lim
k→∞
Sk = lim
k→∞
k
∑
n=0
an.
If the sequence of partial sums fails to converge to a limit we say that the
series diverges.
When the limit of the partial sums Sk exists and is equal to L, we say that the series "con-
verges" to L and we write
∞
∑
n=0
an = L.
For example, we show below that the infinite series 1+
1
2
+
1
4
+
1
8
+ . . . converges to 2.
The series 1+1+1+1+ · · · , for example, clearly increases indefinitely and hence diverges.
Chapter 8: Taylor Series 139
Geometric series
Consider the geometric series 1+ r+ r2+ r3+ · · · with common ratio r.
The kth partial sum of this series is the geometric progression
(8.1a) Sk =
k
∑
n=0
rn = 1+ r+ r2+ r3+ · · ·+ rk.
Now, multiply both sides by r:
(8.1b) r Sk = r+ r
2+ r3+ · · ·+ rk+ rk+1,
and subtract (8.1b) from (8.1a), noting that all terms except the first and the last cancel,
Sk− r Sk = 1− rk+1.
Finally, solving for Sk gives the expression for the sum of the first k terms as
Sk =
1− rk+1
1− r .
When r is a real number such that −1< r < 1, rk+1→ 0 as k→ ∞, and so
1+ r+ r2+ r3+ · · ·= lim
k→∞
(1+ r+ r2+ r3+ · · ·+ rk)
= lim
k→∞
1− rk+1
1− r
=
1
1− r .
However, if |r| > 1, the magnitude of rk+1 increases without bound as k increases, and
lim
k→∞
1− rk+1
1− r does not exist.
When r = 1, the series is the divergent series 1+1+1+ · · · .
When r = −1, we have the series 1− 1+ 1− 1+ 1− 1+ · · · . Consideration of the partial
sums in this case should convince you that this series diverges.
So the geometric series 1+ r+ r2+ r3+ · · · converges to 1
1− r if |r|< 1, and diverges other-
wise.
Example 8.1c Show that the series 1+
1
2
+
1
4
+
1
8
+ . . . converges to 2.
The series may be written in the form 1+
1
2
+
(1
2
)2
+
(1
2
)3
+ . . . which shows that it is a
geometric series with common ratio r =
1
2
and therefore its sum is S=
1
1− r =
1
1−1/2 = 2.
♦
140 MATH 1021 Calculus of One Variable
Warning: It is tempting to think that a series will converge if the terms in the series approach
zero. In general, this is not true. For example, the infinite series
∞
∑
n=1
1
n
= 1+
1
2
+
1
3
+
1
4
+ · · ·
diverges, that is, we can make the sum as large as we like by taking a sufficiently large number
of terms.
8.2 Taylor series
We have been considering series of constant terms which, when they converge, sum to a real
number. In this section we look at series with variable terms which, if they converge, sum to
a function. The importance of the role of infinite series in mathematics is largely due to the
fact that many functions have a representation as a series.
We have, in fact, already seen such a representation. In Section 8.1 we saw the formula for
the sum of a geometric series:
1+ r+ r2+ r3+ r4 · · ·= 1
1− r for |r|< 1.
Now consider this result from a slightly different point of view, by thinking of r as a variable.
Let’s replace r by x to reinforce this view, and turn the formula around, so we have
1
1− x = 1+ x+ x
2+ x3+ x4 · · · for |x|< 1.
Now we have a "power series", 1+ x+ x2+ x3+ x4 · · · , which is equal to the function 1
1− x
for |x|< 1. Notice that this power series is precisely the series we would obtain by extending
indefinitely the Taylor polynomial of
1
1− x .
We can find the Taylor series of any function simply by finding the Taylor polynomial, and
extending it to an infinite number of terms. The function must, of course, have derivatives of
any order at the point about which the polynomial is found.
Taylor series
The "Taylor series" for a function f about x= 0 is
f (0)+ f ′(0)x+
f ′′(0)
2!
x2+
f ′′′(0)
3!
x3+ · · · · · ·=
∞
∑
k=0
f (k)(0)
k!
xk
Taylor series about x= 0 are known asMaclaurin series
Chapter 8: Taylor Series 141
Examples 8.2a These examples use the polynomials we found in the previous chapter. All
the series are about the point x= 0.
i) The Taylor series for cosx: 1− x
2
2!
+
x4
4!
− x
6
6!
+ · · · .
ii) The Taylor series for sinx: x− x
3
3!
+
x5
5!
− x
7
7!
+ · · · .
iii) The Taylor series for ex: 1+ x+
x2
2!
+
x3
3!
+
x4
4!
+ · · · .
Try differentiating, term-by-term, each of the series above. What do you notice?
iv) The Taylor series for
1
1− x : 1+ x+ x
2+ x3+ x4+ · · · .
Try multiplying (1− x) by 1+ x+ x2+ x3+ x4+ · · · .What do you find?
v) The Taylor series for ln(1+ x): x− x
2
2
+
x3
3
− x
4
4
+ · · · .
vi) The Taylor series for coshx: 1+
x2
2!
+
x4
4!
+
x6
6!
+ · · · .
vii) The Taylor series for sinhx: x+
x3
3!
+
x5
5!
+
x7
7!
+ · · · . ♦
Simply knowing the Taylor series of a function f is not particularly useful. We need to know
the values of x for which the series converges to f (x). The answer to this question depends
on the function.
For example, we have seen above that the Taylor series for
1
1− x converges to
1
1− x for values
of x such that |x|< 1.
For some well-behaved functions f the Taylor series converges to f (x) for every value of x.
The functions sinx, cosx and ex all fall into this category. In order to see this, we consider the
behaviour of the remainder term, Rn(x), as n→ ∞.
The remainder term
Recall that the remainder term Rn(x) = f (x)− Pn(x), where Pn(x) is the nth order Taylor
polynomial of f . Rewrite the equation as f (x) = Pn(x)+Rn(x), consider x as fixed at some
particular value, and take the limit as n→ ∞. We therefore have
f (x) = lim
n→∞Pn(x)+ limn→∞Rn(x).
142 MATH 1021 Calculus of One Variable
Hence, if lim
n→∞Rn(x) = 0, then f (x) = limn→∞Pn(x). Now, limn→∞Pn(x) is the Taylor series of f ,
and so we have the following result:
Convergence of Taylor series
If lim
n→∞Rn(x) = 0 for a particular value of x, then the Taylor series of f con-
verges to f at that point.
For the functions sinx, cosx and ex we are able to show that lim
n→∞Rn(x) = 0 for all x.
Example 8.2b Recall that the remainder term Rn(x) is equal to
f (n+1)(c)
(n+1)!
xn+1 for some c
between 0 and x.
For the function f (x) = sinx, every derivative is one of sinx, cosx, −sinx or −cosx. There-
fore, no matter what the value of c, or of n, −1 ≤ f (n+1)(c) ≤ 1, or | f (n+1)(c)| ≤ 1. Hence,
the remainder term for f (x) = sinx is such that
|Rn(x)| ≤
∣∣∣∣ xn+1(n+1)!
∣∣∣∣ .
Now, lim
n→∞
xn+1
(n+1)!
= 0 for any real number x (see the technical aside below). It follows (by
the squeeze law) that lim
n→∞Rn(x) = 0, and so
sinx= x− x
3
3!
+
x5
5!
− x
7
7!
+ · · · for all x ∈ R .
Precisely the same argument applies to the function cosx, and
cosx= 1− x
2
2!
+
x4
4!
− x
6
6!
+ · · · for all x ∈ R .
♦

Technical aside We want to show here that lim
n→∞
xn
n!
= 0 for any real number x. Since x
can be positive or negative, it will be convenient to consider
∣∣∣∣xnn!
∣∣∣∣= |x|nn! .
Choose an integer k such that k > |x|. Then for n> k we can write
|x|n
n!
=
|x|
1
.
|x|
2
. . .
|x|
k
.
|x|
(k+1)
.
|x|
(k+2)
. . .
|x|
(n−1) .
|x|
n
.
Chapter 8: Taylor Series 143
The product of the first k terms is just some finite number; let’s call it K. Since k > |x|, each
of the terms |x|
(k+1)
,
|x|
(k+2)
, . . . ,
|x|
(n−1)
is less than 1. Therefore,
0≤ |x|
n
n!
<
K|x|
n
.
Since lim
n→∞
K|x|
n
= K|x| lim
n→∞
1
n
= 0, it follows (by the squeeze law) that lim
n→∞
|x|n
n!
= 0 and hence
lim
n→∞
xn
n!
= 0 for any real number x.
⊳
Example 8.2c Now consider the remainder term for the function f (x) = ex. Since any
derivative of ex is ex, we have f (n+1)(c) = ec, and
Rn(x) =
ecxn+1
(n+1)!
for some c between 0 and x.
If x is negative, and c is between 0 and x, then 0< ec < 1 and 0< |Rn(x)|< |x|
n+1
(n+1)!
.
Since lim
n→∞
|x|n+1
(n+1)!
= 0 =⇒ lim
n→∞ |Rn(x)|= 0.
If x is positive, 1< ec < ex (since ex is an increasing function) and 0< |Rn(x)|< e
x|x|n+1
(n+1)!
.
Now, lim
n→∞
ex|x|n+1
(n+1)!
= ex lim
n→∞
|x|n+1
(n+1)!
= 0, and once again lim
n→∞ |Rn(x)|= 0.
So the remainder term tends to zero as n→ ∞, and hence the Taylor series for ex converges
to the function ex for all x. That is,
ex = 1+ x+
x2
2!
+
x3
3!
+
x4
4!
+ · · ·=
∞
∑
n=0
xn
n!
for all real x.
Substituting the number 1 for x we obtain a series of constant terms for the number e itself:
Series for the number e
e= 1+1+
1
2!
+
1
3!
+
1
4!
+ · · ·=
∞
∑
n=0
1
n!
The 10th partial sum gives
e≈ 1+1+ 1
2!
+
1
3!
+ · · ·+ 1
10!
≈ 2.7182818,
which is accurate to the number of decimal places shown. ♦
144 MATH 1021 Calculus of One Variable
The series representations of the functions ex, sinx and cosx should be remembered. Here
they are again:
Series of elementary functions
ex = 1+ x+
x2
2!
+
x3
3!
+
x4
4!
+ · · ·
sinx= x− x
3
3!
+
x5
5!
− x
7
7!
+ · · ·
cosx= 1− x
2
2!
+
x4
4!
− x
6
6!
+ · · ·
It is important to realise that these equations are identities that hold for all values of x. In
other words, for each of the equations in the box above, the function on the left is exactly
equal to the sum on the right.
8.3 Euler’s formula
In this section we will briefly discuss series of complex terms. Our treatment of complex
series will not be rigorous because our sole aim is to show how Euler’s formula (which ex-
presses the exponential function in terms of the sine and cosine functions) can be discovered
using infinite series. If you have forgotten about Euler’s formula it would be a good idea to
revise Chapter 2 before reading this section.
The infinite series of complex numbers cn = an+ ibn is defined by
∞
∑
n=0
cn =
∞
∑
n=0
an+ i
∞
∑
n=0
bn.
Provided the two series of real terms on the right-hand side of this equation converge, we say
that the series of complex terms converges.
It can be shown that the exponential series
1+ x+
x2
2!
+
x3
3!
+
x4
4!
+ · · ·
converges whenever the real number x is replaced by a complex number x+ iy. (We are not
going to prove this here; it is beyond the scope of this course and you will just have to accept
it as a fact.) Consequently, we can define the exponential function ez for complex as well as
Chapter 8: Taylor Series 145
real numbers, by the series:
The complex exponential function
ez =
∞
∑
n=0
zn
n!
= 1+ z+
z2
2!
+
z3
3!
+
z4
4!
+ · · ·
Substituting the purely imaginary number z = iθ (where θ is real) into this infinite series
formula for ez, and using the fact that i2 =−1, we obtain
eiθ = 1+ iθ +
(iθ)2
2!
+
(iθ)3
3!
+
(iθ)4
4!
+ · · ·
= 1+ iθ − θ
2
2!
− iθ
3
3!
+
θ 4
4!
+
iθ 5
5!
+ · · ·
=
(
1− θ
2
2!
+
θ 4
4!
− θ
6
6!
+ · · ·
)
+ i
(
θ − θ
3
3!
+
θ 5
5!
− θ
7
7!
+ · · ·
)
.
But for any real θ , we have seen that
cosθ = 1− θ
2
2!
+
θ 4
4!
− θ
6
6!
+ · · · and sinθ = θ − θ
3
3!
+
θ 5
5!
− θ
7
7!
+ · · ·
Hence the expression for eiθ given above reduces to the famous "Euler’s formula".
Euler’s formula eiθ = cosθ + isinθ for all real θ .
We now have two ways to think about the complex exponential function ez: as an infinite
series of powers of z and as a concise formula in terms of the real and imaginary parts of z. If
z= x+ iy then
ez = 1+ z+
z2
2!
+
z3
3!
+
z4
4!
+ · · · = ex(cosy+ isiny).
8.4 The binomial series
One of the most useful series, particularly in applied mathematics, is the "binomial series". It
is the Taylor series of the function (1+x)p, where p is any real number, and is a generalisation
of the binomial theorem that we revise in Chapter 12.
To find the Taylor series, about 0, of (1+ x)p, we first list the derivatives. and evaluate them
at 0:
146 MATH 1021 Calculus of One Variable
f (x) = (1+ x)p
f ′(x) = p(1+ x)p−1
f ′′(x) = p(p−1)(1+ x)p−2
f ′′′(x) = p(p−1)(p−2)(1+ x)p−3
...
f (n)(x) = p(p−1)(p−2) · · ·(p−n+1)(1+ x)p−n
Now evaluate the derivatives at x= 0.
f (0) = 1
f ′(0) = p
f ′′(0) = p(p−1)
f ′′′(0) = p(p−1)(p−2)
...
f (n)(0) = p(p−1)(p−2) · · ·(p−n+1)
The Taylor series for (1+ x)p, also known as the "binomial series" is, therefore:
Binomial series
1+ px+
p(p−1)
2!
x2+
p(p−1)(p−2)
3!
x3+ · · · · · ·
= 1+
∞
∑
n=1
p(p−1)(p−2) · · ·(p−n+1)
n!
xn
⊲ Aside The above series is not always equal to (1+ x)p. In fact, for all values of p, the
series converges to (1+ x)p whenever |x| < 1, and diverges if |x| > 1. The proof of this is
beyond us at this stage. Whether or not the series converges for x=±1 depends on the value
of p. ⊳
Chapter 8: Taylor Series 147
Examples 8.4a
i) When p=−1, we obtain the following series for (1+ x)−1 = 1
1+ x
:
1+(−1)x+ (−1)(−2)
2!
x2+
(−1)(−2)(−3)
3!
x3+
(−1)(−2)(−3)(−4)
4!
x4+ · · ·
= 1− x+ x2− x3+ x4+ · · ·
Compare this series with that for (1− x)−1 in Example 8.2a.
ii) A special case occurs when p is a positive integer. In this case only the first p derivatives
of f (x) are non-zero, since the factor (p− p) = 0 appears in the Taylor series coefficient
of xp+1 and subsequent terms.
For example, with p= 3 the binomial series gives
(1+ x)3 = 1+3x+
3 ·2
2!
x2+
3 ·2 ·1
3!
x3+
3 ·2 ·1 ·0
4!
x4+ · · · .
The coefficient of x4 is zero, as are the coefficients of all higher powers of x. So we
have the familiar binomial formula
(1+ x)3 = 1+3x+3x2+ x3
in this case.
In general, the coefficient of xn in the Taylor series for (1+ x)p is
p(p−1)(p−2) · · ·(p−n+1)
n!
and in the case that p is a positive integer this can also be written as the binomial
coefficient
p!
n!× (p−n)! =
(
p
n
)
,
and the binomial series reduces to the binomial theorem:
(1+ x)p =
p
∑
n=0
(
p
n
)
xn.
iii) Show that the series for
1√
1− x = (1+(−x))
− 12 is
1+
∞
∑
n=1
1 ·3 ·5 · · ·(2n−1)
2 ·4 ·6 · · ·(2n) x
n.
♦
148 MATH 1021 Calculus of One Variable
8.5 A series for the inverse tan function
Sometimes it is not convenient to find the higher order derivatives of a function in order to
find its Taylor series. For example, try finding the first few derivatives of tan−1 x.
You should see that the derivatives are going to be very messy after the third derivative. In
this section we see an alternative method of finding the Taylor series of tan−1 x.
We note that
d
dx
tan−1 x=
1
1+ x2
=⇒
∫
1
1+ x2
dx= tan−1 x+C,
where C is an arbitrary constant.
Now, in Example 8.4a, we saw that the power series for
1
1+ x
is
1− x+ x2− x3+ x4− x5+ · · · .
Indeed, this is just a geometric series, and we know that
1
1+ x
= 1− x+ x2− x3+ x4− x5+ · · · for |x|< 1.
Replacing x by x2, we have
1
1+ x2
= 1− x2+ x4− x6+ x8− x10+ · · · for |x|< 1.
Now integrate this series term-by-term to obtain the following series for tan−1 x. We have
tan−1 x+C = x− x
3
3
+
x5
5
− x
7
7
+
x9
9
− x
11
11
+ · · · for |x|< 1.
Substituting x= 0 into both sides gives C = 0, and so
A series for tan−1 x
tan−1 x= x− x
3
3
+
x5
5
− x
7
7
+
x9
9
− x
11
11
+ · · · for |x|< 1.
What we have done here is to take the infinite series which converges to
1
1+ x2
for |x| < 1,
integrate it term-by-term, and claim that the resulting series converges to the integral of
1
1+ x2
for |x| < 1. It is by no means obvious that this is a legitimate thing to do. It is indeed true,
however, that the series shown above does converge to tan−1 for |x| < 1. What’s more, it
converges for |x| ≤ 1.
Chapter 8: Taylor Series 149
Substituting x= 1 into the equation tan−1 x= x− x
3
3
+
x5
5
− x
7
7
+
x9
9
− x
11
11
+ · · · gives a quite
remarkable formula:
A series for pi/4
tan−1 1=
pi
4
= 1− 1
3
+
1
5
− 1
7
+
1
9
− 1
11
+ · · · · · ·
While this is very nice series for pi/4, it is not an efficient way to compute pi , since one needs
to take a very large number of terms before pi is calculated with any degree of accuracy.
Summary of Chapter 8
• Infinite sequences and infinite series of numbers are briefly dealt with,
the aim being to apply these ideas to extend the definition of Taylor
polynomials to infinite polynomials or Taylor series.
• Taylor series are used to expand the exponential, trigonometric and
other elementary functions.
• Taylor series are then used to prove Euler’s formula
eiθ = cosθ + isinθ .
• Term-by-term integration of the convergent Taylor series of a function
f (x) sometimes leads to the series of the integral of f (x). We used
this result to calculate the Taylor series of tan−1 x from the series of its
derivative: (1+ x)−1 = 1− x+ x2− x3+ x4− x5+ · · · for |x|< 1.
Exercises
8.1 Show that the Taylor series for sinhx converges to sinhx for all real x, by showing that
the remainder term tends to zero as n→ ∞.
8.2 Write the repeating decimal 0.555555 . . . as the sum of a geometric series, and hence
as a rational number m/n.
150 MATH 1021 Calculus of One Variable
8.3 Write the repeating decimal 0.525252 . . . as the sum of a geometric series, and hence
as a rational number m/n.
8.4 Write the following complex numbers in the form reiθ :
a) w= 1− i b) z=√3+ i
8.5 Using the complex numbers z and w from the previous exercise, calculate
a) zw and (zw)12
b) ew in Cartesian form.
c) ez in Cartesian form.
8.6 a) Calculate the binomial series for the function f (x) = (1− x)−1/2 about x= 0.
b) Hence find a series for (1− x2)−1/2 and then a series for sin−1 x by term-by-term
integration.
C H A P T E R 9
The Riemann Integral
In this and the next few chapters we develop the theory of integration along with some of
its applications. One of the main reasons for getting a good understanding of integration is
that a major practical use of mathematics is concerned with building ‘mathematical models’
of physical, biological or financial systems. Typically these models are written in terms of
differential equations, which express relationships between the derivatives of the different
quantities in the system and their solution involves applications of the theory of integration.
9.1 Riemann sums – The area problem
Here we study the area problem which is to calculate the area of the region bounded by a
curve y = f (x) and the x-axis between two points x = a and x = b. The distance problem is
considered in Appendix D.
a) Start with a continuous function f (x) defined on a closed interval [a,b] and for simplic-
ity assume the function is non-negative, so f (x)≥ 0 for all x in the interval.
b) Fix an integer N ≥ 1 and divide the interval [a,b] into N subintervals of equal length
[x0,x1], [x1,x2] . . . [xi−1,xi] . . . [xN−1,xN ]
where x0 = a and xN = b, as shown in Figure 9.1 This is called a partition of the
interval [a,b]. The length of [a,b] is b−a and therefore the length of each subinterval
is
∆x=
b−a
N
.
c) The partition points are constructed as follows:
xi = a+∆x× i for i= 0,1,2 . . .N.
d) Then take the minimum value mi of f (x) on each subinterval, and draw rectangles of
this height based on the subintervals. The result for N = 8 subintervals is shown in
Figure 9.2.
e) Repeat the construction, this time using the maximum value Mi of f (x) on each subin-
terval. This is shown in Figure 9.3.
151
152 MATH 1021 Calculus of One Variable
Figure 9.1: Partition of the interval [a,b]
f (x)
a b
Figure 9.2: Rectangle height = mi.
a b
Figure 9.3: Rectangle height =Mi.
Properties
a) In the case of Figure 9.2 the total area of the rectangles is clearly a lower estimate for
the area under the graph of f (x).
b) Similarly the total shaded area in Figure 9.3 is an upper estimate for this area.
c) When the function increases, the minimum value mi occurs on the left point of the
subinterval [xi−1,xi], that is mi = f (xi−1) and the maximum value Mi occurs on the
right point of the subinterval [xi−1,xi], that is Mi = f (xi).
d) Viceversa, when the function decreases, the minimum valuemi occurs on the right point
of the subinterval [xi−1,xi], that is mi = f (xi) and the maximum valueMi occurs on the
left point of the subinterval [xi−1,xi], that is Mi = f (xi−1).
Chapter 9: The Riemann Integral 153
Lower and Upper Riemann sums
In Figure 9.2 the area of the ith rectangle is (height×base)=mi×∆x. Let LN be the total area
of the smaller rectangles. Then
(9.1a) LN = (m1×∆x)+(m2×∆x)+ · · ·+(mN×∆x) =
N
∑
i=1
mi×∆x .
The number LN is called a Lower Riemann Sum for the function f on the interval [a,b]. It
depends not only on N, but also on f and the interval [a,b].
In Figure 9.3 the area of the ith rectangle is (height×base) = Mi×∆x. Let UN be the total
area of the larger rectangles. Then
(9.1b) UN = (M1×∆x)+(M2×∆x)+ · · ·+(MN×∆x) =
N
∑
i=1
Mi×∆x .
This is called a Upper Riemann Sum for f on [a,b].
9.2 The Riemann integral
The Riemann lower and upper sums give us lower and upper estimates for the area under the
graph of f (x):
LN ≤ AREA UNDER THE GRAPH ≤UN .
Reall that Figures 9.2 and 9.3 correspond to N = 8 subintervals. Figure 9.4 shows the effect
of increasing the number of partition points N, taking first 16 and then 32 subintervals. The
f (x)
n= 16
f (x)
n= 32
Figure 9.4:
shaded area represents the difference between the upper and lower rectangles. The pictures
clearly suggest that the difference approaches zero as the number of intervals is increased. In
fact, careful analysis (using more sophisticated mathematical ideas than we have available at
this stage) confirms this intuition:
As the size of the subintervals is decreased to zero, the upper and lower Riemann sums ap-
proach a common value.
This is the number we call the definite integral or Riemann integral of f (x) over the inter-
val [a,b].
154 MATH 1021 Calculus of One Variable
We can summarize this as follows:
The Riemann Integral
Suppose that f (x) is a continuous function defined on the interval [a,b]. For
each integer N ≥ 1 we can divide [a,b] into N equal subintervals and form
the associated upper and lower Riemann sums. As N → ∞ both the upper
and lower sums approach the same value. This value is called the Riemann
integral (also called the definite integral) of f over the interval [a,b], and is
written as ∫ b
a
f (x) dx .
It is the unique number which satisfies
LN ≤
∫ b
a
f (x) dx≤UN
for all N ≥ 1. Since the area under the graph of f (x) satisfies the same in-
equalities, it must be equal to the integral of f over the interval. See below
for how to interpret the area in the case where f takes negative values.
Since both the upper and lower sums have the same limit as N increases, we
can also write
lim
N→∞
LN = lim
N→∞
UN =
∫ b
a
f (x) dx .
Non-positive Functions
If f (x) takes negative values we must modify the above argument slightly. As before we
divide [a,b] into N equal subintervals and let mi and Mi be the minimum and maximum
values of f on the ith subinterval. Now one or both of these numbers may be negative.
f (x)
a x b
Lower Sum Upper Sum
a x b
Figure 9.5:
Chapter 9: The Riemann Integral 155
If mi (or Mi) is negative the rectangle appears below the axis. The Riemann sum is therefore
equal to the sum of the areas of all the rectangles above the axis, minus the sum of the areas
of all the rectangles below the axis.
It is still true that the lower and upper Riemann sums converge to a common value. The
relation between the definite integral and area continues to hold, except that areas below the
axis count as negative.
NOTE – In this discussion we have always used a subdivision of the interval of integration
into subintervals of equal length. This is sufficient for many practical applications, but from a
theoretical point of view there are some advantages in relaxing this condition. If we subdivide
into (finitely many) subintervals of possibly different lengths the upper and lower Riemann
sums are still defined. It can be proved that both sums converge to the value of the definite
integral, as defined above, as the length of the longest subinterval is decreased towards zero.
9.3 Calculating Riemann sums
The Riemann sum provides a basic method to calculate approximate values of definite inte-
grals from which other, more accurate techniques are derived. Therefore it is important that
we become proficient at calculating Riemann sums.
Recall that if a function f (x) increases with x then the minimum value on an interval is
always at the left endpoint, and the maximum value at the right. If the function is decreasing
on the interval then the minimum value on an interval is always at the right endpoint, and the
maximum value at the left.
To handle the general situation we have to introduce a more general idea of Riemann sum.
The upper and lower Riemann sums will then appear as special cases of this construction. As
usual f (x) is a continuous function defined on the interval [a,b].
Suppose [a,b] is divided into N equal subintervals. For 1 ≤ i ≤ N, let ci be any point in the
ith subinterval and form the sum
(9.3a) ( f (c1)×∆x)+( f (c2)×∆x)+ · · ·+( f (cN)×∆x) =
N
∑
i=1
f (ci)×∆x .
This is the sum of areas of rectangles of width ∆x and height f (ci). Whatever the choice of
the ci, we certainly have
(9.3b) mi ≤ f (ci)≤Mi
since mi andMi are the minimum and maximum values of f on this subinterval. Multiplying
all sides of 9.3b by ∆x, gives
mi ∆x≤ f (ci) ∆x≤Mi ∆x
156 MATH 1021 Calculus of One Variable
and adding up over all subintervals we obtain
N
∑
i=1
mi ∆x≤
N
∑
i=1
f (ci) ∆x≤
N
∑
i=1
Mi ∆x
But the sums
N
∑
i=1
mi ∆x= LN and
N
∑
i=1
Mi ∆x =UN
and therefore we get the inequality:
(9.3c) LN ≤
N
∑
i=1
f (ci)∆x≤UN .
The middle expression here is an example of a (general) Riemann Sum for f (x) on the in-
terval [a,b]. Its value obviously depends on the choice of the ci. By taking N large enough,
we can make LN andUN as close as we like to the value of the definite integral. The inequal-
ities (9.3c) then imply that any Riemann sum must be at least as close to the actual value of
the integral. We illustrate this with an example.
Example 9.3d Use Riemann sums to estimate the integral
∫ 2
1
sinx dx .
Use a partition of [a,b] = [1,2] into N = 20 subintervals and calculate the Riemann sum for
each of the three cases:
a) ci is the left endpoint of the ith subinterval,
b) ci is the right endpoint of the ith subinterval,
c) ci is the midpoint of the ith subinterval.
Solution:
Each subinterval has length
∆x=
b−a
N
=
2−1
20
= 1/20= 0.05.
The partition points are given by
(9.3e) xi = a+∆x× i= 1.0+0.05× i for 0≤ i≤ 20.
The ith subinterval will then be the interval [xi−1,xi]. The three Riemann sums correspond to
the choices
(9.3f) ci = xi−1, ci = xi, ci = xi−1+0.025 .
In each case we have to work out the sum
(9.3g)
20
∑
i=1
sin(ci) ∆x=
(
20
∑
i=1
sin(ci)
)
×0.05 .
Chapter 9: The Riemann Integral 157
This is quite easy with a programmable calculator or by writing a simple computer program.
i xi−1 (1) (2) (3)
1 1.000000 0.841471 0.867423 0.854714
2 1.050000 0.867423 0.891207 0.879590
3 1.100000 0.891207 0.912764 0.902268
4 1.150000 0.912764 0.932039 0.922690
5 1.200000 0.932039 0.948985 0.940806
6 1.250000 0.948985 0.963558 0.956570
7 1.300000 0.963558 0.975723 0.969944
8 1.350000 0.975723 0.985450 0.980893
9 1.400000 0.985450 0.992713 0.989391
10 1.450000 0.992713 0.997495 0.995415
11 1.500000 0.997495 0.999784 0.998952
12 1.550000 0.999784 0.999574 0.999991
13 1.600000 0.999574 0.996865 0.998531
14 1.650000 0.996865 0.991665 0.994576
15 1.700000 0.991665 0.983986 0.988134
16 1.750000 0.983986 0.973848 0.979223
17 1.800000 0.973848 0.961275 0.967864
18 1.850000 0.961275 0.946300 0.954086
19 1.900000 0.946300 0.928960 0.937923
20 1.950000 0.928960 0.909297 0.919416
0.954554 0.957946 0.956549
We calculate the Riemann sums by summing the last three columns and using the for-
mula (9.3g). This gives the values shown in the bottom line of the table.
In this example it is easy to calculate the integral exactly:
∫ 2
1
sinx dx=
∫ 2
1
d
dx
(−cosx) dx= (−cos2)− (−cos1)≈ 0.956449 .
Note that in this example none of the three Riemann sums give the lower or upper Riemann
sum. The maximum value of sinx occurs at x= pi/2, which is inside the range of integration.
Up to this point the function is increasing. After this point it is decreasing. In some cases the
maximum value of the function occurs at the right of the subinterval, other times on the left,
and in one case (at pi/2) inside the subinterval. ♦
Example 9.3h Let the function f (x) = x2 be given on the interval [0,1]. Partition [0,1] into
four equal subintervals and calculate:
a) The lower Riemann sum L4.
b) The upper Riemann sumU4.
c) The exact value of
A=
∫ 1
0
x2 dx
and check that
L4 ≤ A≤U4.
158 MATH 1021 Calculus of One Variable
Solution:
In this problem, a= 0, b= 1, N = 4 therefore
∆x=
b−a
N
= 1/4= 0.25.
The partition points are
xi = a+∆x× i= 0.25× i for i= 0,1,2,3,4
that is, x0 = 0, x1 = 0.25, x2 = 0.50, x3 = 0.75, x4 = 1 and therefore the subintervals are
[0,0.25], [0.25,0.50], [0.50,0.75], [0.75,1].
a) Since f (x) = x2 is increasing in [0,1], the minimum values of f occur at the left point
of the subintervals, therefore the lower sum is
L4 =
[
f (0)+ f (0.25)+ f (0.50)+ f (0.75)
]
∆x
=
[
02+0.252+0.502+0.752
]×0.25= 0.21875.
b) Since f (x) = x2 is increasing in [0,1], the maximum values of f occur at the right point
of the subintervals, therefore the upper sum is
U4 =
[
f (0.25)+ f (0.50)+ f (0.75)+ f (1)
]
∆x
=
[
0.252+0.502+0.752+12
]×0.25= 0.46875.
c) The exact value of the integral is
A=
∫ 1
0
x2 dx=
x3
3
∣∣∣∣∣
1
0
= 1/3≈ 0.33333.
Since
0.21875≤ 0.33333≤ 0.46875 =⇒ L4 ≤ A≤U4.
♦
Example 9.3i Let the function f (x) = ex be given on the interval [a,b] = [0,1] and partition
[0,1] into N equal subintervals. Find the smallest value of N such that the difference
UN−LN ≤ 10−3.
Solution:
Divide the interval [a,b] = [0,1] into N subintervals of equal length
[x0,x1], [x1,x2] . . . [xi−1,xi] . . . [xN−1,xN ]
where x0 = a= 0 and xN = b= 1.
Chapter 9: The Riemann Integral 159
Since f (x) = ex is increasing in [0,1], the minimum values of f occur at the left point of the
subintervals and the maximum values occur at the right point of the subinterval.
Hence the lower Riemann sum is
LN =
[
f (x0)+ f (x1)+ · · ·+ f (xN−2)+ f (xN−1)
]
∆x
and the upper Riemann sum is
UN =
[
f (x1)+ f (x2)+ · · ·+ f (xN−1)+ f (xN)
]
∆x.
Subtracting LN from the expression forUN gives
UN−LN =
[
f (xN)− f (x0)
]
∆x=
[
f (b)− f (a)](b−a)
N
.
In this problem a= 0, b= 1 and f (x) = ex therefore
UN−LN =
[
e1− e0](1−0)
N
≤ 10−3
or
(e−1)
N
≤ 10−3 =⇒ N ≥ 103(e−1)≈ 1718.3.
Hence the smallest integer value of N that satisfiesUN−LN ≤ 10−3 is N = 1719. ♦
9.4 Properties of the Riemann integral
a) If m and M are the minimum and maximum values of f on the inter-
val [a,b], then
(9.4a) m× (b−a)≤
∫ b
a
f (x) dx≤M× (b−a) .
b) If c is a constant, then
(9.4b)
∫ b
a
c f (x) dx= c
∫ b
a
f (x) dx .
c) For functions f and g defined on the interval [a,b],
(9.4c)
∫ b
a
( f (x)+g(x)) dx=
∫ b
a
f (x) dx+
∫ b
a
g(x) dx .
d) If f is defined on the interval [a,c], and b is a point between a and b,
then
(9.4d)
∫ c
a
f (x) dx=
∫ b
a
f (x) dx+
∫ c
b
f (x) dx .
160 MATH 1021 Calculus of One Variable
The first formula (9.4a) is just the relation between the definite integral and the upper and
lower Riemann sums in the case N = 1, so there is a single interval equal to all of [a,b].
To see where the next two equations come from, we subdivide [a,b] into N subintervals. For
each i with 1≤ i≤N we choose a point ci in the ith subinterval. Let SN( f ) and SN(c f ) be the
corresponding Riemann sums for the two functions f (x) and c f (x). It follows immediately
from the definition of the Riemann sum that SN(c f ) = cSN( f ). Then, from the standard
properties of limits,
∫ b
a
c f (x) dx= lim
N→∞
SN(c f ) = lim
N→∞
cSN( f ) = c lim
N→∞
SN( f ) = c
∫ b
a
f (x) dx .
This proves (9.4b). For (9.4c), take N and the ci as before and let SN( f ), SN(g) and SN( f +g)
be the corresponding Riemann sums for f , g and f +g. Then
SN( f +g) =
N
∑
i=1
[ f (ci)+g(ci)]×∆x
=
(
N
∑
i=1
f (ci)×∆x
)
+
(
N
∑
i=1
g(ci)×∆x
)
= SN( f )+SN(g) .
Then ∫ b
a
( f (x)+g(x)) dx= lim
N→∞
SN( f +g)
= lim
N→∞
(SN( f )+SN(g))
= lim
N→∞
SN( f )+ lim
N→∞
SN(g)
=
∫ b
a
f (x) dx+
∫ b
a
g(x) dx .
If we interpret the definite integral as an area, the final formula (9.4d) is just the fact that the
area over the interval [a,c] is the sum of the areas over the two intervals [a,b] and [b,c]. A
formal mathematical proof of this depends on the more general type of Riemann sum (with
subintervals of different lengths) mentioned at the end of the previous chapter.
Reversing the Direction of Integration
So far we have only defined
∫ b
a
f (x)dx in the case where a ≤ b. We can easily extend the
definition to the case a > b in a way which is consistent with the existing definition and
properties. Note that in the definition of the Riemann sum as
(9.4e)
N
∑
i=1
f (ci) ∆x
Chapter 9: The Riemann Integral 161
we have ∆x = (b− a)/N. If a > b we can use the same formula. The only new feature is
that now ∆x is negative. From an algebraic point of view this has no effect on our formulas.
Geometrically it means that in the Riemann sum areas of rectangles above the axis now count
as negative, and areas below the axis count as positive.
The easiest way to see the implication of all this is to go back to the Riemann sum 9.4e and
look at the effect of interchanging a and b. The only difference which this makes to the
formula is to change the sign of ∆x. Hence the Riemann sum also simply changes sign. In
the limit as ∆x→ 0 the same is true of the definite integral, and we conclude that
∫ b
a
f (x) dx=−
∫ a
b
f (x) dx .
One result of this is that the formula (9.4d) now holds whatever the order of the num-
bers a, b, c. For example, starting with a≤ b≤ c and the relation
∫ b
a
+
∫ c
b
=
∫ c
a
we can rearrange to get ∫ b
a
=
∫ c
a
−
∫ c
b
=
∫ c
a
+
∫ b
c
.
This is essentially just (9.4d) again, except the point c no longer lies between a and b.
162 MATH 1021 Calculus of One Variable
Summary of Chapter 9
• Riemann sums were introduced using the calculation of the area under
a curve as a motivation. We assumed a function f (x) that is continuous
on an interval [a,b] and introduced a partition of [a,b] into N equal
subintervals.
• The upper and lower Riemann sums UN and LN provide upper and
lower estimates of the exact area, so that
LN ≤ AREA UNDER THE GRAPH ≤UN .
• The Riemann integral is defined as
lim
N→∞
LN = lim
N→∞
UN =
∫ b
a
f (x) dx .
provided the limit exists.
Exercises
9.1 Partition the interval [1,2] into N subintervals of equal length,
[x0,x1], [x1,x2], . . . , [xN−1, xN ] .
a) Show that a general point in the partition is
xi = 1+
( 1
N
)
i.
b) Show that the maximum value of f (x) =
1
x
on [xi−1,xi] is
Mi = N/(N+ i−1)
and the minimum value is
mi = N/(N+ i).
c) Show that the lower and upper Riemann sums for f (x) = 1/x on [1,2] with N
subintervals are
LN =
1
N+1
+
1
N+2
+ · · ·+ 1
2N
Chapter 9: The Riemann Integral 163
and
UN =
1
N
+
1
N+1
+ · · ·+ 1
2N−1 .
d) Find a value of N such that the difference between the upper and lower sums is
less than 10−6, that is,
UN−LN < 10−6.
9.2 Given that ∫ 1
−3
f (x)dx=−2 ,
∫ 2
1
f (x)dx= 5 ,
∫ 2
−3
g(x)dx= 8 ,
evaluate, where possible:
∫ 2
−3
( f (x)+g(x))dx,
∫ −3
2
g(x)
2
dx,
∫ 2
−3
f (x)g(x)dx .
C H A P T E R 10
Fundamental Theorem of Calculus
In the previous chapter we used the calculation of the area under a curve to motivate the intro-
duction of the definite integral. However, we have not so far discussed the exact mathemati-
cal relation between differentiation and integration that allows us to simplify the calculations.
There are two parts to this relation. One part involves the derivative of an integral and the
other the integral of a derivative. Together these parts make up the Fundamental Theorem
of Calculus.
10.1 Integrals as functions
In the previous chapter we introduced the definite integral
∫ b
a f (x) dx over a fixed interval
[a,b] as a real number representing the area under the curve f (x). In this chapter we will
allow the endpoints to vary so we can look at the definite integral as a function.
Dependence on the endpoint
Let f be a continuous function defined on an interval [a,b]. For any point x in the interval
we have the definite integral of f over the smaller interval [a,x]. Of course, the value of this
integral depends on x, so we can think of it as a function F(x) of x. Formally,
(10.1a) F(x) =
∫ x
a
f (t) dt .
Notice that we have used t as the variable of integration to avoid confusion with the use of x
as an endpoint of the interval of integration. The function F(x) is defined at least for all x in
the interval [a,b]. If f happens to be defined on a larger interval (or perhaps on the whole
real line), we can extend the definition of F accordingly via the formula (10.1a). In such
cases we can even define F(x) for x < a by the same formula. Some properties of F follow
immediately from the definition:
• Changing the starting point a changes F by the addition of a constant. For if we use a1
instead of a, then by (9.4d):∫ x
a1
f (t)dt =
∫ a
a1
f (t)dt+
∫ x
a
f (t)dt =C+F(x) ,
164
Chapter 10: Fundamental Theorem of Calculus 165
where C is the definite integral of f from a1 to a.
• If f is positive at a point x, then F is an increasing function at that point. For, if f (x) is
positive, increasing the range of integration adds a positive quantity to the area under
the graph. Similarly F is a decreasing function where f is negative.
• Intuitively, F(x) ought to be a continuous function. For if we change x to x+ ∆x,
then F(x) changes by the area under the graph between these two points. This area
approaches zero as ∆x→ 0. We will shortly prove the stronger result that F(x) is a
differentiable function of x.
Example 10.1b Sketch the graph of the function
F(x) =
∫ x
0
sin t dt
on the interval [0,2pi].
The graph of sinx on this interval is shown as the solid line in Figure 10.1.
0 2pipi
x
Area F(x)
0
F(x)
sinx
Figure 10.1:
At x = 0 the interval of integration has zero length, so F(0) = 0. Since sinx is positive
on [0,pi] the function F(x) is increasing over this interval. Similarly F(x) is decreasing on
the interval [pi,2pi]. Together these facts show that F(x) attains a maximum at x = pi . Over
the interval [0,2pi] the areas above and below the graph of sinx are exactly equal, so we must
have F(2pi) = 0.
We conclude that the graph of F(x) has the general shape shown by the broken line in Fig-
ure 10.1. Note that this curve has been drawn to show points of inflection where sinx has
stationary points. Can you see any justification for this? ♦
166 MATH 1021 Calculus of One Variable
10.2 The Fundamental Theorem of Calculus I
If we think about the integral itself as a function, we can look at what happens when we try to
differentiate this function. This is the idea behind the first part of the Fundamental Theorem
of Calculus (FTC).
The Fundamental Theorem of Calculus I
Let f (x) be a continuous function defined on an interval [a,b] of the real line
and let F(x) be defined by
(10.2a) F(x) =
∫ x
a
f (t)dt .
Then F(x) is a differentiable function of x and
(10.2b) F ′(x) = f (x)
for all x in the interval.
The significance of the result is that it confirms that every continuous function has an
antiderivative—given by the formula (10.2a). An alternative way to denote the Fundamental
Theorem of Calculus I is
(10.2c)
d
dx
(∫ x
a
f (t)dt
)
= f (x).
Proof of the FTC Part 1
To prove the first part of the Fundamental Theorem we go back to the definition of the deriva-
tive of a function as a limit. Recall that a function F is differentiable at a point x if the limit
of the differential quotient
F(x+h)−F(x)
h
exists as h→ 0. Then the value of the limit is the derivative F ′(x) of F at x. If F is defined
by (10.2a),
F(x+h) =
∫ x+h
a
f (t)dt =
∫ x
a
f (t)dt+
∫ x+h
x
f (t)dt = F(x)+
∫ x+h
x
f (t)dt ,
so
(10.2d) F(x+h)−F(x) =
∫ x+h
x
f (t)dt .
Chapter 10: Fundamental Theorem of Calculus 167
We now have to see what happens as h→ 0. LetM and m be the maximum and minimum of
f (t) for t between x and x+h. Assume h> 0. Then from (9.4a) we have
m×h≤
∫ x+h
x
f (t)dt ≤M×h .
Dividing by h and using (10.2d), we get
(10.2e) m≤ F(x+h)−F(x)
h
≤M .
This inequality can be derived similarily in the case h < 0. By the continuity of f both m
and M must converge to f (x), as h→ 0. But the middle term in the inequality (10.2e) is
sandwiched between m and M. So this must converge to f (x) also. This shows that the limit
exists, and F(x) is differentiable with derivative f (x).
End of proof
The first part of the Fundamental Theorem confirms that differentiation and integration are
‘inverse processes’: we now see that differentiating an integral gives back the original func-
tion. The standard notation for the antiderivative of a function f (x) is
(10.2f)
∫
f (x)dx .
In this form the antiderivative is also called the indefinite integral of f . Note that we can
always add an arbitrary constant without changing the fact that the derivative is f (x). For
example, we write ∫
cosxdx= sinx+C ,
to indicate that any function of the form sinx+C is an antiderivative for cosx.
10.3 The Fundamental Theorem of Calculus II
The Fundamental Theorem of Calculus II
Let F(x) be a function defined on an interval [a,b] of the real line. Suppose
that the derivative of F is defined at each point x of the interval, and that the
resulting function F ′(x) is continuous. Then
(10.3a)
∫ b
a
F ′(x) dx= F(b)−F(a) .
168 MATH 1021 Calculus of One Variable
Recall that the definite integral of F ′(x) is the unique number which lies between the upper
and lower Riemann sums of F ′(x) for all subdivisions of the interval [a,b]. If we can show
that the number F(b)−F(a) has the same property, then (10.3a) is proved.
For this we need theMean Value Theorem of differential calculus introduced in Section 5.6.
This theorem is very important in calculus, so here we rewrite the statement:
Mean Value Theorem
For a function F(x) defined on an interval [a,b] with continuous deriva-
tive F ′(x), there exists a point c somewhere in the interval with the property
that
F ′(c) =
F(b)−F(a)
b−a .
Essentially, this theorem says that the average rate of change over the interval (the ratio of
the change in F(x) to the change in x) is equal to the derivative of F at some point in the
interval.
To prove formula (10.3a) we introduce a partition of the interval [a,b] into N equal subinter-
vals and then apply the Mean Value Theorem to each subinterval of the partition. Label the
partition points as xi, with 0≤ i≤ N, so that
a= x0 ≤ x1 ≤ ·· · ≤ xN−1 ≤ xN = b .
and the ith subinterval is [xi−1,xi]. As usual we let mi, Mi be the minimum and maximum
values of F(x) on this subinterval, and ∆x the length of the subintervals. According to the
Mean Value Theorem there exists a point ci in the ith subinterval where
(10.3b) F ′(ci) =
F(xi)−F(xi−1)
xi− xi−1 .
Rearranging this equation and using the fact that xi− xi−1 = ∆x, we obtain
(10.3c) F(xi)−F(xi−1) = F ′(ci) ∆x.
Adding up over the range 1≤ i≤ N, the left side is just
[F(x1)−F(x0)]+ [F(x2)−F(x1)]+ · · ·
· · ·+[F(xN−1−F(xN−2)]+ [F(xN)−F(xN−1)] .
All the terms except F(x0) and F(xN) appear twice, with opposite signs. Therefore they
cancel out, leaving only
F(xN)−F(x0) = F(b)−F(a) ,
Chapter 10: Fundamental Theorem of Calculus 169
where we have used the fact that xN = b and x0 = a. Now, summing the right side gives the
Riemann sum
N
∑
i=1
F ′(ci) ∆x
for F ′(x) over the interval [a,b]. Finally, equating the left and right sides, we conclude that
N
∑
i=1
F ′(ci) ∆x= F(b)−F(a) .
But we know that any Riemann sum for the specified partition lies between the upper and
lower Riemann sums, that is,
LN ≤
N
∑
i=1
F ′(ci) ∆x≤UN ,
or equivalently,
LN ≤
∫ b
a
F ′(x) dx≤UN ,
and therefore ∫ b
a
F ′(x) dx= F(b)−F(a) .
This completes the proof of this part of the Fundamental Theorem.
Notation The change F(b)−F(a) of a function F over an interval is often denoted by
F(b)−F(a) = [F(x)]ba .
The theorem then appears in the form∫ b
a
F ′(x)dx= [F(x)]ba .
The important thing about this result is that it gives us a potential shortcut to working out a
definite integral. In order to evaluate the integral∫ b
a
f (x) dx
we can look for a function F(x) with the property that F ′(x) = f (x) on the interval [a,b].
According to the fundamental theorem we then have∫ b
a
f (x) dx=
∫ b
a
F ′(x) dx= F(b)−F(a).
The function F(x) is called an antiderivative of f (x). In this way the Fundamental Theorem
of Calculus gives us a link between ‘the area under the curve’ (in terms of Riemann sums)
and ‘reverse differentiation’, that is, finding an antiderivative.
170 MATH 1021 Calculus of One Variable
10.4 Leibniz Integral Rule
We have seen in Section 10.2, equation (10.2c), that the Fundamental Theorem of Calculus I
can be expressed in the form
d
dx
(∫ x
a
f (t)dt
)
= f (x).
For example, the derivative of the function defined by the integral
F(x) =
∫ x
0
sin2(t)dt
is
dF
dx
=
d
dx
(∫ x
0
sin2(t)dt
)
= sin2(x).
If the upper limit is x3 instead of x, then we have to use the chain rule to differentiate a
function of a function,
dF
dx
=
d
dx
(∫ x3
0
sin2(t)dt
)
= sin2(x3)× (3x2).
Leibniz integral rule is a generalization of this problem when both upper and lower limits are
functions of x.
Leibniz integral rule
d
dx
(∫ b(x)
a(x)
f (t)dt
)
= f
[
b(x)
]
.
d
dx
b(x)− f [a(x)]. d
dx
a(x).
Example 10.4a Calculate the derivative of the function F defined by
F(x) =
∫ x5
x2
cos3(t)dt.
Applying Leibniz integral rule gives
d
dx
(∫ x5
x2
cos3(t)dt
)
= cos3(x5).(5x4)− cos3(x2).(2x).
♦
Chapter 10: Fundamental Theorem of Calculus 171
10.5 The natural logarithm and exponential functions
Exponentials and natural logarithms have already made several appearances in this course,
but we have not given any justification of their main properties. We start with a precise
definition of these functions.
The natural logarithm
Note first of all that the function f (x) = 1/x is defined and continuous on the interval 0< x<
∞. According to the Fundamental Theorem of Calculus (and formula (10.2b) in particular)
we can define an antiderivative lnx of f (x) by the formula
lnx=
∫ x
1
1
t
dt .
This formula is valid for all x in the range 0 < x < ∞. Figure 10.2 shows the graph of the
function 1/x and its relation to lnx. Then lnx is the unique antiderivative of 1/x taking the
1 x
Area=lnx
Figure 10.2:
value 0 at x= 1. We take this formula as the definition of the natural logarithm, and use it to
prove the main properties of this function. Some of these properties are immediately obvious:
• ln1= 0,
• lnx> 0 if x> 1,
• lnx< 0 if 0< x< 1.
172 MATH 1021 Calculus of One Variable
NOTE We know that lnx is an antiderivative for 1/x for x > 0. If x < 0 then ln(−x) is
defined, and the chain rule gives
d
dx
ln(−x) =
(
1
−x
)(
d(−x)
dx
)
=
−1
−x =
1
x
.
We can combine the cases x< 0 and x> 0 into a single formula
(10.5a)
d
dx
ln |x|= 1
x
valid for all x 6= 0. Here we use the function |x|, where as usual
|x|=
{
x, if x≥ 0,
−x, if x< 0.
The formula (10.5a) should be used with caution. It is really two separate formulas, one for
x> 0 and the other for x< 0. It does not, for example, allow us to assign a value to a definite
integral over any interval including the point x= 0.
For any fixed positive number a we can define a function g(x) by the formula g(x) = ln(ax).
Then (using the chain rule)
g′(x) =
(
1
ax
)(
d(ax)
dx
)
=
a
ax
=
1
x
.
This shows that g(x) is also an antiderivative of 1/x, and hence differs from the function lnx
itself by a constant:
(10.5b) ln(ax) = ln(x)+C
for all x. We can evaluate the constant C by putting x= 1 in equation (10.5b). We find that
ln(a) = ln(1)+C,
and hence C = ln(a) (since ln(1) = 0). We have proved the formula
(10.5c) ln(ax) = ln(a)+ ln(x) ,
for all a, x> 0. For any a> 0 and integer n≥ 1 we also have the useful formula
ln(an) = ln(a ·a · · · · ·a) = lna+ lna+ · · ·+ lna= n lna .
As usual we define a−n = 1/an, and we also have
ln(an)+ ln(a−n) = ln(an×a−n) = ln1= 0 .
This gives us the formula ln(a−n) =−n lna.
Chapter 10: Fundamental Theorem of Calculus 173
The exponential function
From the definition we have d(lnx)/dx = 1/x > 0 for all x > 0, so lnx is an increasing
function of x. It must therefore be one-to-one on its domain of (0,∞). It can also be shown
that the range of the function lnx is the whole real line, so lnx can take any real value.
In the language of set theory this implies that the function x→ lnx is a bijection from (0,∞)
to the whole real line R. It follows that ln has an inverse function. This inverse is known
as the exponential function, and denoted by exp(x). The domain of exp is the range of ln,
which is R. The range of exp is the domain of ln. This is (0,∞), the set of all positive real
numbers. Figures 10.3 and 10.4 show the graphs of the two functions:
0 1 2 3
0
−1
1
y= lnx
Figure 10.3:
0 1−1
1
2
3
y= expx
Figure 10.4:
Since exp and ln are inverse to each other we have that
ln(exp(x)) = x for all x ∈ R
and
exp(ln(x)) = x for all x ∈ (0,∞).
Equivalently, y = lnx if and only if x = exp(y). The fundamental property of the logarithm,
contained in (10.5c) above, then shows that
ax= exp(ln(ax)) = exp
(
ln(a)+ ln(x)
)
for all a, x ∈ (0,∞).
If we let r = ln(a) and s= ln(x) then a= exp(r) and x= exp(s). It follows that
(10.5d) exp(r)exp(s) = exp(r+ s)
for all r, s ∈ R.
174 MATH 1021 Calculus of One Variable
We can calculate the derivative of exp(x) using standard properties of the derivatives of in-
verse functions. Alternatively we can get the result more directly by using the chain rule to
differentiate the identity
x= ln(exp(x)) .
This gives
1=
1
exp(x)
d
dx
exp(x) ,
which is easily rearranged into the formula
d
dx
exp(x) = exp(x) .
NOTE This shows that the exponential function is a solution to the differential equation
d f
dx
= f (x) .
It can be shown that if a function f is defined and differentiable on the real lineR and satisfies
f (0) = 1 and f ′(x) = f (x), then f (x) = exp(x) for all x. The property that the derivative is the
same as the original function is therefore said to be characteristic of the exponential function.
The General Exponential Function
We have already used the formula
ln(an) = n ln(a) ,
valid for any positive integer n and real number a > 0. Applying the exponential function
gives
an = exp(n ln(a)) .
In this formula we have assumed that n is a positive integer, but the expression on the right
is defined without any such restriction: n can be any real number. This suggests that we use
this formula to define ax for any pair of real numbers a, x (with a> 0), so
ax = exp(x ln(a)) .
For fixed a this is clearly a continuous (and even differentiable) function of x, and coincides
with the usual definition of an when x is a positive integer.
Chapter 10: Fundamental Theorem of Calculus 175
The number e. One feature of the above definition is that it immediately leads us to con-
sider the unique value of a with the property that ln(a) = 1. The existence and uniqueness
of such a number follows from our observation that the logarithm is a one-to-one map from
(0,∞) onto the whole real line. Since the exponential is inverse to the logarithm we see that
a = exp(1). This number is important enough to have its own name—it is called e and has
the approximate value
e≈ 2.7182818284590452354 . . . .
In particular, we see that
ex = exp(x ln(e)) = exp(x) ,
showing that the function exp is just the extension of the exponential powers en of this number
to real values. In fact ex is standard notation for the function exp(x).
Example 10.5e Show that the definition of the general exponential function still satisfies
(ab)c = acbc
for a, b> 0 and all c in R. From the definition, we have
(ab)c = exp(c ln(ab)) (definition)
= exp(c(lna+ lnb)) (by (10.5c)
= exp(c lna+ c lnb)
= exp(c(lna))exp(c(lnb)) (by (10.5d))
= acbc (definition).
♦
176 MATH 1021 Calculus of One Variable
Summary of Chapter 10
• The Fundamental Theorem of Calculus I tells us that
F(x) =
∫ x
a
f (t)dt =⇒ F ′(x) = f (x).
• The Fundamental Theorem of Calculus 2 shows that∫ b
a
F ′(x) dx= F(b)−F(a) .
• The natural logarithm is defined in terms of an integral as
lnx=
∫ x
1
1
t
dt .
The standard properties of lnx follow from this definition.
• The exponential function is defined as the inverse function of the log-
arithm. The standard properties of the exponential follow from this
definition also.
Exercises
10.1 When applying the Fundamental Theorem it is important to check that the conditions
for the theorem are satisfied. In particular, discontinuities in the function or its deriva-
tive can invalidate the formula. Consider the function f (x) = 1/x2. This is not defined
at x = 0 but everywhere else it has antiderivative −1/x. Since f (x) > 0 the integral
over any interval should certainly be positive.
Show that an attempt to apply the Fundamental Theorem over the interval [−1,1] (ig-
noring the difficulties at x= 0) leads to the contradictory result
∫ +1
−1
dx
x2
=−2 .
10.2 Find the derivative with respect to x of the following integrals:
a) F(x) =
∫ x
0
sin3 t dt,
b) F(x) =
∫ x3
0
sin3 t dt.
Chapter 10: Fundamental Theorem of Calculus 177
Hint: Think of the integral as a function of a function and use the chain rule.
10.3 Find the derivative of
∫ x
0
u2eu
2
du. Hence find the derivative of
∫ x2
cosx
u2eu
2
du.
10.4 Use the definition of the general exponential function to verify the following formulas
(for any a> 0).
a) axay = ax+y,
b) a0 = 1,
c) ln(ax) = x ln(a).
d) Show that for x> 0,
d
dx
xa = axa−1.
C H A P T E R 11
Integration Techniques
One way to evaluate a definite integral is to find an antiderivative of the function being in-
tegrated. From a practical point of view there is a big difference between differentiation and
integration. Differentiation is a fairly mechanical process, and follows a simple set of rules.
Finding formulas for antiderivatives is usually much less straightforward, and often depends
on recognising certain patterns in the function to be integrated.
In this chapter we look at the basic rules and then introduce the important methods of integra-
tion by substitution, integration by parts and partial fractions to integrate rational functions.
11.1 Basic rules of integration
In Section 9.4 we looked at some properties of the Riemann integral. The two relevant prop-
erties in this chapter are the so called linearity expressions:
Linearity properties
a) If k is a constant, then
(11.1a)
∫ b
a
k f (x) dx= k
∫ b
a
f (x) dx .
b) For functions f and g defined on the interval [a,b],
(11.1b)
∫ b
a
(
f (x)±g(x)
)
dx=
∫ b
a
f (x) dx±
∫ b
a
g(x) dx .
The expression (11.1a ) is easily extended to sums of more than two terms, so we can inte-
grate a sum of terms by integrating each term individually and adding up the results.
Rules (11.1a) and (11.1b) combined with the results from the Table of Standard Integrals
given in Appendix F make it possible to find antiderivatives in a large number of cases.
Example 11.1c Use the linearity properties and the table in Appendix F to calculate the
178
Chapter 11: Integration Techniques 179
integral: ∫
(ex+3x2+ cosx+4)dx=
∫
ex dx+3
∫
x2 dx
∫
cosxdx+4
∫
dx
= ex+ x3+ sinx+4x+C.
♦
Example 11.1d Find the area between the graphs of f (x) = x2+ 3 and g(x) = 2sinx from
x=−1 to x= 2.
Observe first that f (x) ≥ g(x) for all x in the given range. According to the argument given
in Example 12.3a the area we want is given by the definite integral
∫ 2
−1
( f (x)−g(x))dx=
∫ 2
−1
(
x2+3−2sinx) dx.
Using term-by-term integration and the standard integrals, we find that∫ (
x2+3−2sinx) dx= ∫ x2 dx+3∫ 1dx−2∫ sinxdx
=
x3
3
+3x+2cosx+C .
Therefore the area is[
x3
3
+3x+2cosx
]2
−1
=
(
23
3
+6+2cos(2)
)
−
(
(−1)3
3
+3(−1)+2cos(−1)
)
= 12+2cos(2)−2cos(−1) .
♦
11.2 Integration by substitution
Integration by substitution is based on the chain rule of differentiation that we studied in
Section 5.4. Recall that the chain rule is used to differentiate a composite function or function
of a function and says that if F and u are differentiable functions then
d
dx
F
[
u(x)
]
= F ′
[
u(x)
]
u′(x).
From the viewpoint of integration, this shows that F(u(x)) is an antiderivative of the function
on the right, so ∫
F ′
[
u(x)
]
u′(x)dx= F
[
u(x)
]
+C .
180 MATH 1021 Calculus of One Variable
The problem is clearer if we set f (u) = F ′(u), so that F itself is an antiderivative of f . The
resulting formula is called the change of variable formula. It is an extremely useful tool in
integration problems.
The Change of Variable Formula
(11.2a)
∫
f
[
u(x)
]
u′(x)dx=
∫
f (u)du .
How do we find f and u? Usually the first step is to try to identify the function u(x). The
requirements on u are:
• The expression u′(x)dx should appear as a factor in the integrand,
• The remaining factor should depend on x only via the function u(x).
The following examples illustrate the technique.
Example 11.2b Calculate the indefinite integral
∫
3x2 cos(x3)dx.
Note that the factor 3x2 is the derivative of x3. This means that the integrand comes from
using the chain rule on the function we are looking for. We therefore try the substitution
u= x3.
Differentiating gives du= 3x2 dx and the change of variable formula then becomes∫
3x2 cos(x3)dx=
∫
cos(x3) 3x2 dx
=
∫
cos(u) du= sin(u)+C = sin(x3)+C.
♦
Example 11.2c Find a formula for the indefinite integral
∫
2xdx√
x2+5
.
In this case the presence of the factor 2xdx suggests we try the substitution u = x2+ 5.
Differentiating gives du= 2xdx. The change of variable formula then becomes∫
2xdx√
x2+5
=
∫
(x2+5)−1/2 2x dx
=
∫
u−1/2 du=
u1/2
1/2
+C = 2
√
x2+5+C.
Note that the variable of integration switches from x to u, and we conclude by replacing u by
the function u(x) of x. ♦
Chapter 11: Integration Techniques 181
Example 11.2d Find an antiderivative for the function∫
(tanx)
(
ln(cosx)
)4
dx .
The choice of u is not so clear in this case. However we can write tanx =
sinx
cosx
and so the
integral becomes ∫ ( sinx
cosx
)(
ln(cosx)
)4
dx ,
suggesting the substitution
u= cosx , du=−sinxdx .
Then ∫ ( sinx
cosx
)(
ln(cosx)
)4
dx=
∫
(ln(cosx))4
( sinx
cosx
)
dx
=
∫
(lnu)4
(−1)
u
du .
This looks a bit better, but still needs work. Note the presence of the factor 1/u= d(lnu)/du,
along with the fact that the remaining factors depend on u via lnu. This leads to a second
substitution
v= lnu , dv=
1
u
du ,
giving ∫
(lnu)4
(−1)
u
du=
∫
−v4 dv .
Finally we get the result in terms of x by substituting v(u) for v and then u(x) for u. We can
do both substitutions in one step, replacing v by v(u(x)) = ln(cosx). The final result is
∫
(tanx)
(
ln(cosx)
)4
dx=
−v5
5
+C =
−(ln(cosx))5
5
+C .
♦
Trigonometric substitutions
The change of variables formula (11.2a) can also be applied the other way around. Then we
start with an integral with respect to u and turn it into an integral in x by the substitutions
u→ u(x) , du→ du
dx
dx ,
where u(x) is a suitable function of x. It is very common in this situation to use a trigono-
metric function for u(x). Then it may be possible to simplify the integral using the standard
trigonometric identities. Several ‘standard integrals’ can be evaluated in this way. For exam-
ple, starting with the integral ∫
du√
1−u2
182 MATH 1021 Calculus of One Variable
we observe that the substitution u→ sinx and the identity
cos2 x+ sin2 x= 1
can be used to eliminate the square root. With u→ sinx, we have
√
1−u2→
√
1− sin2 x= cosx , du→ du
dx
= cosxdx .
The integral is transformed as follows:
∫
du√
1−u2 =
∫
cosxdx
cosx
=
∫
dx= x+C .
To recover the result as a function of u, we need to invert the relation u = sinx to get x as a
function of u. But if u= sinx then x= sin−1 u (sometimes also called arcsinu). We therefore
recover the useful formula
∫
du√
1−u2 = sin
−1 u+C .
It is sometimes useful to combine two applications of the change of variables formula into
a single substitution. In this case we need to identify two functions u(x) and v(y) and make
substitutions
u(x)→ v(y) , du
dx
dx→ dv
dy
dy .
For example, the integral ∫
dx
(2x+1)2+1
is transformed by the substitutions
2x+1→ tanθ , 2dx→ sec2θ dθ .
into ∫ 1
2
sec2θ dθ
tan2θ +1
=
1
2
∫
dθ =
θ
2
+C =
1
2
tan−1(2x+1)+C .
Substitution in definite integrals
Any method which applies to finding antiderivatives also applies to the evaluation of definite
integrals. In the case of the substitution method we can avoid changing back the original
variable by a suitable change in the limits of integration. To see how this happens we return
the change of variables formula (11.2a):∫
f
[
u(x)
]
u′(x)dx=
∫
f (u)du .
Chapter 11: Integration Techniques 183
The variable of integration in the formula on the left is x. By contrast, the integral on the right
side of the formula is carried out with respect to u, so we have to change the limits to match.
Therefore the change of variable formula takes the form
∫ b
a
f
[
u(x)
]
u′(x)dx=
∫ u(b)
u(a)
f (u)du .
Example 11.2e Evaluate the definite integral
∫ √pi
0
xsin(x2)dx .
Setting u(x) = x2 gives the substitutions
x2→ u , 2xdx→ du .
Also u(0) = 0 and u(
√
pi) = pi , so the change of variable (and some juggling with the factor
of 2) gives
1
2
∫ √pi
0
2xsin(x2)dx=
1
2
∫ pi
0
sinudu=
1
2
[−cosu]pi
0
=
1− (−1)
2
= 1 .
♦
11.3 Integration by parts
Integration by parts transforms the problem into another integration problem, which may be
easier. The method is based on the product rule for differentiation:
(11.3a) (uv)′ = u′v+uv′,
where u = u(x) and v = v(x) are functions of x. Integrating both sides of (11.3a) yields the
formula
uv=
∫ (
u′ v+uv′
)
dx or uv=
∫ (
u′ v
)
dx+
∫ (
uv′
)
dx.
Rearranging the last expression we obtain the integration by parts formula:
Integration by parts∫ (
uv′
)
dx= uv−
∫ (
u′ v
)
dx.
184 MATH 1021 Calculus of One Variable
The formula is easier to use and to remember if it is written in terms of differentials as
follows (see Appendix C),
(11.3b)
∫
udv= uv−
∫
vdu.
Example 11.3c Use integration by parts to calculate
∫
x sinxdx.
If we differentiate x and integrate sinx we may be able to simplify the integral on the right
hand side. Therefore we try
u= x =⇒ du= dx and dv= sinxdx =⇒ v=−cosx.
Substituting into equation (11.3b), we obtain∫
xsinxdx= uv−
∫
vdu= (−cosx)x−
∫
(−cosx)dx
=−xcosx+
∫
cosxdx=−xcosx+ sinx+C.
♦
Example 11.3d Use integration by parts to calculate
∫
lnxdx .
In this example the choice u = lnx would eliminate the logarithm because du =
1
x
dx. Next,
we take dv= 1dx and therefore v= x. Substituting into equation (11.3b) yields
∫
lnxdx= x lnx−
∫
x× 1
x
dx= x lnx− x+C .
♦
Example 11.3e Use integration by parts to calculate
∫
x ln(x+1)dx.
In this case we try
u= ln(x+1) =⇒ du= 1
x+1
dx and dv= dx =⇒ v= x.
Substituting into equation (11.3b), we obtain∫
x ln(x+1)dx= uv−
∫
vdu.
=
x2
2
ln(x+1)−
∫ (
x2
2
)
1
x+1
dx
=
x2
2
ln(x+1)− 1
2
∫
x2
x+1
dx.
Chapter 11: Integration Techniques 185
Next, to calculate
∫
x2
x+1
dx we divide x2 by x+1 using polynomial long division,
x − 1
x+1 x2
x2 + x
− x
− x − 1
1
Therefore we can write
x2
x+1
=
(x−1)(x+1)+1
x+1
= (x−1)+ 1
x+1
and so
∫
x2
x+1
dx=
∫
(x−1)dx+
∫
dx
x+1
=
x2
2
− x+ ln(x+1).
Collecting all the pieces together, the final answer becomes∫
x ln(x+1)dx=
x2
2
ln(x+1)− 1
2
(
x2− x+ ln(x+1)
)
+C.
♦
Example 11.3f Use integration by parts to calculate
∫
x3 sinxdx.
Here we also let
u= x =⇒ du= dx and dv= sinxdx =⇒ v=−cosx.
Substituting into equation (11.3b), we obtain∫
x3 sinxdx= uv−
∫
vdu
= (−cosx)x3−
∫
(−cosx)3x2 dx
=−x3 cosx+3
∫
x2 cosxdx.
We have reduced the original problem to a simpler one. Repeating the method gives∫
x2 cosxdx= (sinx)x2−
∫
(sinx)2xdx
= x2 sinx−
(
(−cosx)2x−
∫
(−cosx)2dx
)
= x2 sinx+2xcosx−2sinx+C.
So we have shown that∫
x3 sinxdx=−x3 cosx+3x2 sinx+6xcosx−6sinx+C.
♦
186 MATH 1021 Calculus of One Variable
Example 11.3g Use integration by parts to calculate I =
∫
ex sinxdx.
u= sinx =⇒ du= cosxdx and dv= ex dx =⇒ v= ex.
Substituting into equation (11.3b), gives
I = ex sinx−
∫
ex cosxdx.
This is no simpler than the original problem, but applying the same method again gives
I = ex sinx−
(
ex cosx−
∫
ex(−sinx)dx
)
= ex(sinx− cosx)−
∫
ex sinxdx
= ex(sinx− cosx)− I.
We now have an equation for I. Adding I to both sides and dividing by 2 yields
I =
1
2
ex(sinx− cosx)+C.
Differentiate this expression and check that you really do get ex sinx. ♦
Definite integrals
In the case of definite integrals we can express the formula for integration by parts in the form
∫ b
a
udv=
[
uv
]b
a
−
∫ b
a
vdu.
Example 11.3h Use integration by parts to calculate
∫ pi/2
0
x sinxdx.
The indefinite integral was found in Example 11.3c,∫
xsinxdx= uv−
∫
vdu= (−cosx)x−
∫
(−cosx)dx
=−xcosx+
∫
cosxdx=−xcosx+ sinx+C.
Therefore∫ pi/2
0
xsinxdx=
[
uv
]pi/2
0
−
∫ pi/2
0
vdu= [(−cosx)x]pi/20 −
∫ pi/2
0
(−cosx)dx
=
[
− x cosx+ sinx
]pi/2
0
= 1.
♦
Chapter 11: Integration Techniques 187
11.4 Partial fractions
We are often led in applications (population growth, chemical reactions, etc.) to look for
integrals of the form
(11.4a)
∫
2x+1
(x−1)(x−2) dx.
The function r(x) =
2x+1
(x−1)(x−2) is an example of so-called rational functions. Rational
functions are expressions of the form f (x)/g(x), where f (x) and g(x) are polynomials.
In this example, f (x) = 2x+ 1, a polynomial of degree one and g(x) = (x− 1)(x− 2), a
polynomial of degree 2.
Example 11.4b As it stands the integral (11.4a) is difficult to calculate. However, if we split
the rational function r(x) as a sum of simpler functions in the form
2x+1
(x−1)(x−2) =
A
x−1 +
B
x−2 ,
then, it can easily be integrated. This process of spliting a rational function into sums of
simpler functions is called partial fractions.
The way we calculate the constants A and B is as follows. First find the common denominator
by cross-multiplying the right hand side,
2x+1
(x−1)(x−2) =
A
x−1 +
B
x−2 =
A(x−2)+B(x−1)
(x−1)(x−2) .
Now, the denominators on the left and right hand side terms are the same, therefore the
numerators must also be the same,
2x+1= A(x−2)+B(x−1)
for all values of x. Collecting terms on the right hand side, gives
2x+1= (A+B)x+(−2A−B).
Equating coefficients of like powers of x gives the system of equations
A+B= 2, −2A−B= 1,
and solving simultaneously gives a=−3 and b= 5. Therefore
2x+1
(x−1)(x−2) =
−3
x−1 +
5
x−2 .
The integral (11.4a) can now be calculated,∫
2x+1
(x−1)(x−2) dx=
∫ −3
x−1 dx+
∫
5
x−2 dx
=−3ln(x−1)+5ln(x−2)+C.
♦
188 MATH 1021 Calculus of One Variable
The same method works if there are three or more linear factors in the denominator.
Example 11.4c Find the partial fraction form for the function
2x2−1
x(x−1)(x+1) .
Note that the degree of the denominator is 3, while the degree of the numerator is 2. We need
to find the a, b and c such that
2x2−1
x(x−1)(x+1) =
a
x
+
b
x−1 +
c
x+1
.
There is a handy shortcut method, as follows. First find the common denominator and equate
numerators as before. But instead of equating coefficients of like powers of x and then solving
a simultaneous system, we let x= 0, x= 1 and x=−1 succesively as follows.
2x2−1= a(x−1)(x+1)+bx(x+1)+ cx(x−1).
To find a, put x = 0. Then two of the summands on the righthand side vanish. We find that
a= 1.
To find b, put x= 1. Two summands vanish and we find that b= 1
2
.
Finally, to find c we put x=−1, the summands involving x+1 vanish and we get c= 1
2
. So
we have
2x2−1
x(x−1)(x+1) =
1
x
+
1/2
x−1 +
1/2
x+1
.
♦
Important Rule
Partial Fractions applies to rational functions where the degree of the numer-
ator is strictly less than the degree of the denominator. If this condition
does not hold (as in the example 11.4d below) we have to divide the numerator
by the denominator using polynomial (long) division.
Example 11.4d
(11.4e)
x4+2x+2
x3− x2+ x−1 .
In this example the numerator (the upper term) has degree 4 and the denominator (the lower
term) has degree 3, therefore we have to use polynomial division first. The division leaves a
Chapter 11: Integration Techniques 189
remainder term which has degree less than the denominator:
x + 1
x3− x2+ x−1 x4 + 2x + 2
x4 − x3 + x2 − x
x3 − x2 + 3x + 2
x3 − x2 + x − 1
2x + 3
In each step we subtract off a multiple of the denominator, with the aim of eliminating the
highest power of x remaining. The process stops when the remainder has degree less than the
degree of the denominator. In this case the quotient is x+ 1 and the remainder 2x+ 3. That
is, we have shown that
x4+2x+2= (x+1)× (x3− x2+ x−1)+(2x+3)
or equivalently,
x4+2x+2
x3− x2+ x−1 =
(x+1)× (x3− x2+ x−1)+(2x+3)
x3− x2+ x−1
= x+1+
2x+3
x3− x2+ x−1 .
The polynomial term (x+1 in our example) is easy to integrate directly, and so the remaining
task is the integration of a rational function in which the degree of the numerator is less than
the degree of the denominator. This is where the partial fraction method comes in. First
factorise the denominator. In our example,
2x+3
x3− x2+ x−1 =
2x+3
(x2+1)(x−1) .
Now the task is to break the given rational function up into several rational functions, one for
each of the factors of the denominator. In each the degree of the numerator should be less
than the degree of the denominator. So in our case we need to find constants a, b and c such
that
2x+3
(x2+1)(x−1) =
ax+b
x2+1
+
c
x−1 .
The basic method for doing this is to clear denominators and equate coefficients. Thus,
multiplying through by (x2+1)(x−1) gives
(11.4f) 2x+3= (ax+b)(x−1)+ c(x2+1) = (a+ c)x2+(b−a)x+(c−b),
and we deduce that a+ c = 0, b− a = 2 and c− b = 3. We find that a = −5
2
, b = −1
2
and
c= 5
2
. By now we have found that
∫
x4+2x+2
x3− x2+ x−1 dx=
∫
(x+1)dx− 5
2
∫
x
x2+1
dx− 1
2
∫
1
x2+1
dx+
5
2
∫
dx
x−1 .
190 MATH 1021 Calculus of One Variable
The second integral on the right hand side requires the substitution v = x2+ 1, after which
the answer is found to be 1
2
ln(x2+1). The third integral is a standard integral, tan−1(x). The
fourth is easily found (after the straightforward substitution u = x− 1 to be ln |x− 1|. The
final answer is∫
x4+2x+2
x3− x2+ x−1 dx=
∫
(x+1)dx− 5
2
∫
x
x2+1
dx− 1
2
∫
1
x2+1
dx+
5
2
∫
dx
x−1 .
=
x2
2
+ x− 5
4
ln(x2+1)− 1
2
tan−1 x+
5
2
ln |x−1|+C.
♦
We will not attempt to give a full account of the theory of partial fractions here. There are
several different cases, depending on whether there are repeated factors in the denominator,
and whether the factors have degree one or two. The most important case is definitely the
case of distinct factors of degree one that we looked at.
Summary of Chapter 11
• Integration by substitution is a useful method of integration obtained
from the change of variable formula which is a consequence of the
chain rule for differentiation.
The change of variables formula can also be applied to definite inte-
grals. In this case it is necessary to change the limits of integration as
well as the variable of integration.
• Integration by parts is a powerful integration technique obtained from
the product rule for differentiation. The method involves separating the
function to be integrated into two factors and transforming the problem
into a different integral. In this process one factor is differentiated and
the other is integrated.
• Partial fraction expansions is a method for integrating rational func-
tions, that is, functions of the form P(x)/Q(x) where P(x) and Q(x) are
polynomials.
• Reduction formulas may be obtained when the integral involves an
integer parameter n. In this case, it may be possible to relate the inte-
gral to other instances of the same integral with smaller values of the
parameter.
Chapter 11: Integration Techniques 191
Exercises
11.1 Use substitutions to evaluate the following integrals:
a)
∫
x−1 logxdx,
b)
∫
e
√
x dx,
c)
∫
e2x√
ex+1
dx,
d)
∫ √
1+ x2
x4
dx .
11.2 Find
∫
2sinxcosxdx
a) by using the substitution u= sinx,
b) by using the substitution v= cosx,
c) by using the formula from trigonometry for sin2x.
Check that your three answers are consistent with each other.
11.3 Show that the substitution u= tanx leads to the formula∫
du
1+u2
= tan−1 u+C ,
where tan−1 u (or arctanu) is the inverse function to the tangent.
11.4 Use partial fraction expansions to find the following integrals:
a)
∫
dx
(x−1)(x−3) dx
b)
∫
x2 dx
(x−1)(x−3) .
11.5 Find a reduction formula for the indefinite integral
In =
∫
dx
(1+ x2)n
.
HINT: Take dv/dx= 1.
11.6 For a continuous function f (x) defined for x ≥ a the integral from a to ∞ is defined to
be the limit ∫ ∞
a
f (x)dx= lim
N→∞
∫ N
a
f (x)dx
(assuming the limit exists). By obtaining a suitable reduction formula, show that
∫ ∞
0
xne−x dx= n!
192 MATH 1021 Calculus of One Variable
for all integers n ≥ 0. This suggests the possibility of extending the definition of the
factorial function to non-integer values of the argument. This is one motivation for the
definition of the Gamma Function Γ(s) by the formula
Γ(s) =
∫ ∞
0
xs−1e−x dx ,
where s is any real number > 1. Although xs−1 is not defined at x = 0 when s < 1 we
can still set ∫ ∞
0
xs−1e−x dx= lim
a→0
∫ ∞
a
xs−1e−x dx
to extend the definition of Γ(s) to all s> 0. Then it is possible to check that Γ(n+1) =
n! (for integer n) and Γ(s+1) = sΓ(s) for all s> 0.
11.7 Verify the formula ∫ 1
0
xm(1− x)n dx= m!n!
(m+n+1)!
,
where m, n are integers ≥ 0.
C H A P T E R 12
Applications of Integration
In this chapter we apply a similar method to the calculation of area under a curve that we
used in Chapter 10 to calculate
a) The length of a curve,
b) The area between two curves,
c) Volumes of solids of revolution.
For example, for solids of revolution, we divide the volume into smaller sections whose
volumes are easy to calculate and then add them all up to obtain Riemann sums. By taking
the limit as the number of sections tends to infinity, we obtain a definite integral that can
be evaluated using the Fundamental Theorem of Calculus. Before that, however, we look at
more advanced techniques of integration.
12.1 Further integration techniques
Powers of trigonometric functions
Since some of the applications of sinθ and cosθ in terms of complex exponentials usually
involve the binomial theorem, we now revise it briefly.
Binomial theorem
The expression x+ y is called a binomial expression, and the binomial theorem is a general-
isation of the familiar formula (x+ y)2 = x2+ 2xy+ y2. It states that for all x, y and for all
integers n≥ 0,
(x+ y)n =
(
n
0
)
xn+
(
n
1
)
xn−1y+ · · ·+
(
n
r
)
xn−ryr+ · · ·+
(
n
n
)
yn.
Defining
nCr =
(
n
r
)
=
n!
r!(n− r)! ,
the binomial theorem becomes
(x+ y)n =n C0x
n+nC1x
n−1y+ · · ·+nCrxn−ryr+ · · ·+nCnyn.
The numbers
(
n
0
)
,
(
n
1
)
, . . . ,
(
n
n
)
are called the binomial coefficients.
193
194 MATH 1021 Calculus of One Variable
Thus, for example, with n= 4 we have
If n= 4 =⇒ (x+ y)4 = x4+4x3y+6x2y2+4xy3+ y4,
and
if n= 5 =⇒ (x+ y)5 = x5+5x4y+10x3y2+10x2y3+5xy4+ y5.
The binomial coefficients can also be found in the rows of Pascal’s Triangle, part of which is
shown below:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
Note that each row gives the binomial coefficients for a particular value of n in the expansion
of (x+ y)n. For example, in the expansion of (x+ y)3 the coefficients are 1,3,3,1.
Another interesting feature of Pascal’s triangle is that each entry is the sum of the two entries
in the row above it. In the last row shown above, 21 = 15+ 6, 35 = 20+ 15, and so on. So
you can add on as many additional rows as you require, remembering that each row begins
and ends with a 1, to enlarge the triangle to any n value you please.
Example 12.1a Calculate cos3θ using the binomial theorem with n = 3 (binomial coeffi-
cients 1, 3, 3, 1) and the fact that cosθ =
eiθ + e−iθ
2
.
cos3θ =
(eiθ + e−iθ
2
)3
=
1
8
(e3iθ +3e2iθ e−iθ +3eiθ e−2iθ + e−3iθ ) (binomial theorem)
=
1
8
(e3iθ +3eiθ +3e−iθ + e−3iθ ) (simplify exponents)
=
1
4
(1
2
(e3iθ + e−3iθ )+
3
2
(eiθ + e−iθ )
)
(collect conjugate terms)
=
1
4
(cos3θ +3cosθ)
♦
This formula for cos3θ may also be calculated simply using standard trigonometric identities,
but the working is not as elegant as this approach. The same technique can be used to find
expressions for any positive power of cosθ and sinθ , which can then be used in integration
problems.
Chapter 12: Applications of Integration 195
Example 12.1b Find
∫
cos3θ dθ .
We have ∫
cos3θ dθ =
∫
1
4
(cos3θ +3cosθ)dθ
=
1
4
∫
cos3θ dθ +
3
4
∫
cosθ dθ
=
1
12
sin3θ +
3
4
sinθ +C,
where C is an arbitrary constant. ♦
Example 12.1c Find a formula for sin4θ in terms of cos4θ and cos2θ .
Using the binomial theorem with n= 4 and the expression for sinθ in terms of exponentials,
we obtain
sin4θ =
(eiθ − e−iθ
2i
)4
=
1
16i4
(
e4iθ −4e3iθe−iθ +6e2iθe−2iθ −4eiθe−3iθ + e−4iθ
)
=
1
16
(
e4iθ + e−4iθ −4(e2iθ + e−2iθ )+6
)
=
1
8
(1
2
(e4iθ + e−4iθ )−4× 1
2
(e2iθ + e−2iθ )+3
)
=
1
8
(
cos4θ −4cos2θ +3
)
.
♦
So far we have shown how to obtain formulas for powers of sinθ and cosθ in terms of cosines
of multiples of θ , but we can also reverse the process to find formulas for expressions like
cos(nθ) and sin(nθ). In these cases we revert to the non-exponential polar form and apply
both de Moivre’s theorem and the binomial theorem to (cosθ + isinθ)n.
Example 12.1d Find formulas for cos4θ and sin4θ in terms of powers of sinθ and cosθ .
cos4θ + isin4θ =
(
cosθ + isinθ
)4
= cos4θ +4cos3θ
(
isinθ
)
+6cos2θ
(
i2 sin2θ
)
+4cosθ
(
i3 sin3θ
)
+ i4 sin4θ
= cos4θ −6cos2θ sin2θ + sin4θ + i(4cos3θ sinθ −4cosθ sin3θ)
196 MATH 1021 Calculus of One Variable
Equating the real parts on both sides of the equation gives
cos4θ = cos4θ −6cos2θ sin2θ + sin4θ
and equating the imaginary parts gives
sin4θ = 4cos3θ sinθ −4cosθ sin3θ .
♦
Reduction formulas
1) Indefinite integrals
In example 11.3f we found an antiderivative for the function x3 sinx by repeated application of
the integration by parts formula. It is sometimes useful to make this process more systematic.
For example, if we want to evaluate ∫
x10ex dx
we can use integration by parts (with u = x10) to reduce to a similar problem involving x9.
Repeating the process another nine times eventually eliminates the power of x altogether.
This promises to be very tedious. We can reduce the effort by performing the integration for
a general power xn of x. For n≥ 1 set
In =
∫
xnex dx.
Integration by parts with u= xn and dv/dx= ex gives
In = x
nex−
∫
nxn−1ex dx= xnex−nIn−1.
The formula In = x
nex− nIn−1 relating In and In−1 is an example of a reduction formula.
Starting with I0 we can apply this formula with n = 1, 2, 3, 4 . . . to generate successive In
directly. For example
I0 = e
x,
I1 = xe
x− ex,
I2 = x
2ex−2(xex− ex)
= (x2−2x+2)ex,
I3 = x
3ex−3(x2−2x+2)ex
= (x3−3x2+6x−6)ex.
In each case we get the general solution by adding on the usual arbitrary constant.
Chapter 12: Applications of Integration 197
n= 1
0 pi/2 n= 20 pi/2
n= 3
0 pi/2
Figure 12.1:
2) Definite integrals
We can apply the same technique to definite integrals. An interesting example involves the
integral
an =
∫ pi/2
0
(sinx)n dx
for a general positive integer n. Since 0≤ sinx≤ 1 for x in the range 0≤ x≤ pi/2 we have
0≤ (sinx)n+1 ≤ (sinx)n ≤ 1
for these values of x. Therefore∫ pi/2
0
(sinx)n+1 dx≤
∫ pi/2
0
(sinx)n dx
also, showing that the numbers an form a decreasing sequence. Figure 12.1 shows the cases
1≤ n≤ 3; here an is the shaded area under the curve.
For small values of n the integral is easy to evaluate. For example,
a0 =
∫ pi/2
0
dx=
pi
2
,
while
a1 =
∫ pi/2
0
sinxdx=
[−cosx]pi/2
0
= 1.
For n > 1 we can try to use integration by parts to get a reduction formula. For this we first
have to identify the factors u and dv/dx in (sinx)n. If dv/dx = sinx then v = −cosx and
u= (sinx)n−1. This gives
an =
[−(sinx)n−1 cosx]pi/2
0
+
∫ pi/2
0
(n−1)(sinx)n−2(cosx)2 dx.
Since sinx= 0 when x= 0 and cosx= 0 when x= pi/2, the first term on the right is zero. We
can use the familiar formula cos2 x= 1− sin2 x to rewrite the remaining term:
an =
∫ pi/2
0
(n−1)(sinx)n−2(1− (sinx)2)dx
=
∫ pi/2
0
(n−1)(sinx)n−2−
∫ pi/2
0
(n−1)(sinx)n dx
= (n−1)an−2− (n−1)an.
198 MATH 1021 Calculus of One Variable
This immediately rearranges to give
nan = (n−1)an−2 or an = n−1
n
an−2
for all n> 1. Note that this reduction formula gives two sequences of numbers, one for n even
and the other for n odd. For example
a0 =
pi
2
a2 =
pi
2
× 1
2
a4 =
pi
2
× 1
2
× 3
4
a6 =
pi
2
× 1
2
× 3
4
× 5
6
...
a2n =
pi
2
× 1
2
× 3
4
× 5
6
×·· ·× 2n−1
2n
,
while the odd sequence looks like
a1 = 1
a3 =
2
3
a5 =
2
3
× 4
5
a7 =
2
3
× 4
5
× 6
7
...
a2n+1 =
2
3
× 4
5
× 6
7
×·· ·× 2n
2n+1
.
12.2 Length of a curve
Consider a function f (x) that is continuous on an interval [a,b]. In order to calculate the length
of the curve from x= a to x= bwe partition the interval [a,b] into N equal subintervals. Label
the division points as xi, with 0≤ i≤ N, so that the typical ith subinterval is [xi−1,xi] and
a= x0 ≤ x1 ≤ ·· · ≤ xN−1 ≤ xN = b .
Figure 12.2 shows a section of the curve on the typical subinterval [xi−1,xi] of length ∆x =
xi− xi−1. If ∆x is small, the length of the curve between xi−1 and xi, denoted by ∆s, can
be approximated by the length of the straight line segment, as shown in the figure. By the
Pythagorean theorem the length of this segment is given by
(12.2a) ∆s=
√
∆x2+∆y2 = ∆x
√
1+
(∆y
∆x
)2
.
Chapter 12: Applications of Integration 199

Figure 12.2:
Also, if ∆x is small,
(12.2b)
∆y
∆x
≈ f ′(xi−1).
Inserting (12.2b) into (12.2a) and adding the subscript i, we have
∆si =
√
1+
[
f ′(xi−1)
]2
∆xi,
and therefore, the length of the curve from a to b is given approximatelly by the Riemann
sum
s≈
N
∑
i=1
∆si =
N
∑
i=1
√
1+
[
f ′(xi−1)
]2
∆xi.
Taking the limit as N→ ∞ the Riemann sum becomes the integral
(12.2c) s=
∫ b
a
√
1+
[
f ′(x)
]2
dx.
This integral gives the length of the curve f (x) between a and b and it is often written as
s=
∫ s
0
ds,
where
ds=
√
1+
[
f ′(x)
]2
dx
is called the arc length differential.
200 MATH 1021 Calculus of One Variable
Example 12.2d Use the integral expression (12.2c) to calculate the length of the straight
line y= x between x= 0 and x= 1.
We easily see, using elementary geometry, that the length of this segment of straight line is√
2. However, we will use formula (12.2c) to illustrate the method.
In this case f (x) = x and therefore f ′(x) = 1. The arc length differential is
ds=
√
1+
[
f ′(x)
]2
dx=
√
1+(1)2 dx=
√
2dx
and the total length is
s=
∫ s
0
ds=
∫ 1
0
√
2 dx=
√
2.
♦
Example 12.2e Calculate the circumference of a circle of unit radius x2+ y2 = 1 using the
formula for the arc length (12.2c).
We calculate the length of the upper semicircle y=
√
1− x2 and multiply by 2.
In this case f ′(x) =− x√
1− x2 , therefore
s=
∫ b
a
√
1+
[
f ′(x)
]2
dx=
∫ 1
−1
√
1+
x2
1− x2 dx=
∫ 1
−1
dx√
1− x2 dx.
Now make the substitution x= sinθ .
When x=−1 =⇒ θ =−pi/2 and when x= 1 =⇒ θ = pi/2. Therefore
s=
∫ pi/2
−pi/2
cosθ dθ
cosθ
=
∫ pi/2
−pi/2
dθ = pi/2+pi/2= pi.
Therefore, the circumference of the unit circle is equal to 2pi as expected. ♦
12.3 Area between two curves
Here we illustrate the method of Riemann sums to the calculation of area between two curves.
Example 12.3a Find a formula for the area between two curves y= f (x) and y= g(x) over
the interval a≤ x≤ b.
We suppose g(x)≤ f (x) in this interval. For a particular value of x we consider a thin strip of
width ∆x at the given value of x and lying between the two graphs. The length of the strip is
approximately f (x)−g(x). See figure 12.3. This gives an estimate of [ f (x)−g(x)]∆x for the
area of the strip. The Riemann sum for the total area is a sum of such terms. To indicate that
Chapter 12: Applications of Integration 201
∆x
f (x)
g(x)
x
Figure 12.3:
we shrink the length of the intervals towards zero we replace ∆x by dx and the summation
sign Σ by the sign
∫ b
a for the definite integral. Then the formula
∑ [ f (x)−g(x)]∆x
for the Riemann sum becomes the definite integral∫ b
a
[ f (x)−g(x)] dx.
♦
12.4 Solids of revolution
The disk method
Suppose we have a continuous function f (x)> 0 defined on the interval a≤ x≤ b. Consider
the region bounded by the x-axis, the lines x= a and x= b, and the graph of f . Now imagine
that this region is rotated around the x-axis, sweeping out a 3-dimensional solid of revolution.
We look at the problem of working out the volume of this solid. We have already seen that
we can approximate the area of the indicated region by a Riemann sum. Figure 12.4 shows
the area in question, as well as a typical rectangle for a Riemann lower sum on the interval.
If [a,b] is partitioned into N equal subintervals, then each subinterval has length
∆x=
b−a
N
.
As usual we letmi andMi be the minimum andmaximum values of f (x) on the ith subinterval.
If we take the rectangle of height mi based on this subinterval and rotate it around the x-axis
202 MATH 1021 Calculus of One Variable
x
f (x)
Figure 12.4:
it sweeps out a disk of thickness ∆x and radius mi. If we do the same for all rectangles
contributing to the Riemann lower sum we get a solid of revolution made up of coaxial disks
with a total volume of
N
∑
i=1
pim2i ×∆x .
Figure 12.5 shows the disk generated by rotation on a single rectangle.
Figure 12.5:
Since each disk is contained in the original solid of revolution, this gives us a lower estimate
for the volume of this solid. Replacing mi by Mi, the maximum value of f on the ith subin-
terval, gives an upper estimate. If V is the volume we are trying to calculate, then
N
∑
i=1
pim2i ×∆x≤V ≤
N
∑
i=1
piM2i ×∆x .
But the first and last expressions here are just lower and upper Riemann sums for the function
F(x) = pi f (x)2.
Chapter 12: Applications of Integration 203
Since the definite integral is the only number that satisfies all these inequalities, we conclude
that
(12.4a) V =
∫ b
a
pi f (x)2 dx .
If we can find an antiderivative G(x) for the function g(x) = pi f (x)2 then we can calculate the
volume directly as
V = G(b)−G(a).
For obvious reasons, this method of calculating volumes is sometimes called the disk
method.
A Simpler way to obtain the formula
For a solid of revolution we can get the formula for volume more directly by finding the
relation between a small change ∆x in x and the corresponding small change ∆V in volume.
Between points x and x+∆x the volume of the solid of rotation is approximately that of a
disk of radius f (x) and thickness ∆x. We write this as the formula
∆V ≈ pi f (x)2∆x ,
where ∆V is the contribution to the total volume coming from the interval between
x and x+∆x and ‘≈’ means ‘approximately equal’. Figure 12.6 shows the dimensions of the
disk in question.
∆x
f (x)
Figure 12.6:
Imagine the whole solid sliced up into thin disks of thickness ∆x. The total volume V is then
the sum of volumes of the individual disks, so
TOTAL VOLUME V = ∑∆V ≈∑pi f (x)2∆x .
204 MATH 1021 Calculus of One Variable
As the number of subintervals increases and the thickness of the disks is decreased towards
zero the error in the approximation also goes towards zero. The terms on the right are Rie-
mann sums converging to the definite integral of 12.4a. Therefore
V = lim
∆x→0∑pi f (x)
2×∆x=
∫ b
a
pi f (x)2 dx ,
as before. We will use this type of argument again in the examples which follow.
Example 12.4b Find a formula for the volume of the solid generated by rotating the graph
of the function
y= sinx, 0≤ x≤ 3
about the x-axis. Do not attempt to get a numerical answer; just express the answer as a
definite integral.
We give an abbreviated version of the preceding argument, which may help you reconstruct
the formula.
Slice the area under the graph of sinx on the interval [0,3] into thin vertical strips. A typical
such strip has width ∆x and approximate height sinx. Rotation of this strip about the x-axis
generates a disk of approximate radius sinx and thickness ∆x. So
Volume of disk≈ pi(sinx)2∆x .
As the strips are made narrower and narrower the sum of their volumes (over the interval
0≤ x≤ 3) converges towards the integral
∫ 3
0
pi(sinx)2 dx.
♦
The shell method
We have already looked at one method for calculation of the volumes of solids of revolution
using approximation by coaxial disks. Here we derive another method using approximation
by cylindrical shells.
Consider the region below the graph of the (positive-valued) function f (x) and above the x-
axis between x = a and x = b. We now look at the problem of finding the volume swept out
when this region is rotated around the vertical axis x= 0 instead of the horizontal x-axis.
This time, rotating a thin rectangle as shown in Figure 12.7 about the y-axis generates a
cylindrical shell of approximate height f (x) and thickness ∆x. In order to work out the total
volume, we need an approximate formula for the volume of the solid lying between radii x
and x+∆x. Since ∆x is small, the cylinder can be thought of a thin shell. We can imagine
cutting through it and flattening it out into a rectangle of width 2pix (the circumference of
Chapter 12: Applications of Integration 205
xa b
f (x)
Figure 12.7:
the cylinder) and height f (x) (the height of the cylinder). Since the thickness is ∆x this
contributes an amount
∆V ≈ 2pix× f (x)×∆x
to the total volume. Summing up over the subintervals of a partition of [a,b] gives
V ≈∑ 2pix f (x) ∆x .
Letting ∆x→ 0 gives the formula
VOLUME V = 2pi
∫ b
a
x f (x) dx .
In contrast to the disk method used previously, this is called the shell method.
It should be noted that the disk and shell methods have been used here to find the volume
of different solids. Sometimes, though, either method can be used to find the volume of the
same solid. Both methods will give the same result if correctly applied, but is often the case
that one method will lead to easier calculations than the other.
Example 12.4c Find a definite integral for the volume of a donut with circular cross-section,
inner hole of radius r and an outer radius of R.
In order to simplify the calculations we introduce the constants
a=
R− r
2
and b=
R+ r
2
.
These correspond to the dimensions shown in Figure 12.8, which shows a cross-section
through the donut. In order to apply the shell method we take a cylindrical shell of radius x
206 MATH 1021 Calculus of One Variable
2
√
a2− (x−b)2b
a
x
R r
Figure 12.8:
and thickness ∆x around the donut’s axis of symmetry. Simple trigonometry shows that the
height of this shell is 2
√
a2− (x−b)2. To estimate the volume of this cylinder imagine that
it is cut and rolled out into a rectangle of dimensions 2
√
a2− (x−b)2 by 2pix. Since the
thickness is ∆x, this contributes approximately
4pix
√
a2− (x−b)2 ∆x
to the total volume. We need to do this for r ≤ x ≤ R to get the total volume. Converting to
an integral we end up with the formula
∫ R
r
4pix
√
a2− (x−b)2 dx
for the total volume. ♦
Chapter 12: Applications of Integration 207
Summary of Chapter 12
• Applications of integration are usually based on the idea of decom-
posing some quantity into small pieces, finding a simple formula for
each of the pieces, and summing the results into a Riemann sum ap-
proximating the total quantity. Taking the limit as the size of the pieces
shrinks to zero gives the required quantity as a definite integral.
• The length of a curve was calculated by subdividing it into small sec-
tions called the arc length differential and then adding up all the small-
sections to construct a Riemman sum. Finally, taking the limit as the
number of sections tend to infinity, we obtained a definite integral that
gives the total length of the curve.
• Area between two curves and volumes of solids of revolution were
also calculated by subdividing them into thin disks or shells and then
taking the limit too obtain a definite integral.
Exercises
12.1 Consider the part of the hyperbola x2− y2 = a2 in the first quadrant and between the
lines x= a and x= b, where 0< a< b. Find the volume obtained by rotating this curve
a) About the x-axis,
b) about the y-axis.
12.2 Let R be the region in the first quadrant bounded by the coordinate axes and the parabola
y= 4−x2. Use the disc method to calculate the volume of the solid formed by revolving
R about the y-axis. Use the shell method to check your answer.
Appendix A
Formal Definition of Limits
The informal definition of limit given in Chapter 4 is quite clear in an intuitive sense; however,
it is not very precise. Example 4.1c shows that it may lead to the wrong guess and therefore
we need a more rigourous approach.
A mathematical way of saying that we can make f (x) “as close as we like to ℓ” is to say that
whenever ε > 0 is a “small” positive number then we can always ensure that f (x) is between
ℓ+ ε and ℓ− ε; that is ℓ− ε < f (x)< ℓ+ ε , or in concise mathematical notation,
| f (x)− ℓ|< ε.
The informal definition says that this should be true whenever x is “sufficiently close” to ℓ;
in other words, whenever c− δ < x < c+ δ for another “small” positive number δ ; that is,
|x− c| < δ . In fact, since the informal definition states that x is close to c but not equal to
c, the condition we actually require is 0 < |x− c| < δ . Putting all of this together gives the
following mathematically precise definition of a limit.
Formal (rigorous) definition of limit
Suppose that ℓ is a real number. Then the limit of f (x) as x approaches c is
equal to ℓ if for each number ε > 0 there exists a number δ > 0 such that
| f (x)− ℓ|< ε whenever 0< |x− c|< δ .
Since the ε , δ notation has become standard, this more precise version of the definition of
limit is also known as the ε−δ definition. The roles of ε and δ are illustrated in the following
diagram:
c
ℓ y= f (x)
c−
δ
c+
δ
ℓ+ε
ℓ−ε
This picture tells us what we need to do in order to test whether or not ℓ is the limit of f (x) as
x approaches c. For each ε > 0 we consider ℓ± ε , on the y–axis, and then look at the graph
208
Chapter A: Formal Definition of Limits 209
to see how small we need to make δ , on the x–axis, so that ℓ− ε < f (x) < ℓ+ ε whenever
c−δ < x< c+δ and x 6= c.
Another way of stating the formal definition is that given a number ε > 0, there exists a
number δ > 0 such that if 0 < |x− c| < δ then | f (x)− ℓ| < ε . It is important to understand
that the condition | f (x)− ℓ|< ε must be satisfied for values of x to the right of c on the real
number line (that is, when c< x< c+δ ) and also to the left of c (that is, when c−δ < x< c).
Thus the definition of “limit” is a two-sided one and a consequence of this is that if f (x) has
a limit as x approaches c, this limit is unique; that is, there is only one limit as x approaches c
regardless of whether it approaches from the left or the right.
Notice also that the value of δ will depend on ε and also on f (x); in general, there will be
a different value of δ for each value of ε and sometimes we emphazise this fact by writing
δ = δ (ε).
Example A.0a Use the formal definition to prove that, if f (x) = 4x− 6 then lim
x→2
f (x) = 2
(hence, in this example c= 2 and ℓ= 2).
This example is quite straight forward if we think about it intuitively. The purpose, however,
is to show how a formal proof is constructed. A simple problem is the best place to start.
The proof is done in two stages: 1) Guess a value for δ and 2) Show that the chosen δ works:
1) Guessing a value for δ
Let ε be a given positive number. We want to find a number δ such that
if 0< |x− c|< δ then | f (x)− ℓ|< ε
or, in our case,
if 0< |x−2|< δ then |(4x−6)−2|< ε
Now we have that | f (x)− ℓ|= |(4x−6)−2|= |4x−8|= 4|x−2|. Therefore we want:
if 0< |x−2|< δ then 4|x−2|< ε
or, equivalently,
if 0< |x−2|< δ then |x−2|< ε
4
.
This suggests that we should choose δ =
ε
4
.
2) Show that δ works
Given ε > 0, choose δ =
ε
4
. If 0< |x−2|< δ , then
| f (x)− ℓ|= |(4x−6)−2|= |4x−8|= 4|x−2| ≤ 4δ = 4
(ε
4
)
= ε.
That is, we have found a δ =
ε
4
so that for any ε ,
if 0< |x−2|< δ then |(4x−6)−2|< ε.
Therefore by the formal definition of limit, we have limx→2
(
4x−6)= 2.
♦
210 MATH 1021 Calculus of One Variable
Limits at infinity – Horizontal asymptotes
The definition of limx→c f (x) given in Section 4.1 tells us about the expected behaviour of
f (x) as x approaches the finite number c. We can also ask how f (x) behaves as x becomes
arbitrarily large and positive (x→ ∞) and arbitrarily large and negative (x→−∞).
Looking at the graph of the function f (x) =
1
x
below, it is intuitively clear that limx→∞
1
x
=
lim
x→−∞
1
x
= 0.
y= 1
x
Based on these observations, the precise definition of a limit as x→±∞ can be formulated
as follows:
Limits at infinity
Suppose that ℓ is a real number. Then the limit of f (x), as x approaches ∞, is
equal to ℓ if for each ε > 0 there exists a number N > 0 such that
| f (x)− ℓ|< ε whenever x> N.
In this case we write limx→∞ f (x) = ℓ.
Similarly, the limit of f (x), as x approaches −∞, is equal to ℓ if for
each ε > 0 there exists a number N > 0 such that
| f (x)− ℓ|< ε whenever x<−N.
In this case we write limx→−∞ f (x) = ℓ.
Referring back to the graph of the function f (x) =
1
x
we observe that the curve gets closer to
the x-axis (the line y= 0) when x→±∞. In this case we say that the line y= 0 is a horizontal
Chapter A: Formal Definition of Limits 211
asymptote of the curve y=
1
x
. In general, we have the following definition:
Horizontal asymptotes
The line y= L is called a horizontal asymptote of the curve y= f (x) if either
lim
x→∞ f (x) = L or limx→−∞ f (x) = L
As with limits as x→ c, we do not really want to use the formal definitions given above in
order to calculate limits at infinity; luckily, the limit laws we have already discussed also hold
for limits at infinity.
Appendix B
Geometric proof that limx→0 sinx/x= 1
Here we use a geometric argument to show that lim
x→0
sinx
x
= 1. This is the limit we calculated
in Example 4.1b using a more heuristic approach.
Intuitively, this seems unlikely because
1
x
becomes very large as x approaches 0, so we might
guess that
sinx
x
must also get very large. However, this limit is in fact equal to 1. The
problem with our intuition is that even though
1
x
does become very large as x approaches
0, sinx simultaneously becomes very small; what happens is that these two effects exactly
cancel out.
Just in case you are not convinced, Figure B.1 below shows again the graph of y=
sinx
x
with
the gap at x= 0.
y=
sinx
x
Figure B.1:
To show that lim
x→0
sinx
x
= 1 we are going to use the squeeze law.
Before we begin we need to recall some facts about angles measured in radians. From school
you know that the area of a unit circle is pi units squared. You should also know that the area
of a sector of the unit circle is
x
2
square units if its inner angle is x radians — the area of the
entire circle corresponds to the case x= 2pi .
x
Area =
x
2
square units
212
Chapter B: Geometric proof that limx→0 sinx/x= 1 213
In order to calculate lim
x→0
sinx
x
suppose that x is a non–zero angle measured in radians and
consider the following picture of a circle of radius 1 in the Cartesian plane.
x
1
sinx
tanx
O
A
B
C
D
Now, the area of the triangle AOD is less than or equal to the area of the sector AOD; in turn,
this sector has area less than or equal to the area of the triangle COD. We now compute these
areas. First consider the triangle AOD. Its base, the line OD, has length 1, and its height is
sinx (the length of the line AB). Therefore, AOD has area
1
2
sinx. Next, as noted above, the
area of the sector AOD is
x
2
. Finally, the length of the line CD is tanx, and hence the area of
COD is
1
2
tanx. Combining these equations we see that
1
2
sinx≤ 1
2
x≤ 1
2
tanx =⇒ sinx≤ x≤ tanx
=⇒ 1≤ x
sinx
≤ 1
cosx
(if sinx> 0),
=⇒ 1≥ sinx
x
≥ cosx,
where the last line follows by taking reciprocals. Remember when taking reciprocals of a
set of inequalities where all terms have the same sign, we reverse the inequalities. We have
assumed that x≥ 0 here so sinx≥ 0; you might like to write out the argument for x< 0.
Rewriting the last inequality, we have
cosx≤ sinx
x
≤ 1.
Now lim
x→0
1 = 1 and lim
x→0
cosx = cos0 = 1 because cosx is continuous. Therefore by the
squeeze law, lim
x→0
sinx
x
exists and equals 1, as claimed.
Appendix C
Linear approximations and differentials
Linear approximations
A differentiable function f (x) can be approximated, close to the point x = a, by the tangent
to the curve at that point. The equation of this tangent line is y = f (a)+ f ′(a)(x−a) which
can be written as the linear function
L(x) = f (a)+ f ′(a)(x−a),
called the linearisation of the function f at a. The idea is to approximate the function f near
x= a by its linearisation, that is,
f (x) ≈ L(x) near x= a.
Example C.0a Find the linearisation of the function f (x) =
√
x+3 at x = 1 and use it to
calculate approximations of the numbers
√
3.98 and
√
4.05.
To calculate approximations of the given numbers, we first need to choose a point near x= 1
to evaluate the function. Writing 3.98 = 0.98+ 3 gives us the clue that we need to choose
x= 0.98. Similarly, writing 4.05= 1.05+3 means that x= 1.05.
Now, the linearisation of f at x= 1 is
L(x) = f (1)+ f ′(1)(x−1)
= 2+
1
4
(x−1)
=
7
4
+
x
4
.
So we can approximate f (x) as follows:
f (x) =
√
x+3 ≈ 7
4
+
x
4
when x is near 1.
Therefore, √
3.98=
√
0.98+3 ≈ 7
4
+
0.98
4
= 1.995
and √
4.05=
√
1.05+3 ≈ 7
4
+
1.05
4
= 2.0125.
Because x = 0.98 and x = 1.05 are near x = 1, the approximations are reasonably good.
If fact, using a calculator we get the exact values
√
3.98 = 1.99499 and
√
4.05 = 2.01246
which, after rounding to four decimal places, coincide with the approximate values. ♦
214
Chapter C: Linear approximations and differentials 215
Differentials
In this section we introduce the concept of differentials and use them to find linear ap-
proximations to differentiable functions. Recall the alternative notation of the definition of
derivative as a function given in Section 5.2,
f ′(x) = lim
∆x→0
∆y
∆x
where ∆y= f (x+∆x)− f (x).
From the meaning of limit we see that for ∆x “small”, the ratio
∆y
∆x
is close to the derivative
f ′(x), that is,
∆y
∆x
≈ f ′(x) and therefore ∆y≈ f ′(x) ∆x.
The last equation tells us that if x changes by a small amount ∆x, then y will change by
approximately the amount ∆y ≈ f ′(x) ∆x. This motivates the introduction of the concept of
differential.
Differential
Let y = f (x) where f is a differentiable function and let ∆x be any nonzero
real number. Then
a) The differential dx is a variable given by dx= ∆x.
b) The differential dy of the function f is another function given by
dy= f ′(x)dx with alternative notation d f = f ′(x)dx.
The differential of a function of one variable is itself a function of two variables; both x and
dx are needed to evaluate the differential.
Note that if the definitions seem somewhat artificial it is because they have been introduced
so that we can manipulate the symbols dx and dy and treat the notation for the derivative
dy
dx
as a “ratio”, since in terms of differentials we have
dy
dx
=
f ′(x)dx
dx
= f ′(x).
Example C.0b Calculate the differential d f of the function f (x) = x3+5x2.
Since f ′(x) = (3x2+10x) then d f = f ′(x)dx= (3x2+10x)dx. ♦
Example C.0c Write down the differential d f of f (x) = 2xcosx and find an expression for
the differential in terms of dx when x= 0.
Since f ′(x) = 2cosx−2xsinx then d f = f ′(x)dx= (2cosx−2xsinx)dx and when x= 0 the
differential becomes d f = 2dx. ♦
216 MATH 1021 Calculus of One Variable
Relationship between the increment ∆y and the differential dy
Refering to the figure below, consider the function y = f (x) and let x0 be a fixed number
where f ′(x0) exists. For any value of ∆x, we have dx= ∆x. The equation
dy= f ′(x0) dx
is the equation of the tangent to the curve with slope f ′(x0) in the coordinates (dx,dy).
Observe carefully in the diagram that ∆y= f (x0+∆x)− f (x0) changes along the curve
y= f (x) while dy changes along the tangent line. Therefore, assuming small values of
dx= ∆x, the values of dy and ∆y become closer together, and we can say that
∆y≈ dy.
From the diagram we also note that f (x0+∆x) = f (x0)+∆y and using the approximation
∆y≈ dy we obtain f (x0+∆x) = f (x0)+∆y≈ f (x0)+dy. That is,
(C.0d) f (x0+∆x)≈ f (x0)+dy.
What this relation is telling us is that, if we have the value of a function f (x0) at a point x0, to
find an approximate value at a nearby point x0+∆x we only have to find dy= f
′(x0)∆x using
the simpler equation of the tangent line.
x
y
y= f (x)
f (x0+∆x)
∆y
f (x0)
x0 ∆x x0+∆x
dx= ∆x
dy
∆y
Example C.0e
Given the function f (x) =
√
3x+4, use differentials to find an approximate value for f (7.1).
We first need to choose a point to evaluate the differential. We can see that f (7) is easy to
evaluate since f (7) = 5, therefore a sensible choice is x0 = 7. Then the differential in x is
dx= ∆x= 7.1−7= 0.1 and the derivative
f ′(x) =
3
2
√
3x+4
and so f ′(7) =
3
10
= 0.3.
Hence
dy= f ′(x0)dx= 0.3×0.1= 0.03.
Chapter C: Linear approximations and differentials 217
This differential is the approximate change in f (x) between x = 7 and x = 7.1 and so an
approximation for f (7.1), as given by equation (C.0d), is
f (7.1) = f (x0+∆x) = f (7+0.1)≈ f (7)+dy= 5+0.03= 5.03.
Using a calculator gives f (7.1) = 5.02991 to six significant figures and so the approximation
using differentials is accurate to three significant figures. ♦
Example C.0f The volume of a sphere is given by V = 4
3
pir3 where r is the radius of the
sphere. By approximately how much is the volume of a sphere of 6cm radius reduced if 0.25
cm is shaved off the radius of the sphere?
Here we seek to find the differential of V at r = 6. The differential in r is dr =−0.25, since
we reduce the radius from 6 to 5.75. The differential inV is dV =
dV
dr
dr where the derivative
is evaluated at r = 6.
dV
dr
= 4pir2 = 144pi when r = 6.
So dV = 144pi ×−0.25 = −36pi ≈ −113.1. Therefore the volume of the sphere is reduced
by approximately 113.1 cm3 if 0.25 cm is shaved off its radius. ♦
Example C.0g Find the approximate error in the area of a circle of radius 30, if there is an
error of ±0.2 cm in the measurement of its radius.
Let A= pir2 be the area of the circle, where r is its radius. The error is approximated by the
differential of A: dA=
dA
dr
dr = 2pirdr.
Here r = 30 and |dr| = 0.2. We use the absolute value of dr here because it is possible that
the measurement is underestimated (dr < 0) or overestimated (dr > 0). We need only find
the magnitude of the error in A and so we seek |dA|.
|dA|=
∣∣∣∣dAdr dr
∣∣∣∣=
∣∣∣∣dAdr
∣∣∣∣ |dr|= 60pi×0.2= 12pi ≈ 37.7
Therefore the estimated maximum error in the area of the circle is 38 cm2. ♦
Example C.0h Relative error – Sometimes errors are given as a percentage or a fraction
of the measurement. This is known as the "relative error". These can be calculated using
differentials with only a little more work.
For example: if the relative error in the measurement of the radius of a circle is 5%, find the
relative error in the circle’s area.
If the relative error of the radius is 5%, or 0.05 as a decimal, we have
|dr|
r
= 0.05. We seek
to find
|dA|
A
:
|dA|
A
=
∣∣dA
dr
dr
∣∣
A
=
|2pirdr|
pir2
=
∣∣∣∣2pirdrpir2
∣∣∣∣=
∣∣∣∣2drr
∣∣∣∣= 2
∣∣∣∣drr
∣∣∣∣= 2×0.05= 0.1.
218 MATH 1021 Calculus of One Variable
Therefore, the error in the area of the circle is 0.1, as a decimal, or 10% as a percentage. As
you can see from the example, the aim is to rearrange the expression for
|dA|
A
until the term
dr
r
appears on the right hand side. ♦
Appendix D
The Distance Problem
Imagine a car accelerating along the road over a period of 10 seconds. Suppose the car starts
with a speed of 5m/sec and ends up with a speed of 32.5m/sec. What can we say about the
distance travelled? We can make a rough estimate as follows. Since the car is accelerating the
speed is always increasing. In particular, the speed is always between 5m/sec and 32.5m/sec.
Over the period of 10 seconds the car therefore travels at least 5×10= 50 metres, but no more
than 32.5×10= 325 metres. We can write these two inequalities as:
50m≤ DISTANCE TRAVELLED ≤ 375m .
This is a very rough estimate indeed. We can do much better if we know more about the
velocity at intervening points of time. For example, suppose that we measure the velocity
every two seconds. We can present the results as a table, which might look like the following:
Time (sec) 0 2 4 6 8 10
Velocity (m/sec) 5 14.5 22 27.5 31 32.5
The minimum and maximum velocities over the first two seconds are 5m/sec and 14.5m/sec.
Therefore the distance travelled in this period is between 5×2= 10m and 14.5×2= 29m.
Applying this to each interval in turn and adding up over all five intervals, we get a lower
estimate of
(D.0a) (5×2)+(14.5×2)+(22×2)+(27.5×2)+(31.0×2) = 200m,
and an upper estimate of
(D.0b) (14.5×2)+(22×2)+(27.5×2)+(31×2)+(32.5×2) = 255m.
The gap between the two estimates is now much smaller, with a maximum possible error of
255−200= 55m. It is very instructive to draw a graph of velocity against time and use it to
interpret these calculations.
This is done in Figure D.1 below, where the curved line shows the actual velocity of the car
plotted against the time t. On each 2 second interval along the t-axis the height of the dark
rectangle is equal to the minimum velocity on that interval. Since the velocity is increas-
ing, this always occurs on the left endpoint of each subinterval. Thus the first rectangle has
219
220 MATH 1021 Calculus of One Variable
height 5, the second height 14.5, and so on. The total height of the dark and light rectangles
together is equal to the maximum velocity on each interval. For an increasing function this
will occur at the right endpoint.
We relate this geometrical construction to the distance travelled by introducing the idea of
area. The width of each rectangle is a time interval and the height corresponds to our estimate
of velocity over the same interval. Therefore the product WIDTH×HEIGHT gives the distance
travelled during the interval, assuming the velocity is constant and equal to the height of the
rectangle. Of course this product is also just the area of the rectangle. Therefore the lower
estimate for the velocity given by (D.0a) is just the sum of the areas of the dark rectangles.
Similarly the expression (D.0b) is the total area of the dark and light rectangles taken together.
0 2 4 6 8 10
0
10
20
30
time t
v
el
o
ci
ty
m
/s
ec
.
Difference in Area
Figure D.1:
The difference between these upper and lower estimates of the distance is then equal to the
sum of the areas of the light rectangles. In Figure D.1 the light rectangles have been copied
over to the right of the diagram and stacked together, in order to better visualize their total
area. In fact it is easy to see that the composite rectangle has dimensions (32.5−5)×2, with
a total area of 55, in agreement with our earlier calculation.
With more data on the car’s speed we can improve accuracy further still. Suppose we record
the speed twice as often, so every second:
Time (sec) 0 1 2 3 4 5 6 7 8 9 10
Velocity (m/sec) 5 10 14.5 18.5 22 25 27.5 29.5 31 32 32.5
This gives us 10 intervals instead of 5, and we can again use the lowest and highest speeds
on each interval to estimate the distance travelled. This is shown graphically in Figure D.2.
Chapter D: The Distance Problem 221
0 2 4 6 8 10
0
10
20
30
time t
v
el
o
ci
ty
m
/s
ec
.
Difference in Area
Figure D.2:
As before the areas of the shaded rectangles give lower and upper estimates for the distance
travelled, and difference between these two estimates is equal to the total area of the light
rectangles. Comparison with Figure D.1 shows that this difference is now much smaller (in
fact it is equal to half its previous value). Adding up the upper and lower estimates of distance
over each 1 second interval gives us inequalities:
215m≤ DISTANCE TRAVELLED ≤ 242.5m .
The maximum possible error is now 242.5−215= 27.5m. Of course, we can continue in the
same way, using shorter and shorter subintervals. Figure D.3 shows the result of measuring
the velocity every 0.5 seconds.
0 2 4 6 8 10
0
10
20
30
time t
v
el
o
ci
ty
m
/s
ec
.
Difference in Area
Figure D.3:
The difference between the upper and lower estimates is again equal to the area of the rect-
angle drawn at the side of the figure. It should be clear by now that by taking small enough
steps, we can make this area as small as we like.
222 MATH 1021 Calculus of One Variable
In mathematical terms this means that both the upper and lower estimates approach a common
limit as the size of the steps shrinks towards zero. There is an obvious relation between this
limit and area under the curved line. In each case the lower estimate is the sum of areas of
rectangles which lie inside this curve. The upper estimate is a sum of areas of rectangles
which enclose the curve. The area under the curve, like the total distance, therefore also
lies between these upper and lower estimates. Since the upper and lower estimates have a
common limit, there is only one number with this property. We conclude that
TOTAL DISTANCE = AREA UNDER THE CURVE.
There are two new concepts here. First, we have a way of estimating total distance travelled
from a knowledge of velocity. By ‘sampling’ the velocity sufficiently frequently, we can make
this estimate as accurate as we wish. Second, we see that the total distance travelled is the
same as the area under the graph of velocity plotted against time. Note that the argument does
not depend on having an algebraic formula for the velocity, and it does not use any differential
calculus. The only place where we use the fact that velocity is rate of change of distance with
time is in the formula DISTANCE = VELOCITY× TIME for motion with constant velocity. In
the next section we generalize this argument into a purely mathematical construction which
we can apply to any continuous function.
Appendix E
Growth Rates
One of the important properties of a function f (x) is its growth rate—how does f (x) behave
as x→∞. According to the definitions of the previous section we have xa = exp(a lnx). From
the behaviour of the exponential and logarithm functions we deduce that
lim
x→∞x
a =


0, if a< 0,
1, if a= 0,
∞, if a> 0.
But the limit of a function f (x) as x→ ∞ is not the only interesting feature. Also significant
are the relative growth rates of different functions. We say that a function f (x) grows faster
than g(x) as x→ ∞ if
lim
x→∞
f (x)
g(x)
= ∞ .
This is equivalent to the condition g(x)/ f (x)→ 0 as x→ ∞. For example, if a > b then xa
grows faster than xb, since
lim
x→∞
xa
xb
= lim
x→∞x
a−b = ∞
(since a−b> 0).
How does the function lnx fit into this picture? In fact lnx grows more slowly than any
positive power of x. This is not obvious from the graph of lnx. Figure E.1 shows the graph of
lnx along with the graphs of xa for the cases a = 0.5 and a = 0.2. At least up to x = 10 the
logarithm seems to be grower faster than x0.2. But the power function eventually overtakes it,
as the following argument shows.
Lemma 1 For all x> 0 and b> 0 we have the inequality
ln(xb)< xb , or lnx<
xb
b
.
Proof. This is certainly true for 0 < x ≤ 1, since lnx ≤ 0 and xb > 0 for these values of x.
Define
F(x) =
xb
b
− lnx .
223
224 MATH 1021 Calculus of One Variable
0 2 4 6 8 10
0
1
2
3
x0.5
x0.2
lnx
Figure E.1:
Then F(1) = 1/b> 0 and
F ′(x) = xb−1− 1
x
=
xb−1
x
.
Since b > 0 this shows that F ′(x) ≥ 0 for all x ≥ 1. Since F is positive at x = 1 and non-
decreasing for x≥ 1, we conclude that F(x)> 0 for all such x.
Theorem 1 The function lnx grows more slowly than any positive power of x: for any a> 0
we have
lim
x→∞
lnx
xa
= 0 .
Proof. We use Lemma 1 for a particular value of b. Given a > 0 we can choose b with
0 < b < a. We could take b = a/2, for example. Then, for all x > 0, the previous lemma
gives
lnx
xa
<
xb
bxa
=
xb−a
b
.
Since b−a< 0 we see that xb−a→ 0 as x→ ∞.
Appendix F
Table of Standard Integrals
1.
∫
xn dx=
xn+1
n+1
+C (n 6=−1)
2.
∫
dx
x
= log |x|+C
3.
∫
ex dx= ex+C
4.
∫
sinxdx=−cosx+C
5.
∫
cosx,dx= sinx+C
6.
∫
sec2 xdx= tanx+C
7.
∫
cosec2 xdx=−cotx+C
8.
∫
sinhxdx= coshx+C
9.
∫
coshxdx= sinhx+C
10.
∫
dx√
a2− x2 = sin
−1 x
a
+C
11.
∫
dx
a2+ x2
=
1
a
tan−1
x
a
+C
12.
∫
dx√
x2+a2
= sinh−1
x
a
+C
13.
∫
dx√
x2−a2 = cosh
−1 x
a
+C (x> a> 0)
225
Appendix G
Answers to Selected Exercises
Chapter 1
1. (a) 7+ i (c) 5−12i (e) −i
(g) −(30/61)+(36/61)i (i) (3/13)−2i/13
(k) i (m) (3−5i)/2
2. w+ z= 8+16i, z−w= 2+8i, zw=−33+56i, z/w= (63+16i)/25
3. (a) (x2− y2)/(x2+ y2) (c) 3x2y− y3 (e) (x2+ y2)3 (g) (x2− y2)/(x2+ y2)2
4. (a) −1/2 (c) 1 (e) 8/17
5. (a) y=−1±2i (c) t = −1±
√
5
2
8. z must be real.
Chapter 2
1. (a) 1, −√3, 2, −pi/3 (c) 2, 2√3, 4, pi/3 (e) −1, 0, 1, pi
2. −4i= 4(cos(−pi/2)+ isin(−pi/2)) , −2+2i= 2√2(cos(3pi/4)+ isin(3pi/4)),
1− i=√2(cos(−pi/4)+ isin(−pi/4)) . (a) 4i (c) 8i
3. (a) 1 (c) 1
4. (a)
√
2e
pi
4 i (c) 2e−
pi
2 i
5. 4096
6. (3pi−8)/32
7. cos6θ = cos6θ −15cos4θ sin2θ +15cos2θ sin4θ − sin6θ .
8. z= i(
pi
2
+2kpi) for all k ∈ Z.
9. The roots are z= 2i and z= 3+ i. Note that p(z) is not a polynomial with real coefficients,
so we do not expect the non–real roots to occur in complex conjugate pairs.
10. 2±3i, 1, −2
11. 1± i, −1, 4
226
Chapter G: Answers to Selected Exercises 227
Chapter 3
1. (b) and (c)
2. (a) only
4. (a) R (c) R (e) {x ∈ R | x≥ 2} (g) R
5. (a) [0,∞) (c) [ln2,∞) (e) [0,1] (g) [cos1,1]
6. (a) f ◦g is defined for all real x; g◦ f is defined for all real x such that x≥ 0.
Chapter 4
1. (a) 7 (c) 0
2. (a) 3 (c) 2
3. (a) 0 (c) 0
4. (a) 3/4 (c) −1 (e) does not exist
5. (a) 0
6. (a) Limit is 0.
7. (a) −4 (c) 3
Chapter 5
1. (a) ex+5 (c) ex(x+1) (e) 99(x+1)98
(g) −sin tecos t (i) 2t tan(1− t2) (k) cos(sin(sinx)cos(sinx)cosx
2. (a) f ◦ f ′ =−x2, f ′ ◦ f =−x2 (c) f ◦ f ′ = 2, f ′ ◦ f = 0
3. (a) −x2/y2 (c) tanx tany (e) −2√y/x
4. (a) y= 2− x
Chapter 6
1. (a) 21x2−2− x−2
(c) −2xsin(x2)cos(cos(x2))
(d) 3x2 sinx+ x3 cosx+ excosx(cosx− xsinx)
2. y= 4x+1
3. (a) 9e−3,9e3
(c) 2,6
5. e−1
6. 1
228 MATH 1021 Calculus of One Variable
7. (a) 1 (c) 1 (e) 1 (g) −1
(i) −1
2
Chapter 7
1. (a) The constant approximation coinciding with sinx at x = 0 is identically zero. The
‘error’ is therefore equal to sinx itself.
(b) The tangent line at x= 0 is just the graph of x itself, so the error is x− sinx.
2. x+
x3
3
+
2x5
15
+
17x7
315
+
62x9
2835
3. 1− x2+ x
4
2
4. You get the same polynomial back again, at least if you take the Taylor polynomial of
degree equal to the degree of the given polynomial.
5. (a) x− 1
2
x2+
1
3
x3− 1
4
x3 (c) x+
x3
3!
+
x5
5!
+
x7
7!
(e) 1+
x
2
− x
2
8
+
x3
16
6. (x−1)3
Chapter 8
2.
5
9
3.
52
99
4. (a) w=
√
2e
pi
4 i
5. (a) zw= 2
√
2e−i
pi
12 , (zw)12 = (−2)18 (c) ez ≈ 3.05+4.76i
6. (a) 1+
1
2
x+
3×1
4×2x
2+
5×3×1
6×4×2x
3+
7×5×3×1
8×6×4×2x
4+ · · · .
(b) A series for sin−1 x is x+
1
2
x3
3
+
1.3
22.2!
x5
5!
+
1.3.5
233!
x7
7
+ · · ·
Chapter 9
1. (d) Any N > 5×105 (e.g. N = 500001) works.
2.
∫ 2
−3
( f (x)+g(x))dx= 11,
∫ −3
2
g(x)
2
dx=−4.
∫ 2
−3
f (x)g(x)dx cannot be evaluated with the given information.
Chapter 10
2. (b) 3x2 sin3(x3)
Chapter G: Answers to Selected Exercises 229
3.
d
dx
∫ x2
cosx
u2eu
2
du= 2x5ex
4
+ sinxcos2(x)ecos
2(x)
Chapter 11
1. (a)
(logx)2
2
+C (c)
2
3
(ex−2)√ex+1
2. −cos(2x)
2
4.
1
2
(log(3− x)− log(1− x))
5. 2nIn+1 = (2n−1) In+ x(1+ x2)−n
Chapter 12
1. (a)
pi
3
(2a3−3a2b+b3) (b) 2pi
3
(b2−a2) 32
2. 8pi