Apostila University of Aberdeen

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Department of
Mathematical Sciences
MA1002 Calculus
Integral Calculus
Dr John Pulham
ii
September 13, 1999, Version 1.2
Copyright  1999 by Ian Craw, John Pulham and the University of Aberdeen
All rights reserved.
Additional copies may be obtained from:
Department of Mathematical Sciences
University of Aberdeen
Aberdeen AB9 2TY
DSN: mth199-101465-0
Contents
1 Integration 1
1.1 The Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2 Doing Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.3 Some Properties of the Inde�nite Integral . . . . . . . . . . . . . . . . . . 4
1.4 *Impossible Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.5 The De�nite Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.5.1 Di�erentiating an Integral . . . . . . . . . . . . . . . . . . . . . . . 6
1.5.2 Properties of the De�nite Integral . . . . . . . . . . . . . . . . . . 7
1.5.3 In�nite Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
2 Applications of Integration 11
2.1 Integrals and Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
2.2 Intervals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
2.3 Further Properties of the De�nite Integral . . . . . . . . . . . . . . . . . . 15
2.4 Another View of the De�nite Integral . . . . . . . . . . . . . . . . . . . . 19
2.5 Further Applications of the De�nite Integral . . . . . . . . . . . . . . . . . 20
2.5.1 The Length of a Curve . . . . . . . . . . . . . . . . . . . . . . . . . 20
2.5.2 Volumes of Revolution . . . . . . . . . . . . . . . . . . . . . . . . . 22
2.5.3 Area of Surface of Revolution . . . . . . . . . . . . . . . . . . . . . 24
2.5.4 *Area in polar coordinates . . . . . . . . . . . . . . . . . . . . . . . 25
2.6 *Numerical Approximation to De�nite Integrals . . . . . . . . . . . . . . . 27
2.7 *Estimating the value of e . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
3 Methods of Integration 29
3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
3.2 Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
3.2.1 Substitution in De�nite Integrals . . . . . . . . . . . . . . . . . . . 32
3.3 The de�nite integral
∫
b
a
dx
x
. . . . . . . . . . . . . . . . . . . . . . . . . . 33
3.4 Integration by Parts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
3.4.1 *The Gamma Function . . . . . . . . . . . . . . . . . . . . . . . . . 37
3.5 Quadratics and Trigonometric Substitutions . . . . . . . . . . . . . . . . . 41
3.6 Integration of Rational Functions and Partial Fractions . . . . . . . . . . . 44
iii
iv CONTENTS
3.6.1 *Further Details . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
4 Differential Equations 49
4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
4.2 Separable Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
4.2.1 The Malthus Equation . . . . . . . . . . . . . . . . . . . . . . . . . 52
4.2.2 *The Logistic Equation . . . . . . . . . . . . . . . . . . . . . . . . . 53
4.3 Generalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
4.4 *Linear First-Order Equations . . . . . . . . . . . . . . . . . . . . . . . . . 54
Appendices 59
List of Figures
2.1 The area between a curve and the x-axis. . . . . . . . . . . . . . . . . . . . 11
2.2 Bounding an area between two rectangles. . . . . . . . . . . . . . . . . . . 12
2.3 The area between two curves. . . . . . . . . . . . . . . . . . . . . . . . . . 14
2.4 Another area. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
2.5 Estimating an area. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
2.6 Dividing up the area under a curve . . . . . . . . . . . . . . . . . . . . . . 19
2.7 One of the rectangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
2.8 Straight line approximation . . . . . . . . . . . . . . . . . . . . . . . . . . 20
2.9 A volume of revolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
2.10 The volume of a slice. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
2.11 Volume of rotation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
2.12 Describing an area in polar co-ordinates. . . . . . . . . . . . . . . . . . . . 25
2.13 The area of a very small segment. . . . . . . . . . . . . . . . . . . . . . . . 25
2.14 Approximating by a trapezium . . . . . . . . . . . . . . . . . . . . . . . . 27
2.15 The graph of ex lies between tangent and chord. . . . . . . . . . . . . . . . 28
4.1 Solutions of the Logistic Equation. . . . . . . . . . . . . . . . . . . . . . . 54
v
vi LIST OF FIGURES
How to use these Notes
The notes are divided into
� theory and explanation
� worked examples
� exercises
Most of the theory will also be presented in lectures. The exceptions are the sections
marked with a �. These are ‘extras’ and are not an examinable part of the course. They
are not necessarily more di�cult than the o�cial bits|it’s just that they are not in the
syllabus. Read them if you are interested. The more you read of everything the more you
are likely to understand.
Please study the worked examples carefully because that is usually the best way to
grasp the theory. There is also a lot to be said for treating some of the worked examples as
exercises in the �rst instance, trying to solve them yourself and only then reading through
my solution. It is usually rather silly to try reading mechanically through a worked example
when you have not yet got into your head what the example is really about.
The exercises come in two varieties|unstarred and starred. The starred exercises are
either a bit more di�cult than the others or else are less directly relevant to the course.
The more exercises you try the better, but do not get worried if you have not attempted
many of the starred questions. The standard of the examination questions is based on the
unstarred questions.
There are some solutions and some answers at the back of the book. Use these sensibly.
Make a serious attempt at a problem before looking up the answer or you will just be
wasting your time.
Chapter 1
Integration
1.1 The Integral
There are a number of di�erent ways to approach the Integral Calculus. We will start with
the least useful approach, which has the merit of being very simple to understand. We will
then look at a more complicated approach which is actually much more useful.
Our �rst approach is going to be to say that Integration is just the opposite (or inverse
operation) to Di�erentiation. If f(x) di�erentiates to give F (x) then, by de�nition, F (x)
integrates to give f(x). We have been calling F (x) the derivative of f(x) and we will now
call f(x) an Indefinite Integral of F (x). We use the notation
f(x) =
∫
F (x) dx
(This rather complicated notation derives from the other approach to integration that we
will be considering. Note that both the
∫
and the dx are part of the notation. If you like,
the
∫
says that we are dealing with an integral and the dx tells us the name of the variable
under consideration.)
As a simple example, sin(x) di�erentiates to give cos(x), so cos(x) integrates to give
sin(x). Or ∫
cos(x) dx = sin(x).
Now we notice a slight problem. sin(x) di�erentiates to give cos(x), but so do sin(x) + 1
and sin(x)− 34.23. In that case which of these is the integral of cos(x)? We seem to have
an ambiguity. This is true, and it is an ambiguity that will not go away. That is why I
carefully referred to f(x) as an inde�nite integral in the previous paragraph.
How much ambiguity do we have to deal with? Well,suppose that f1(x) and f2(x) are
both inde�nite integrals of F (x). That simply means that both f1(x) and f2(x) di�erentiate
to give F (x). But if f 01(x) = f
0
2(x) then, by the rules of di�erentiation the derivative of
f1(x)− f2(x) must be zero. So f1(x)− f2(x) must be a constant. So the only ambiguity is
that we can add an arbitrary constant to our integral.
1
2 CHAPTER 1. INTEGRATION
Many authors make this explicit by writing something like∫
cos(x) dx = sin(x) + C
where C stands for the arbitrary constant that we can add in. You may have been brought
up to do this all the time. You are welcome to do it if that is what you are used to. I will
tend to leave it out|simply because I know that it is always there (though there will come
a time later in the course when I will have to be more careful).
Let me now gather together some terminology. Consider the formula∫
cos(x) dx = sin(x) + C
The function cos(x) is called the Integrand, the variable x is called the variable of
integration, the constant C is called the Constant of Integration and sin(x) is called
the Indefinite Integral or Anti-Derivative.
1.2 Doing Integrals
Now that we know what integrals are the next problem is to �nd out how to do them.
This turns out to be a heavy problem. For di�erentiation we had a tidy set of rules that
allowed us to work out the derivative of just about any function that we cared to write
down|the procedure is basically mechanical and can be done quite well by computers.
There is nothing like this for integration, in general. Even (apparently) simple integrals
can be very di�cult or downright impossible to work out. More of this later. Integration
is more of a skill than a routine.
One approach that can work sometimes in simple cases is just ‘spotting the answer’. If
you can spot a function which di�erentiates to give your function then you have found an
integral. Look at the following simple examples.
d
dx
(
1
n + 1
xn+1) = xn so
∫
xn dx =
1
n + 1
xn+1 (n 6= −1)
d
dx
sin x = cos x so
∫
cos x dx = sin x
d
dx
cos x = − sin x so
∫
sin x dx = − cos x
d
dx
eax = aeax so
∫
eax dx =
1
a
eax
d
dx
lnx =
1
x
so
∫
dx
x
= ln x
Let me do something a little more complicated. We have seen that
∫
cos(x) dx = sin(x).
What then is the value of
∫
cos(2x) dx?
1.2. DOING INTEGRALS 3
We want to know what function di�erentiates to give cos(2x). Well, if you di�erentiate
sin(2x) you get 2 cos(2x), which is not quite right because of the factor 2. That is easily
adjusted for and we see that the required integral is sin(2x)/2.
Note that this simple approach does not work with an integral like
∫
cos(x2) dx. The
obvious function to play with seems to be sin(x2) but the derivative of this is 2x cos(x2)
(damn the chain rule) which is nowhere near the required answer (and no, you can’t go
dividing by x!). Our simple approach has failed.
Your work on di�erentiation will allow you to check that the following short table is
correct. These integrals should be known.
4 CHAPTER 1. INTEGRATION
f(x)
∫
f(x) dx
xa a 6= 1 1
a+1
xa+1
sin(x) − cos(x)
cos(x) sin(x)
ex ex
1
x
ln(x)
1
1+x2
arctan x
1p
1−x2 arcsin x
There is a table of integrals at the end of this booklet. There are books in the library
that devote hundreds of pages to listing integrals.
Most of the ‘techniques’ for working out integrals are just methods for changing one
integral into another in the hope that you will eventually come to an integral that you can
‘spot’ or �nd in the tables.
Let me �nish this section by raising (quietly) yet another di�culty with the inde�nite
integral|it really is plagued with problems. Consider the statement∫
dx
x
= ln(x)
which is in the above table. This looks correct because we have already found that the
derivative of ln(x) is 1/x. But look a little closer. The integrand (1/x) makes sense for
any value of x other than 0. The integral (ln(x)) makes no sense at all if x � 0. What we
have in this formula is really an integral for 1/x valid for x > 0. We have had to limit the
‘domain of de�nition’ of the integrand in order to make sense. This problem will have to
be coped with later on.
1.3 Some Properties of the Indefinite Integral
These are immediate consequences of the corresponding properties of derivatives. In each
equation there is really an arbitrary constant of integration hanging around.
Let f(x) and g(x) be functions and λ a constant. Then∫
λf(x) dx = λ
∫
f(x) dx∫
f(x) + g(x) dx =
∫
f(x) dx +
∫
g(x) dx
1.4. *IMPOSSIBLE INTEGRALS 5
provided that the integrals exist; we have not yet shown that there necessarily exists a
function F (x) that di�erentiates to give a speci�ed function f(x).
These rules may allow us to reduce an integral to the point where we can spot the
answer. ∫
2x + ex dx = 2
∫
x dx +
∫
ex dx = x2 + ex∫
sin x + 2 cos x dx =
∫
sin x dx + 2
∫
cos x dx = − cos x + 2 sin x
1.4 *Impossible Integrals
This is an aside. Ignore it unless interested.
You will frequently be told that some integrals are ‘impossible’ | they cannot be done. Common
examples are such seemingly respectable integrals as∫
ex
2
dx;
∫
sin(x2) dx;
∫
dxp
x5 + 1
dx
This does not mean that, for example, there is no function that di�erentiates to give ex
2
; there certainly
is. The problem is that you do not know how to write it down. The answer cannot be expressed in terms
of ‘elementary functions’.
Let me give another example to make this clearer. We have seen that∫
dx
1 + x2
= arctan(x)
That is simple enough. But suppose I had not done the inverse trig functions. Then you would never have
met arctan(x), so I would have to say to you that, though the integral could certainly be done, it could not
be done in terms of functions known to you. The same kind of thing applies to the ‘impossible’ examples
given above.
It is a distressing fact that the vast majority of integrals are, in this sense, impossible. When I write
exercise sheets on di�erentiation I can get away with writing functions down at random, knowing full well
that they can be di�erentiated. When writing exercises on integration I have to tread very carefully indeed
and check everything out before I set anything.
In some ways it is more surprising that derivatives can be done. It just happens to be the case that
the derivatives of all the elementary functions can be expressed in terms of elementary functions. There is
no logical reason for this|it is pure luck. That’s a lie actually, there are good historical reasons why our
idea of ‘elementary’ for functions leads to simple derivatives.
1.5 The Definite Integral
Suppose that F (x) is an inde�nite integral of f(x), i.e. F 0(x) = f(x). The Definite
Integral ∫ b
a
f(x) dx
where a and b are numbers, is de�ned to be the number F (b)− F (a):
∫ b
a
f(x) dx = [F (x)]ba = F (b)− F (a)
6 CHAPTER 1. INTEGRATION
The choice of inde�nite integral (choice of constant of integration) does not matter|the
constant of integration cancels out.
a and b are called the Limits of Integration. a is called the Lower Limit and b is
called the Upper Limit.
Note the ‘square bracket’ notation. [f(x)]ba stands for f(b) − f(a). Some books use
f(x)jba, but this can be confusing in complicated expressions because it does not tell you
where the expression starts.
This may all seem a strange idea at this stage. The justi�cation for de�nite integrals
will come quite soon.
Let me show you a few examples to show you that the idea is quite simple.
Example 1.1.
∫ 2
1
x2 dx
Now ∫
x2 dx =
x3
3
So ∫ 2
1
x2 dx =
[
x3
3
]2
1
=
8
3
− 1
3
=
7
3
Example 1.2.
∫ 2
−2
ex dx
Now ∫
ex dx = ex
So ∫ 2
−2
ex dx = [ex]2−2= e
2 − e−2
Example 1.3.
∫ x
0
t2 dt
Just as usual, ∫ x
0
t2 dt =
[
1
3
t3
]x
0
=
1
3
x3
Notice that, in this case, the answer is a function of x rather than a number.
1.5.1 Differentiating an Integral
Consider the function
g(x) =
∫ x
a
f(t) dt
where a is a constant and we are thinking of the upper limit of the integral as a variable.
If F (x) is an inde�nite integral for f(x) then
g(x) = F (x)− F (a)
1.5. THE DEFINITE INTEGRAL 7
So, by de�nition of F (x),
g0(x) = F 0(x) = f(x)
This gives the result
d
dx
∫ x
a
f(t) dt = f(x)
(Note that we have only proved it on the assumption that an inde�nite integral exists for
f(x).)
1.5.2 Properties of the Definite Integral
If we assume that the function f(x) has an integral then∫ b
a
f(x) dx = −
∫ a
b
f(x) dx (F (b)− F (a) = −(F (a)− F (b)))
∫ b
a
f(x) dx +
∫ c
b
f(x) dx =
∫ c
a
f(x) dx ((F (b)− F (a)) + (F (c)−F (b)) = F (c)− F (a))
1.5.3 Infinite Limits
Have a quick look at this section, but don’t get worried by it.
We have seen what we mean by a de�nite integral where both limits of integration are
finite. In many important applications of integration you will come across cases where one
or more of the limits of integration is in�nite. There are three possibilities:∫ 1
a
f(x) dx
∫ a
−1
f(x) dx
∫ 1
−1
f(x) dx
The interpretation of these is that they are to be taken as limiting cases of the corresponding
integrals with �nite limits. For example,∫ 1
a
f(x) dx = lim
b!1
∫ b
a
f(x) dx
∫ a
−1
f(x) dx = lim
b!−1
∫ a
b
f(x) dx
Here is a simple example: consider the integral∫ 1
1
dx
x2
Now, for any �nite positive value of b,∫ b
1
dx
x2
=
[−1
x
]b
1
= 1− 1
b
8 CHAPTER 1. INTEGRATION
Now we look at the limiting behaviour as b !1. In the limit 1/b ! 0 and so the value of
the integral tends to 1. Therefore ∫ 1
1
dx
x2
= 1.
Of course, in most cases the limits will not exist, so the integral will not exist either.
For example, ∫ 1
0
x dx = lim
a!1
1
2
a2
This limit does not exist, so the integral does not exist, though you might get away with
saying that it is ‘in�nite’.
More seriously, consider the integral∫ 1
0
cos x dx = lim
a!1
sin a
Here the limit does not exist either, nor does it do anything so straightforward as going o�
to in�nity|it just oscillates to and fro between −1 and 1.
Finally, let me note a small problem. The integral
∫1
−1 f(x) dx is the limiting case of
∫ b
a
f(x) dx as
a ! −1 and b !1. This need not be the same thing as
lim
a!1
∫ a
−a
f(x) dx:
It is best to think of such doubly-in�nite limits in two stages: �rst let the top limit tend to in�nity to get∫1
a
and then let the bottom limit tend to −1 to get the answer. The reason for such care is shown by the
integral of sin(x). For any �nite value of a we have
∫ a
−a sin(x) dx = 0 (check). If we took the limit a !1
in this we would get ∫ 1
−1
sin(x) dx = 0
but this is actually nonsense. If, on the other hand, you start by studying
∫1
a sin(x) dx you will see at
once that no meaningful limiting value exists.
Questions 1 (Hints and solutions start on page 62.)
Q 1.1. Evaluate the following integrals∫
x2 dx,
∫
cos x dx,
∫
3x2 − 2x dx
∫
2 sin x dx
∫
4x3 − 3x2 + 1 dx,
∫
3 sinx− 2 cosx dx,
∫
2ex dx,
∫
2 dx
x∫ 1
0
x2 + x dx,
∫ pi
0
2 sinx− cos x dx,
∫ 1
−1
3ex − x dx
1.5. THE DEFINITE INTEGRAL 9
Q 1.2. The following integrals are minor variations on standard ones. Try to spot the
answer (and then check by di�erentiating).∫
cos(2x + 3) dx,
∫
e3x−4 dx,
∫
dx
2 + 3x
,
∫
dx
4 + 9x2∫
dx
5− 3x,
∫
dxp
4− 9x2 ,
∫
dx
cos2 2x
,
∫
dx
ex
Q 1.3. Use the trig identities cos 2x = 2 cos2 x−1 = 1−2 sin2 x to evaluate ∫ sin2 x dx and∫
cos2 x dx.
Q 1.4. Evaluate the following de�nite integrals∫ 1
0
x3 dx,
∫ 3
−2
(x2 − 3x + 1) dx,
∫ pi
0
sin(2x) dx,
∫ 4
2
dx
x
∫ 1
0
dx
1 + x2
,
∫ 1/2
0
dxp
1− x2 ,
∫ 2
1
ex/2dx,
∫ 4
1
dx
2x + 3∫ 1
0
e−2x dx,
∫ 1
0
dx
1 + 4x2
,
∫ x
1
dt
t
, (x > 0)
∫ a
−a
(3x7 − 4x5)15 dx,
∫ pi/4
0
sec2 x dx,
∫ ex2
1
du
u
.
10 CHAPTER 1. INTEGRATION
Chapter 2
Applications of Integration
2.1 Integrals and Area
It is an odd fact that the basic ideas of integration predate those of di�erentiation by nearly
2000 years. Archimedes was quite good at ‘doing integrals’, albeit in a rather di�erent
context. He was interested in the problem of calculating the areas of curved regions. The
ancient Egyptians and Babylonians were quite good at �nding the areas of straight-sided
regions (basic surveying|much needed in Egypt because of the regular flooding of the
Nile). Curved regions are an altogether di�erent problem and the ancient Ancients knew
little more than that the area of a circle is pir2. More complicated regions need much more
complicated ideas. It is not enough just to know the area of a triangle.
Archimedes, in an remarkable anticipation of the methods of the Calculus, tried to adopt
a ‘limiting values’ approach to calculating areas. His idea was that you could approximate
a curved region arbitrarily closely by straight-sided regions and could therefore hope to
obtain the area of the curved region as a limiting case of the (calculable) areas of the
approximations. This has since become one of the most basic ways of de�ning integrals
because, as we will see in this section, there is a close connection between the idea of an
integral (which we just introduced as the opposite of di�erentiation) and the idea of an
area.
a b
x
A(b)
y = f(x)
Figure 2.1: The area between a curve and the x-axis.
11
12 CHAPTER 2. APPLICATIONS OF INTEGRATION
Consider the graph of y = f(x) between x = a and x = b. We want to calculate the
area between this curve and the x-axis shown in Fig. 2.1.
In what follows I am going to regard a as �xed and denote the area between a and b by
A(b).
Now consider a point slightly to the right of b with coordinate b + h.
bb b + hb + h
h
M
m
f(x)
Figure 2.2: Bounding an area between two rectangles.
Then
A(b + h) = A(b) + shaded area
Let me draw a bigger picture of the shaded area. M is the highest point on the graph on
this interval and m is the lowest. By looking at the two rectangles in the diagram it is clear
that the shaded area has value between hm and hM . So, by above,
mh � A(b + h)− A(b) � Mh
or
m � A(b + h)− A(b)
h
� M
(we have h > 0). Note that the central term is a Newton Quotient.
Now let h shrink down to zero. If f is continuous it is clear that as h ! 0 both m and
M tend to f(b). The Newton Quotient is sandwiched between them, so
f(b) = lim
h!0
A(b + h)−A(b)
h
= A0(b)
So f is the derivative of A. Therefore A is an inde�nite integral of f . So∫ b
a
f(x) dx = [A(x)]ba = A(b)− A(a)
but, obviously, A(a) = 0. So we get the fundamental result that∫ b
a
f(x) dx = A(b)
2.1. INTEGRALS AND AREA 13
So the area between the graph and the x-axis between x = a and x = b is given by the
formula
area =
∫ b
a
f(x) dx
Warning: I drew the picture conveniently with the graph above the axis. If f(x) goes
negative then the ‘area’ calculated by the integral also goes negative. You have to be careful
about this as we will see in a moment.
Example 2.1. What is the area between the graph of f(x) = x2 and the x-axis between
x = 0 and x = 2?
On this range the graph never goes below the x-axis, so by the above formula the area
is given by
area =
∫ 2
0
x2 dx =
[
1
3
x3
]2
0
=
8
3
So the area is 8/3 (square units).
Example 2.2. What is the area between the graphof f(x) = sin x and the x-axis between
x = 0 and x = pi?
On this range the graph never goes below the x-axis, so
area =
∫ pi
0
sin x dx = [− cos x]pi0 = −(−1) + 1 = 2
Example 2.3. What is the area between the graph of f(x) = x2−1 and the x-axis between
x = 0 and x = 2?
Here we have to be careful. f(x) is negative between x = 0 and x = 1 and positive
between x = 1 and x = 2. If we proceed mindlessly we get
area =
∫ 2
0
(x2 − 1) dx =
[
1
3
x3 − x
]2
0
=
8
3
− 2 = 2
3
This is the wrong answer, if you interpret area in the usual sense.
Let me now do this properly. The integral between x = 0 and x = 1 has value∫ 1
0
(x2 − 1) dx = 1
3
− 1 = −2
3
This answer is negative because the graph is below the axis. The actual area is 2
3
.
Now do the integral between x = 1 and x = 2.∫ 2
1
(x2 − 1) dx = (8
3
− 2)− (1
3
− 1) = 2
3
+
2
3
=
4
3
This gives the second part of the area.
So the total area is
2
3
+
4
3
= 2
14 CHAPTER 2. APPLICATIONS OF INTEGRATION
(1,1)
y = x2
y = x3
x
y
Figure 2.3: The area between two curves.
1
-1
y=x
y = x3
x
y
Figure 2.4: Another area.
Example 2.4. What is the area of the region between the two graphs y = x2 and y = x3
between x = 0 and x = 1?
This type of question is best answered by �nding the areas between these graphs and
the x-axis and subtracting the results. It is sensible to start with a sketch like Fig 2.3.
The area between y = x2 and the x-axis is
∫ 1
0
x2 dx =
[
1
3
x3
]1
0
=
1
3
The area between y = x3 and the x-axis is
∫ 1
0
x3 dx =
[
1
4
x4
]1
0
=
1
4
Now for a little care: the graph of y = x2 always lies above the graph of y = x3 on the
range 0 < x < 1. So the area between the graphs is the di�erence of the two results that
we have just calculated:
area =
1
3
− 1
4
=
1
12
Example 2.5. What is the area enclosed by the graphs of y = x3 and y = x?
Here we again need to draw a picture to get our bearings | see Fig 2.4. The graphs
cross when x3 = x, i.e. x = 0, x = �1. The enclosed area comes in two identical parts,
from x = −1 to x = 0 and from x = 0 to x = 1. Compute the second and double it.
Area =
∫ 1
0
x dx−
∫ 1
0
x3 dx =
1
4
So the total area enclosed is 1/2.
2.2. INTERVALS 15
2.2 Intervals
Before going on to the next section it is worth introducing some standard notation for
intervals. An Interval on the real line is all the points between two points (apart from a
special case to be done in a moment). These intervals come in a number of types, depending
on whether the end-points are included in the range or not.
An Open Interval is one in which the end-points are not included. e.g. 0 < x < 1.
We use the notation (a, b) for the open interval with end-points a < b
(a, b) = fx 2 R : a < x < bg
A Closed Interval is one in which both end points are included. e.g. 0 � x � 1. We
use the notation [a, b] for the open interval with end-points a � b
[a, b] = fx 2 R : a � x � bg
We also use the notations (a, b] for a < x � b and [a, b) for a � x < b.
The other types of interval are those that stretch to in�nity in one direction or another.
For these we use the following notations
(−1, a) = fx 2 R : x < ag
(−1, a] = fx 2 R : x � ag
(a,1) = fx 2 R : a < xg
[a,1) = fx 2 R : a � xg
(−1,1) = R
Note that things like [−1, a) do not make sense. ‘In�nity’ is not a number!
2.3 Further Properties of the Definite Integral
The ‘area’ interpretation of the de�nite integral gives us the following extra properties:
if f(x) � 0 on [a, b] then
∫ b
a
f(x) dx � 0
(If the graph does not go below the x-axis the integral is not going to be negative.)
if f(x) � g(x) on [a, b] then
∫ b
a
f(x) dx �
∫ b
a
g(x) dx
(If the graph of f never goes below the graph of g then the area under the f graph is greater
than that under the g graph. That explanation makes sense if both graphs are above the
x-axis, but the result is true in general as you can see by using the �rst property on the
function h(x) = f(x)− g(x).)
if m � f(x) � M on [a, b] then m(b− a) �
∫ b
a
f(x) dx � M(b− a)
16 CHAPTER 2. APPLICATIONS OF INTEGRATION
(Think of m as the constant function g(x) = m and use the earlier results, together with∫ b
a
m dx = m(b− a). Similarly for M .)
Finally, a property that is often quite useful∣∣∣∣
∫ b
a
f(x) dx
∣∣∣∣ �
∫ b
a
jf(x)j dx
Questions 2 (Hints and solutions start on page 62.)
Q 2.1. Work out the area between the following graphs and the x-axis on the given ranges:
y = x4 on [1, 2]
y = ex on [−1, 1]
y = cos x on [−pi
2
,
pi
2
]
y = 2x2 + x on [0, 1]
y = x2 − x on [0, 2]
y = x2 − 3x + 2 on [0, 3]
y = x3 on [−2, 1]
y = e−x on [0,1)
Q 2.2. Find the area enclosed between the graphs of y = x2 and y = x+2 by �rst sketching
the graphs to see what is going on and then working out the points at which the two graphs
meet before doing an integration.
Q 2.3. Find where the graphs of y = x(1 − x) and y = x2 cross and then �nd the area
enclosed between the two graphs.
Q 2.4. Find the area enclosed by the graphs y = ex, y = x2 and the lines x = 0 and x = 1.
Q 2.5. Consider the area enclosed between the graph of y = 1− x2 and the x-axis. Which
line parallel to the x-axis divides this area into two equal parts?
*Q 2.6. Consider the function b(x) de�ned by
b(x) =


0 x < −1
1 −1 � x � 1
0 1 < x
2.3. FURTHER PROPERTIES OF THE DEFINITE INTEGRAL 17
Remembering the basic fact that the de�nite integral gives the area under a graph
(taking due account of signs), draw the graph of the function b1(x) de�ned by
b1(x) =
∫ x
0
b(t) dt
for all values of x.
If you managed that you can now go on to sketch the graph of
b2(x) =
∫ x
0
b1(t) dt
*Q 2.7. Suppose that f(x) and g(x) are two functions and consider the integral
I(t) =
∫ b
a
(f(x) + tg(x))2 dx.
whose value depends on t. Because of the square, this integral cannot be negative: I(t) � 0.
Now expand out the square and show that you get I(t) = At2 +2Bt+C where A, B, C are
constants. Use the fact that I(t) cannot be negative, together with what you know about
quadratics, to show that you must have B2 � AC and that this becomes
(∫ b
a
f(x)g(x) dx
)2
�
∫ b
a
f(x)2 dx �
∫ b
a
g(x)2 dx (Cauchy-Schwartz Inequality)
This is a very useful inequality in more advanced work.
*Q 2.8. This exercise relates to the method that Archimedes used to �nd the area of the
region between a parabola (let’s say y = x2) and a chord of the parabola. He did this
nearly 2000 years before the calculus was o�cially invented.
Let P (a, a2) and Q(b, b2) be two points on the graph y = x2 with b > a. Let R be the
point on the graph where the slope of the graph is equal to the slope of the chord PQ.
Show that R is the point on the graph with x-coordinate (a + b)/2. Why is PQR the
biggest triangle with base PQ and with third vertex on the graph between P and Q? The
area of the triangle PQR is (b − a)3/8 (as you may be able to check). Now �nd the area
of the ‘parabola segment’ bounded by the parabola and the chord and show that it comes
out as 4/3 times the area of the triangle.
*Q 2.9. This question is about estimating the value of pi. Draw the graph of y = sin x for
0 � x � pi/2. Now work out the area between this graph and the x-axis. By looking at
your picture can you show that 2 < pi < 4. Not very precise, but it’s a start. By looking at
the area of the graph between x = 0 and x = pi/4 can you show that pi < 8(
p
2− 1) and,
by using the fact that sin x � x on this range, that pi2 > 32(1− 1/p2)?
18 CHAPTER 2. APPLICATIONS OF INTEGRATION
1 2 3 4 5 n n+1
x
y
y = 1/x
Figure 2.5: Estimating an area.
*Q 2.10.The Harmonic Numbers H1, H2, H3, . . . are de�ned by
Hn = 1 +
1
2
+
1
3
+
1
4
+
1
5
+ � � �+ 1
n
So, for example, H1 = 1, H2 = 3/2, H3 = 11/6, etc.
Study the Fig. 2.5 which shows the graph of y = 1/x:
Remember that the area under the graph between x = 1 and x = n is given by the
integral
∫ n
1
dx/x = ln(n). Add up the areas of the rectangles and show that
UPPER = 1 +
1
2
+
1
3
+ � � �+ 1
n− 1 > ln(n)
LOWER =
1
2
+
1
3
+
1
4
+ � � �+ 1
n
< ln(n)
Deduce from this that for n > 1,
ln(n) +
1
n
< Hn < ln(n) + 1
Notice that this implies that Hn !1 as n !1, rather surprisingly. Roughly what value
of n do you need to get Hn = 10? What about Hn = 20 or Hn = 100?
The above calculation narrowed the value of Hn down to a range of length about 1. We
can do a lot better than this. The main reason for the roughness in the approximation was
that the �rst few rectangles �tted the graph very badly. So suppose we start the exercise
further along. By applying the above process to the graph on the range N � x � n show
that
ln(n) +
1
n
+ AN − 1
N
< Hn < ln(n) + AN
where AN = HN − ln(N). Suppose I tell you that H1000 = 7.4854708606 to 10 decimal
places. What does the above inequality become and how accurately can you tell me the
value of H1000000 ?
2.4. ANOTHER VIEW OF THE DEFINITE INTEGRAL 19
2.4 Another View of the Definite Integral
Consider the interval [a, b]. Let f(x) be a continuous function de�ned on this interval. Let
us think once more of the problem of �nding the area under the graph of f(x) between
x = a and x = b.
y = f(x)
a b
xk
δx
Figure 2.6: Dividing up the area under a curve
We will adopt an approach that is much more elementary (and much older) than our
previous method. Divide the interval [a, b] up into a large number of small parts. For
convenience we will take them all to be of the same width, but that is not very important.
Now use this subdivision to break up the area into thin strips as shown.
Denote the subdivision points by x0, x1, . . . , xn where xk = a+kδx and δx is the width
of each strip δx = (b− a)/n.
���
���
���
���
���
���
y = f(x)
f(xk)
xk xk+1
Figure 2.7: One of the rectangles
We can get an approximation to the area under the
graph by adding up the areas of the n rectangles shown
in the picture. The rectangle on the base [xk, xk+1] has
area f(xk)δx. So
approximate area =
n−1∑
k=0
f(xk)δx
Now we use the same kind of argument that we used
when inventing the derivative. As n gets bigger and
bigger we expect the sum of the areas of the rectangles
to get closer and closer to the true area under the graph.
We would hope that if we took the limit as n !1 the
sum would tend to the true area as its limiting value.
We will assume that this is true. So
area under graph = lim
n!1
n−1∑
k=0
f(xk)δx
But we already know that this area is given by the de�nite integral. So, putting our two
20 CHAPTER 2. APPLICATIONS OF INTEGRATION
results together, we get (in the case of a continuous function)
∫ b
a
f(x) dx = lim
n!1
n−1∑
k=0
f(xk)δx
We can in fact take this as a de�nition of the de�nite integral (the Riemann Integral), and
normally do so in more advanced work.
This interpretation of the de�nite integral is the one that is most useful in applications,
as we will soon see.
2.5 Further Applications of the Definite Integral
Our new interpretation of the de�nite integral is very useful. It shows us how to turn
a lot of di�erent problems into integration problems. We introduced it in the context of
�nding areas, but it is vastly more general than that. Almost anything that comes into
the category of ‘break it into small bits, approximate on the bits and then add up the bits’
yields a de�nite integral of some form.
In this section I will look at a few applications. You may meet others in other subjects.
2.5.1 The Length of a Curve
Suppose we want to calculate the length of the graph of y = f(x) between x = a and
x = b.
Subdivide [a, b] as before into n small parts. Look at the graph on one of these parts.
The idea is to get an approximation to the length of the graph on this part, in the same
way that we used the rectangle approximation when �nding the area. Then we add up the
approximations and take the limit as n !1 so as to produce an integral which gives the
true length.
P
Q
δx
δy
δs
Figure 2.8: Straight line ap-
proximation
What approximation do we use?
The obvious approach is to use the length of the chord
PQ as an approximation to the length of the graph between
P and Q. In the notation of Fig. 2.8 this length is
δs =
√
δx2 + δy2
which we can write more conveniently as
δs =
√
1 +
(
δy
δx
)2
δx
We now have the approximation to the length which I will
write crudely as ∑√
1 +
(
δy
δx
)2
δx
2.5. FURTHER APPLICATIONS OF THE DEFINITE INTEGRAL 21
Our new interpretation of the de�nite integral tells us that, as n ! 1 this tends to the
value of the de�nite integral ∫ b
a
√
1 +
(
dy
dx
)2
dx
(the δy/δx is really the Newton Quotient and tends to dy/dx in the limit.)
So we have obtained the formula
s =
∫ b
a
√
1 +
(
dy
dx
)2
dx
for the length of the graph of y = f(x) between x = a and x = b.
Note: if the curve is given parametrically by x = x(t) and y = y(t) then a very similar
argument gives us the formula
s =
∫ b
a
√
_x2 + _y2 dt
for the length of the curve between t = a and t = b.
Example 2.6. What is the length of the graph y = cosh(x) between x = 0 and x = a?
In this case dy/dx = sinh(x) so√
1 +
(
dy
dx
)2
=
√
1 + sinh2(x) = cosh x
So the length is
s =
∫ a
0
cosh x dx = [sinh x]a0 = sinh a
Example 2.7. As a check on whether we are making sense, let’s work out the length of a
part of a straight line. Let y = mx + c be a straight line. Let us use the formula to �nd its
length between x = a and x = b.
y0 = m so
√
1 +
(
dy
dx
)2
=
p
1 + m2
So
s =
∫ b
a
p
1 + m2 dx = (b− a)
p
1 + m2 =
b− a
cos φ
where φ is the angle that the line makes with the x-axis (m = tan φ). You can easily check
that this is indeed the right answer.
Example 2.8. What is the circumference of a circle of radius r?
Let the circle be given by the equation x2 + y2 = r2. We will get the circumference by
working out the length of the top semicircle and doubling the result.
22 CHAPTER 2. APPLICATIONS OF INTEGRATION
On the top half y =
p
r2 − x2. So
dy
dx
=
−xp
r2 − x2
and √
1 +
(
dy
dx
)2
=
rp
r2 − x2
So the length of the semicircle is given by
l =
∫ r
−r
r dxp
r2 − x2
This is actually a fairly standard integral and you will �nd it in most tables (we will see
a method for doing it later on). The solution is
∫
dx/
p
r2 − x2 = arcsin(x/r). You should
check this by di�erentiating the RHS.
l = [r arcsin(x/r)]r−r = r(arcsin(1)− arcsin(−1)) = rpi
So the total circumference of the circle is twice this, i.e.
circumference = 2pir
2.5.2 Volumes of Revolution
Take the graph of y = f(x) on the interval [a, b] and spin it round the x-axis so as to
produce what is known as a solid of revolution as shown in Fig. 2.9. We want to get a
formula for the volume of this solid.
y = f(x)
a
b
x
y
Figure 2.9: A volume of revolution
The method is almost exactly the same as in the previous examples. Think of the
interval [a, b] being subdivided into lots of little bits. Now look at one of the bits and try
to get an approximation for the volume of the ‘thin slice’ of the solid obtained by rotating
the piece of the graph on this interval.
2.5. FURTHER APPLICATIONS OF THE DEFINITE INTEGRAL 23
y δx
Figure 2.10: The volumeof
a slice.
In the notation of the diagram, the thin slice of the solid is
virtually a cylinder of radius y and thickness δx. The volume
of a cylinder is the product of its height and the area of its
base. So we get the approximation
δV = piy2 δx
for the volume of the slice.
The approximation to the total volume can then be writ-
ten as
V �
∑
piy2 δx
Now take the limit as n !1 and get
Volume =
∫ b
a
piy2 dx
Example 2.9. Cone
Take the line y = mx on [0, h] and spin it round the x-axis so as to produce a cone of
height h and ‘semi-angle’ α, where tanα = m.
The volume of this cone is, by our formula,
V =
∫ h
0
pim2x2 dx =
[
pim2
1
3
x3
]h
0
=
1
3
pim2h3
If R is the radius of the base of the cone then m = tanα = R/h so, with a bit of
rearranging, we get
V =
1
3
piR2h =
1
3
base � height
Example 2.10. Sphere
Take the semicircle y =
p
r2 − x2 on [−r, r] and spin it round the x-axis. We get, as
our solid of revolution, a Sphere of radius r.
Our formula tells us that the volume of this sphere is
V =
∫ r
−r
pi(r2 − x2) dx =
[
pir2x− 1
3
pix3
]r
−r
=
4
3
pir3
So the volume of a sphere of radius r is
V =
4
3
pir3
Example 2.11. Consider the funnel formed by taking the curve y = 1/x and rotating it
round the x-axis on the interval [1, a], as shown in Fig. 2.11
The volume of this funnel is
V =
∫ a
1
pi
1
x2
dx =
[
−pi
x
]a
1
= pi(1− 1
a
)
24 CHAPTER 2. APPLICATIONS OF INTEGRATION
1
a
Figure 2.11: Volume of rotation.
Now notice that, as a !1, this volume tends to the �nite value pi (because pi/a ! 0).
We write ∫ 1
1
pi
x2
dx = lim
a!1
∫ a
1
pi
x2
dx = lim
a!1
(
pi − pi
a
)
= pi
2.5.3 Area of Surface of Revolution
I’m going to be brief here and just give you the formula.
Suppose we take the graph y = f(x) on the range [a, b] and rotate it around the x-axis
as before. Then the Surface Area of the surface formed by this is given by
area =
∫ b
a
2piy
√
1 +
(
dy
dx
)2
dx
Example 2.12. Sphere
As before, we obtain our sphere by rotating the semicircle
y =
p
R2 − x2 − R � x � R
around the x-axis.
For this curve (
dy
dx
)
=
−xp
R2 − x2
So √
1 +
(
dy
dx
)2
=
√
1 +
x2
R2 − x2 =
Rp
R2 − x2
So the area is given by
Area =
∫ R
−R
2pi
p
R2 − x2 Rp
R2 − x2 dx = 2piR
∫ R
−R
dx = 4piR2
So a sphere of radius R has area 4piR2.
2.5. FURTHER APPLICATIONS OF THE DEFINITE INTEGRAL 25
2.5.4 *Area in polar coordinates
Read this after you �nd out about polar coordinates, if you don’t know about them already.
Suppose we have a curve given in polar coordinate by r = f(θ). How do we �nd the
area of the region bounded by the curve and the radial lines θ = a and θ = b?
O
r = f(θ)
a b x
y
Figure 2.12: Describing an area in polar
co-ordinates.
O
δθ
θ
r = f(θ)
f(θ)
Figure 2.13: The area of a very small
segment.
This is another example of the ‘thin slices’ approach, though a bit di�erent to previous
ones. We imagine the θ range from a to b being subdivided into lots of small parts of width
δθ and the area consequently being divided up into lots of thin pie slices. Any one of these
slices, at angle θ, is approximately a triangle with sides f(θ) and included angle δθ. The
area of such a triangle is 1
2
f(θ)2 sin(δθ). But, for very small angles, sin(x) � x. So we
approximate the area of the pie-slice by 1
2
f(θ)2δθ.
Now, ‘adding up’ these thin slice areas and passing over to the limit we get the formula
area =
1
2
∫ b
a
f(θ)2 dθ
for the area of the region. This is usually written more simply as
area =
1
2
∫ b
a
r2 dθ
Questions 3 (Hints and solutions start on page 63.)
Q 2.11. Calculate the volume generated by rotating the following graphs around the x-axis.
y = 2x on [0, 1]
y = cos 2x on [−pi
2
,
pi
2
]
y = e−3x on [0, a]
(For the second one, remember that cos 2x = 2 cos2 x− 1.)
26 CHAPTER 2. APPLICATIONS OF INTEGRATION
Q 2.12. Find the volume of the solid that is produced by taking the area bounded by the
lines y = x, y = 2x and x = 2 and rotating it about the x-axis.
Q 2.13. What is the volume of the solid obtained by rotating the graph of y = x2 around
the y-axis (I said y-axis) between y = 0 and y = 4?
Q 2.14. Consider the curve (ellipse) given by
x2
a2
+
y2
b2
= 1
Find the volume of the solid produced by rotating this about the x-axis.
Q 2.15. A standard application of integration is in �nding average values. The average
value of the function f(x) on the range a � x � b is given by
µ =
1
(b− a)
∫ b
a
f(x) dx.
Can you see how this formula is derived in terms of ‘thin slices’? What is the average value
of f(x) = x for 0 � x � 1? What is the average value of f(x) = sin(x) for 0 � x � pi and
for 0 � x � 2pi? Explore a few other averages.
P is the point (1, 0) on the circle x2 + y2 = 1. Show that the distance from P to the
point Q on the circle given by (cos θ, sin θ) is
p
2− 2 cos θ and that the average distance
from P to other points of the circle is 4/pi (you may need to remember the trig formula
1− cos 2x = 2 sin2 x).
*Q 2.16. If I roll a penny along the x-axis in the (x, y)-plane then a point on the edge of
the penny traces out a curve called a cycloid.
If the penny has radius r then the cycloid can be parametrised as
x(θ) = rθ − r sin θ, y(θ) = r − r cos θ.
(You can try to derive these equations yourself | they just need a bit of trigonometry, θ
is the angle through which the penny has turned.) What is the height of each arch of the
cycloid and where does the cycloid meet the x-axis?
The area of one arch can be found by using the chain rule as follows:
∫
y dx =
∫
y dx
dθ
dθ.
Show that the area of each arch is 3pir2 and that the length of each arch is 8r.
*Q 2.17. I have a sphere of radius R. Using an apple corer of radius r, I punch a symmet-
rically placed cylindrical hole through the sphere. How much of the sphere is left?
P.S. I’m a bit short on examples of lengths and surface areas simply because the square
roots in the formulas tend to produce very nasty integrals—way beyond first year compet-
ence.
2.6. *NUMERICAL APPROXIMATION TO DEFINITE INTEGRALS 27
2.6 *Numerical Approximation to Definite Integrals
If a and b are numbers then the value of
∫ b
a
f(x) dx is a number. In many applications we are not interested
in the precise value of this number|a good approximation will do. We should be grateful for this because,
in practice, very few integrals can be evaluated explicitly so there is no hope of getting an exact answer.
B
C
A D
f(x)
xk xk+1h
Figure 2.14: The area of below
the curve is approximately the
area of the trapezium.
How do we approximate the value without being able to do
the integral? The basic idea is our interpretation of the de�nite
integral as an area (allowing for signs). You could get an approx-
imation to the value of the integral simply by drawing the graph
of f(x) on graph paper and counting squares below the graph.
This is crude and is a lot of work for a not very accurate result.
Our thin strips approach to areas gives us a better idea.
Choose an integer N and subdivide the interval [a; b] into N equal
subdivisions, each of length h = (b− a)=N . Call the subdivision
points a = x0; x1; x2; : : : ; xN = b. Let the values of f(x) at these
points be f(xi) = yi.
If we look at one strip we can see that a reasonable approx-
imation to its area is given by the area of the ‘trapezium’ ABCD.
The smaller h becomes the better the approximation should be.
This area is given by h
2 (yk + yk+1). Adding up all these pieces
we get∫ b
a
f(x) dx � h
2
(y0 + y1) +
h
2
(y1 +y2) + � � �+ h2 (yN−1 + yN)
This can be tidied up the produce what is known as the Trapezium Rule.∫ b
a
f(x) dx � h
2
(y0 + 2y1 + 2y2 + 2y3 + � � �+ 2yN−1 + yN)
The only real virtue of this method is that it is easy the derive (and is rather better than counting squares).
Its accuracy turns out to be roughly proportional to 1=N2, at least if h is quite small and the graph is not
too crazy. This means that doubling the number of strips should about quarter the error.
Let me write down without derivation a much better method known as Simpson’s Rule. Same notation
as before only this time the number of strips has to be even. We will take the number of strips to be 2N ,
so h = (b− a)=(2N). Simpson’s Rule says that∫ b
a
f(x) dx =
h
3
(y0 + 4y1 + 2y2 + 4y3 + 2y4 + � � �+ 2yN−2 + 4yN−1 + yN )
Note that the pattern of the coe�cients is 1; 4; 2; 4; 2; 4; 2; : : : ; 2; 4; 1. This method has accuracy roughly
proportional to 1=N4, on the same assumptions as before.
In practice these methods are applied by using them repeatedly with increasing values of N until the
results seem to have stabilised at the required level of accuracy. Very much a job for computers.
2.7 *Estimating the value of e
This is a footnote. We can use our ideas about integrals and areas to get an idea about the value of the
constant e. I quoted a value earlier, but did not deduce it.
Consider the graph of y = ex between x = 0 and x = a > 0 shown in Fig. 2.15. We know enough
about ex and its derivatives to be able to say that, on this interval, the graph lies above the tangent line
at (0; 1) and below the chord ED.
28 CHAPTER 2. APPLICATIONS OF INTEGRATION
This says that
area ABCE <
∫ a
0
ex dx < area ABDE
Now ABCE has area a + a2=2 and ABDE has area a(1 + ea)=2. The integral has value ea − 1. So
a +
1
2
a2 < ea − 1 < 1
2
a +
1
2
aea
From these two inequalities we get, for 0 < a < 2,
1 + a +
1
2
a2 < ea <
2 + a
2− a
The outer values can be calculated. Put a = 1 and get 5=2 < e < 3. True, but not very precise.
BA
E
C
D
1
y = ex
x
y
Figure 2.15: The graph of ex lies
between tangent and chord.
Now put a = 1=n in the formula:
1 +
1
n
+
1
2n2
< e1/n <
2n + 1
2n− 1
Finally, raise everything to the nth power:
(1 +
1
n
+
1
2n2
)n < e <
(
2n + 1
2n− 1
)n
This is now a good result. The outer expressions can still, in
principle, be evaluated by arithmetic. The larger a value we take
for n a more precise result we get.
If n = 10 we get 2:7141 < e < 2:7206 (getting close). If
n = 1000 we get (with the help of a machine)
2:71828138 < e < 2:7182821
Larger values of n will produce correspondingly better results.
Chapter 3
Methods of Integration
3.1 Introduction
So far our only way to work out
∫
f(x) dx has been, in e�ect, to ‘spot the answer’|i.e.
notice the function F (x) which, when di�erentiated, gives f(x).
This method will not help us very much with integrals like∫
2x− 3p
x2 + 5x + 7
dx
The purpose of this section is to introduce you to some techniques that can be used to
evaluate some integrals. Most integrals cannot be evaluated in terms of functions that you
know about.
The methods that I will show you are mostly techniques for converting one integral into
another one in the hope that the new integral is more familiar.
3.2 Substitution
This is really the reverse process to the chain rule in di�erentiation.
Consider an integral like
∫
e2x+3 dx. Suppose that we make the substitution y = 2x + 3
which de�nes a new variable y in terms of x. Then the integral turns into∫
ey dx
This is not much use to us as it stands because of the mixture of x’s and y’s. We still have
to do something about the dx. Well, by a perverted form of the Chain Rule:
dy =
dy
dx
dx and dx =
dx
dy
dy note:
dy
dx
=
1
dx
dy
(Don’t take these too seriously, the process can be justi�ed in other ways. I’m just telling
you what to do.)
29
30 CHAPTER 3. METHODS OF INTEGRATION
In our example, dy = 2dx, so dx = 1
2
dy and our integral becomes
∫
ey
1
2
dy =
1
2
ey.
Changing back to x’s we get ∫
e2x+3 dx =
1
2
e2x+3
That is all there is to the method of substitution. The main problem in using it is in
deciding what substitution to make. There are very few rules for this. It is mainly a matter
of experience and common sense (not to mention good luck).
The best way to get you used to the method is to work through some examples. In the
�rst few I will just plonk down the substitution without explaining why I chose it. Later
on I will try to describe why I pick the substitution that I use.
Summary: Decide on a substitution. Change the integrand into a formula in terms of
the new variable. Change the ‘dx’ into something involving the new variable by means of
the chain rule. Now you have a new integral to do.
Example 3.1.
∫ p
3− 5x dx
Make the substitution y = 3−5x. Then the chain rule gives, rather trivially, dy = −5dx
and the integral becomes∫ p
y
(
−1
5
dy
)
= −1
5
∫ p
y dy = − 2
15
y3/2
That is the answer in terms of y. Now convert the answer back to x and get∫ p
3− 5xdx = − 2
15
(3− 5x)3/2
Example 3.2.
∫
x2 + 1
x + 1
dx
Put y = x + 1. Then dy = dx and the integral becomes∫
(y − 1)2 + 1
y
dy =
∫
(y − 2 + 2
y
) dy =
1
2
y2 − 2y + 2 ln y
or ∫
x2 + 1
x + 1
dx =
1
2
(x + 1)2 − 2(x + 1) + 2 ln(x + 1)
Example 3.3.
∫
dx
1 + x2
Put x = tan y (substitution the other way round this time). Then by the chain rule
dx = sec2 y dy and the integral converts into∫
sec2 y dy
1 + tan2 y
=
∫
sec2 y dy
sec2 y
=
∫
dy = y
where we have used the identity 1 + tan2 x = sec2 x.So, converting back to x’s,∫
dx
1 + x2
= arctanx
3.2. SUBSTITUTION 31
Example 3.4.
∫
xe2x
2
dx
Put y = 2x2, then dy = 4x dx and the integral becomes∫
1
4
ey dy =
1
4
ey
So ∫
xe2x
2
dx =
1
4
e2x
2
Many integrals (especially those in examinations etc.) have the following form:∫
f(g(x))g0(x) dx
and these can be simpli�ed by the substitution y = g(x) because then We get dy = g0(x) dx
and the integral becomes ∫
f(y) dy
which may be easier to handle.
The general idea here is to look to see if the derivative of a nasty bit in the integral
appears as an independent term that can be grouped with the dx. A number of the above
examples were of this form.
Let me do a number of simultaneous examples so as to make the point
∫
xp
x2 + 2
dx
∫
x sin(x2) dx
∫
e
p
x
p
x
dx
Rearrange slightly∫
1p
x2 + 2
(x dx)
∫
sin(x2) (x dx)
∫
e
p
x
(
dxp
x
)
Now for the substitution
y = x2 + 2 y = x2 y =
p
x
dy = 2(x dx) dy = 2(x dx) dy =
1
2
(
dxp
x
)
1
2
∫
1p
y
dy
1
2
∫
sin(y) dy 2
∫
ey dy
p
y −1
2
cos(y) 2ey
p
x2 + 2 −1
2
cos(x2) 2e
p
x
32 CHAPTER 3. METHODS OF INTEGRATION
Example 3.5.
∫
1
x
p
ln x
dx
This looks rather intimidating at �rst sight, particularly the
p
lnx bit. But we notice
(or I notice) that there is a 1/x outside the square root and 1/x is the derivative of ln x.
So put y = ln x. Then dy = dx/x and the integral turns into∫
1p
y
dy = 2
p
y = 2
√
ln(x)
Example 3.6. Finally
∫
e2x
(3 + ex)3
dx.
This looks rather grim, but all you have to notice is that e2x = exex. Put y = ex + 3.
Then dy = ex dx and the integral becomes∫
y − 3
y3
dy = −1
y
+
3
2y2
= − 1
3 + ex
+
3
2(1 + ex)2
3.2.1 Substitution in Definite Integrals
In one sense this is nothing new. If we want to work out
∫ b
a
f(x) dx we work out the
inde�nite integral F (x) =
∫
f(x) dx and then∫ b
a
f(x) dx = F (b)− F (a)
There is another approachto the problem which becomes more signi�cant later. The
idea is to use the substitution to change the de�nite integral itself. The process is very
simple. We change the integrand and the dx just as before and we also change the limits of
integration to their corresponding values in terms of the new variable of integration. (There
are situations in which this needs a little care, but we will not worry too much about that
at the moment.)
Example 3.7. I =
∫ 2
1
sin(4x− 3) dx
Suppose we make the substitution y = 4x − 3. Then the integrand becomes sin y and
the dx becomes 1
4
dy. The �nal step is to change the limits. When x = 1 we have y = 1.
When x = 2 we have y = 5. So
I =
1
4
∫ 5
1
sin y dy =
[
−1
4
cos y
]5
1
=
1
4
(cos(1)− cos(5))
Example 3.8.
∫ pi/2
0
x cos(x2) dx
Put y = x2, so dy = 2x dx. As x goes from 0 to pi/2 the value of y goes from 0 to pi2/4.
So the integral transforms into
1
2
∫ pi2/4
0
cos(y) dy =
[
1
2
sin(y)
]pi2/4
0
=
1
2
sin
pi2
4
3.3. THE DEFINITE INTEGRAL
∫
B
A
DX
X
33
Example 3.9.
∫ e2
e
dx
x ln x
Put y = ln x. Then dy = dx/x and as x goes from e to e2, y goes from 1 to 2. So the
integral becomes ∫ 2
1
dy
y
= [ln y]21 = ln(2)
3.3 The definite integral
∫ b
a
dx
x
This needs care. As I pointed out in an earlier chapter, saying that∫
dx
x
= ln x
does not work unless x > 0
Now
∫ −2
−3
dx
x
is a perfectly sensible integral, as you will realise if you draw the graph,
but ∫ −2
−3
dx
x
= [ln(x)]−2−3 = ln(−2)− ln(−3)
is nonsense, since you cannot take the log of a negative. (Unfortunately, if we carry on
with our nonsense and write ln(−2) − ln(−3) = ln((−2)/(−3)) = ln(2/3) we end up with
the right answer!)
The form that we really need for the de�nite integral is this:
∫ b
a
dx
x
= ln(b/a)
provided that a and b have the same sign! If they don’t then the integral is not de�ned.
It is a good exercise for you to check that this answer is actually correct.
Questions 4 (Hints and solutions start on page 64.)
Q 3.1. Use the suggested substitutions to evaluate the following integrals.∫
(3x− 1)8 dx (y = 3x− 1),
∫
e1−4x dx (y = 1− 4x),
∫
x2 sin(x3) dx (y = x3),
∫
x
p
2x2 − 1 dx (y = 2x2 − 1)∫ p
1− x2 dx (x = sin y),
∫
xp
x2 + 1
dx (y = x2 + 1)
34 CHAPTER 3. METHODS OF INTEGRATION
∫
sin6 x cos x dx (y = sin x),
∫
tan4 x sec2 x dx (y = tan x)
Q 3.2. Use substitution to evaluate the following integrals. This time you have to guess
the substitution. ∫
(5x− 3)7 dx,
∫
e6x−7 dx,
∫
x3 cos(x4) dx
∫
xp
1− x2 dx,
∫
x2
p
x3 − 1 dx,
∫
cos x
sin4 x
dx∫
xe−x
2
dx,
∫
cos4 x sin x dx,
∫
cos4(2x) sin(2x) dx∫
x + 1p
x2 + 2x + 3
dx,
∫
dxp
2x− 1 ,
∫
ln t
t
dt
Q 3.3. Use the trig formulas sin(a) cos(b) = (sin(a + b) + sin(a − b))/2, sin(a) sin(b) =
(cos(a− b)− cos(a + b))/2 and cos(a) cos(b) = (cos(a + b) + cos(a− b))/2 to show that in
n and m are positive integers then
∫ 2pi
0
sin(nx) sin(mx) dx =
{
0 n 6= m
pi n = m
,
∫ 2pi
0
sin(nx) cos(mx) dx = 0
∫ 2pi
0
cos(nx) cos(mx) dx =
{
0 n 6= m
pi n = m
*Q 3.4. A function f(x) is said to be periodic with period T > 0 if f(x + T ) = f(x)
for any value of x. For example, sine and cosine are periodic with period 2pi. Show that∫ T
0
f(x) dx =
∫ 2T
T
f(x) dx. More generally, show that for any value of a,∫ a+T
a
f(x) dx =
∫ T
0
f(x) dx.
*Q 3.5. Check, once more, that if we put t = tan(x/2) then
sin x =
2t
1 + t2
cos x =
1− t2
1 + t2
dx
dt
=
2
1 + t2
This gives us a convenient substitution for turning a trigonometric integral into an algebraic
one. Use it on the following integrals∫
dx
1 + cos x
,
∫
dx
1 + sin x
,
∫
dx
sin x
3.4. INTEGRATION BY PARTS 35
3.4 Integration by Parts
This is a consequence of the product rule for di�erentiation.
(uv)0 = u0v + uv0
Integrate this to get ∫
(uv)0 dx =
∫
u0v dx +
∫
uv0 dx
But
∫
(uv)0 dx = uv. So, rearranging things a bit,
∫
u0v dx = uv −
∫
uv0 dx
This is the Integration by Parts formula.
Again, this is a method for converting one integral into another one in the hope that
the new one is easier to handle.
The method is usually appropriate in situations where the integrand is clearly the
product of two, often rather di�erent, terms, one of which can be integrated (u0 ! u) and
the other di�erentiated (v ! v0).
Example 3.10.
∫
tet dt
The integrand is obviously a product of t and et. As it happens we can both di�erentiate
and integrate each factor. So we get a choice.
Let u0 = et and v = t. Then u = et and v0 = 1. So the formula says that∫
tet dt = tet −
∫
1.et dt = tet − et
If we had made the opposite choice we would have got∫
tet dt =
t2
2
et − 1
2
∫
t2et dt
which is making things worse rather than better.
Example 3.11. We can handle de�nite integrals just as well as inde�nite ones.∫ 1
0
tet dt =
[
tet
]1
0
−
∫ 1
0
et dt = (e− 0)− (e− 1) = 1
Notice that we evaluate the �rst term on the RHS in the usual way for de�nite integrals.
Example 3.12.
∫
x cos x dx
Once more, the integrand is obviously the product of x and cos x, which are rather
di�erent beasts. Once more we know how to di�erentiate and integrate each factor so we
36 CHAPTER 3. METHODS OF INTEGRATION
have to make a choice. If we integrate the x we will get x2/2 which will probably make
things worse. So let us take u0 = cos x and v = x. Then u = sin x and v0 = 1. So∫
x cos x dx = x sin x−
∫
1. sinx dx = x sin x + cos x
Example 3.13. I =
∫
ex sin x dx
This turns out to be rather more complicated. Suppose we take u0 = ex and v = sin x.
Then u = ex and v0 = cos x. So we get
I = ex sin x−
∫
ex cos x dx
which does not seem to help because we don’t know the integral on the RHS either. But
that integral, call it J , is also of the ‘parts’ type so, putting u0 = ex and v = cos x we get
J = ex cos x−
∫
ex(− sin x) dx = ex cos x + I
This looks as though we have gone full-circle to no avail. But we haven’t. What we have
got is
I = ex sin x− J = ex sin x− (ex cos x + I) = ex(sin x− cos x)− I
So
2I = ex(sin x− cos x)
and ∫
ex sin x dx =
1
2
ex(sin x− cos x)
That answers the question.
We have also found, for free, that∫
ex cos x dx =
1
2
ex(sin x + cos x)
Example 3.14.
∫
ln x dx
This is a famous one. It does not look like a parts integral because there aren’t two
parts! However the dirty trick in this case is to regard the integrand as 1. lnx and then
take u0 = 1 and v = ln x. Then u = x and v0 = 1/x. So∫
ln x dx = x lnx−
∫
x.
1
x
dx = x ln x− x
Example 3.15.
∫
x arctan x dx
3.4. INTEGRATION BY PARTS 37
Once more, this is an obvious candidate for parts. Take u0 = x and v = arctanx (we
don’t have much choice this time). Then u = 1
2
x2 and v0 = 1/(1 + x2). So we get∫
x arctan x dx =
x2
2
arctanx− 1
2
∫
x2
1 + x2
dx
which has changed our problem into that of integrating∫
x2
1 + x2
dx
We can handle this by writing
x2
1 + x2
=
1 + x2 − 1
1 + x2
= 1− 1
1 + x2
So ∫
x2
1 + x2
dx =
∫
dx−
∫
dx
1 + x2
= x− arctan x
So, �nally, we get∫
x arctanx dx =
x2
2
arctan x− 1
2
(x− arctanx) = 1
2
(x2 + 1) arctanx− 1
2
x
3.4.1 *The Gamma Function
This is an interesting application of integration by parts. Let’s have a look at the integral
In =
∫ 1
0
xne−x dx
where n is some non-negative whole number.
This seems to be a candidate for integration by parts. Di�erentiate the xn and integrate the e−x. This
will give us
In =
[−xne−x]1
0
+ n
∫ 1
0
xn−1e−x dx
It is an important property of theexponential function that, whatever the value of n, xne−x ! 0 as x !1
(ex grows far faster than any power of x). So, if n > 0 the �rst term on the RHS vanishes at both ends.
The other term on the RHS involves an integral which is just the same as our original one with n replaced
by n− 1. So we have
In = n:In−1 for n � 1
You grumble that this has not evaluated the integral|it has just turned it into another similar integral.
True, but that does actually make it easy to evaluate the original integral. The point is that the above
formula is valid for any positive integer n. So, climbing up the ladder, we have
I1 = 1:I0
I2 = 2:I1 = 2:1:I0
I3 = 3:I2 = 3:2:1:I0
I4 = 4:I3 = 4:3:2:1:I0
38 CHAPTER 3. METHODS OF INTEGRATION
|and so on. You will probably believe me if I say that the overall result is that In = n!:I0.
This has reduced everything to the single problem of evaluating
I0 =
∫ 1
0
x0e−x dx = 1
(as you should check).
So we have obtained the rather nice result that∫ 1
0
xne−x dx = n!
Now for a generalisation. We had a positive integer n in the integrand as a power. There was actually
nothing in our integration procedure (the integration by parts) that relied on the fact that n was a whole
number. We could just as well look at the integral
f(a) =
∫ 1
0
xae−x dx
where a is any positive value. Doing integration by parts on this will give
f(a) = a:f(a− 1) if a > 1
as before. If a is a whole number then f(a) = a!, but f(a) makes perfectly good sense for any positive
value a. So we have produced a kind of generalisation of the factorial function. We can now talk about the
factorial of 12 or �. This may sound a rather silly thing to do, but it does actually have wide application.
Traditionally, we do not work with the function f(a) but with the function Γ(a) = f(a − 1). This is
known as the Gamma Function,
Γ(a) =
∫ 1
0
xa−1e−x dx; Γ(a + 1) = a:Γ(a)
for a > 0. Of course, we have Γ(1) = Γ(2) = 1.
To satisfy idle curiosity I can tell you that Γ(12 ) =
p
�
Questions 5 (Hints and solutions start on page 65.)
Q 3.6. Use integration by parts to evaluate the following integrals.∫
x sin(2x) dx,
∫
x3 ln x dx,
∫
xe3x−1 dx
Now do the following by parts | where you may have to use parts more than once.∫
x2 sin(x) dx,
∫
x2e3x−1 dx,
∫
ln2 x dx
∫
ex sin x dx,
∫
arctan(x) dx,
∫
x2e−x dx∫
x3e−x dx,
∫
x4e−x dx
∫
x5e−x dx
3.4. INTEGRATION BY PARTS 39
Q 3.7. This is mainly of interest to Statisticians.
A probability density is a function p(x) that is never negative and satis�es the condition∫ 1
−1
p(x) dx = 1. A random variable X has probability density p(x) if the probability that
X takes a value between a and b is given by
Pr(a � X � b) =
∫ b
a
p(x) dx
where a < b.
The mean value of X is de�ned to be
µ =
∫ 1
−1
x.p(x) dx
The variance of X is de�ned to be
ν =
∫ 1
−1
(x− µ)2p(x) dx
and the standard deviation σ is given by σ2 = ν.
Here are three functions. Check that they are all probability densities.
p1(x) =
{
1
b−a a � x � b
0 otherwise
p2(x) =
{
θe−θx x � 0
0 x < 0
(θ > 0)
p3(x) =
{
1
x2
x � 1
0 x < 1
Show that p1(x) (a uniform distribution) has mean
1
2
(a + b) and variance 1
12
(b− a)2.
Show that p2(x) (a Poisson distribution) has mean 1/θ and variance 1/θ
2.
Show that p3(x) does not have a �nite mean.
Show that the formula for the variance can be rearranged to give
ν =
∫ 1
−1
x2p(x) dx− µ2.
If the random variable X has the Poisson distribution p2(x) show that the probability
that the value of X is greater than the mean value is 1/e.
40 CHAPTER 3. METHODS OF INTEGRATION
*Q 3.8. This is a follow up to some integrals that you did in an earlier exercise. Let
In =
∫ 1
0
xnex dx
for n a whole number � 0. So, for example,
I6 =
∫ 1
0
x6ex dx and I9 =
∫ 1
0
x9ex dx
By using integration by parts, integrating the exponential and di�erentiating the power,
prove the general result that if n > 0
In = e− nIn−1
This is what is known as a Reduction Formula, it gives us the value of In in terms of the
value of In−1. What use is that? Well, it is easy to evaluate I0 | do so. The formula now
tells us the value of I1 and, using it once more, the value of I2 and so on. By repeated
use of the formula, and no further integration, we can get the value of In for and positive
integer n. Work out I4.
Go through the same process for the integral
∫ 1
0
xne−x dx and work out I3.
Now look at the integral
In =
∫ pi
0
xn sin x dx n � 0
By doing integration by parts on this twice over, integrating the trig function and di�er-
entiating the power, show that
In = pi
n − n(n− 1)In−2 n > 1
This is slightly more complicated in that it relates In to In+2 rather than to In+1. This
means that you go up in steps of 2. If you think about it you will realise that you now
need two starting points: I0 and I1. Work out both of these and then work out I3 and I4.
*Q 3.9. Consider
I(n, m) =
∫ 1
0
xn(1− x)m dx
where n and m are integers � 0. By making the substitution y = 1 − x, show that
I(n, m) = I(m, n). Show that
I(n, 0) = I(0, n) =
1
n + 1
Use integration by parts to show that
I(n, m) =
m
n + 1
I(n + 1, m− 1)
Deduce from this that
I(n, m) =
n! m!
(n + m + 1)!
3.5. QUADRATICS AND TRIGONOMETRIC SUBSTITUTIONS 41
3.5 Quadratics and Trigonometric Substitutions
You often �nd terms like
p
x2 − 3x + 5 in integrals. This section suggests a way of dealing
with them that might work.
The basic idea is to use variations on the formulas
1− sin2 x = cos2 x 1 + tan2 x = sec2 x sec2 x− 1 = tan2 x
cosh2 x− 1 = sinh2 x
as suggestions for substitutions.
Example 3.16.
∫
dxp
1− x2
Make the substitution x = sin y. Then dx = cos y dy and
p
1− x2 = cos y (if we are
not being fussy about signs).
So our integral becomes ∫
cos y dy
cos y
=
∫
dy = y = arcsin x
Example 3.17.
∫
dx
1 + x2
Make the substitution x = tan(y). Then 1 + x2 = 1 + tan2(y) = sec2(y) and dx =
sec2(y) dy. So with this substitution the integral just boils down to
∫
dy = y. Turning
back to x we get the answer arctan(x).
Example 3.18.
∫
dx
2 + x2
This is where we have to start being a little bit cleverer. The
expression in the denominator is similar to the 1 + x2 that we dealt with in the previous
example. We take the hint and try to turn it into exactly that form. Start by making the
substitution x =
p
2 y. Then the denominator becomes 2 + 2y2. We can take a 2 out from
the denominator and get the integral (check)
1p
2
∫
dy
1 + y2
This is now just the same as our previous example and integrates to give arctan(y)/
p
2.
So the answer for our original integral is
1p
2
arctan
(
xp
2
)
Example 3.19.
∫
dxp
3x2 − 2
Let’s use a hyperbolic substitution for a change. First it would help if we tidied the
integral up a bit. The ‘standard form’ that we have in mind in this case for the expression
under the square root is y2 − 1. We want to make a substitution that will produce this
42 CHAPTER 3. METHODS OF INTEGRATION
form. Then we can go ahead and use a cosh substitution. So put x =
√
2/3 y. Then the
integral becomes
1p
3
∫
dy√
y2 − 1
Now make the substitution y = cosh z. Then dy = sinh z dz and the integral becomes
1p
3
∫
sinh z dz
sinh z
=
1p
3
∫
dz =
zp
3
=
cosh−1 yp
3
and �nally back to x:
cosh−1 yp
3
=
1p
3
cosh−1
√
3
2
x
Gulp.
As a quick summary of this:
Handle
p
1− x2 by the substitution x = sin y
Handle
p
1 + x2 by the substitution x = tan y
Handle
p
x2 − 1 by the substitution x = cosh y
Note that there are other possibilities, like x = sec y in the x2 − 1 case.
The next problem is to handlemore general quadratics, like ax2 + bx + c. We handle
these by using the process known as ‘completing the square’ to turn the square root into
one of the cases that we have already dealt with. This process goes as follows.
ax2 + bx + c = a(x2 +
b
a
x) + c = a(x +
b
2a
)2 + (c− b
2
4a
)
So the substitution y = x + b/2a will produce the change from ax2 + bx + c to
ay2 + �
where
� =
4ac− b2
4a
What you do next depends entirely on the signs of a and �.
Example 3.20.
∫
dxp
x2 + x + 1
Complete the square on the quadratic:
x2 + x + 1 = (x +
1
2
)2 +
3
4
So put y = x + 1
2
and turn the integral into∫
dy√
y2 + 3
4
This will now yield to the substitution y =
√
3
4
tan z.
3.5. QUADRATICS AND TRIGONOMETRIC SUBSTITUTIONS 43
Example 3.21.
∫
dxp
2x2 + 5x− 1
Complete the square on the quadratic:
2x2 + 5x− 1 = 2(x2 + 5
2
x)− 1 = 2(x + 5
4
)2 − 50
16
− 1 = 2(x + 5
4
)2 − 66
16
So put y = x + 5
4
and turn the integral into
∫
dy√
2y2 − 66
16
=
√
16
66
∫
dy√
a2y2 − 1
where a =
√
32
66
. Now put y = 1
a
cosh z and proceed.
Example 3.22.
∫
2x + 3p
2− x− x2 dx
First handle the square root by completing the square
2− x− x2 = 2− (x2 + x) = 2− ((x + 1
2
)2 − 1
4
) =
9
4
− (x + 1
2
)2
So start by putting y = x + 1
2
and get∫
2y + 2√
9
4
− y2
dy
This is almost in standard form. Put y = 3
2
sin θ. Then the integral becomes∫
(3 sin θ + 2)3
2
cos θ dθ
3
2
cos θ
=
∫
3 sin θ + 2 dθ = 2θ − 3 cos θ
That’s the answer, but it would be nice to have it in terms of x!
First we step back to y:
θ = arcsin(
2
3
y) and cos θ =
√
(1− 4
9
y2)
So the value of the integral becomes
2 arcsin(
2
3
y)− 3
√
1− 4
9
y2
Finally, go back to x and get the expression
2 arcsin
(
2x + 1
3
)
− 2
p
2− x− x2
If you care to di�erentiate this you should �nd that you get back to the original integrand.
44 CHAPTER 3. METHODS OF INTEGRATION
Questions 6 (Hints and solutions start on page 67.)
Q 3.10. Rewrite each of the following in one of the forms X2 + A2, X2 − A2, A2 − X2,
−X2 − A2 or �X2 by completing the square.
9x2 + 6x + 5, 4x2 − 4x− 3, 9x2 − 12x + 4
4x− 2x2, 2− 4x− 4x2, x2 + x + 2
Now work out the integrals∫
dxp
2− 4x− 4x2 ,
∫
dxp
4x− 2x2 ,
∫
dxp
9x2 + 6x + 5
3.6 Integration of Rational Functions and Partial Frac-
tions
We are now going to study integrals of the form∫
p(x)
q(x)
dx
where p and q are polynomials.
We will not study the general case. Instead, we will start by assuming the following:
(1) The denominator q can be factored as a product of linear factors (ax + b), all
di�erent.
(2) The degree of the numerator is less than that of the denominator.
(Note: (1) needs a bit of care. We do not count (x+1) and (2x+2) as di�erent factors.
It would be more accurate to say that no two factors should be proportional.)
As examples:∫
dx
(x + 1)(x + 2)
,
∫
2x + 3
(3x− 4)(1− 2x) dx
∫
x2
(x + 1)(x− 1)(2x− 3) dx
The following integrals do NOT satisfy the conditions∫
x2
(x + 1)(x + 2)
dx degrees wrong
∫
(x + 1)
(x− 1)2(x + 2) dx square factor∫
(x + 1)
x2 + x + 1
dx cannot factorise denominator
3.6. INTEGRATION OF RATIONAL FUNCTIONS AND PARTIAL FRACTIONS 45
To do this kind of integral we need the following result from algebra:
Any expression
fracp(x)(x + b1)(x + b2) � � � (x + bk)
where p(x) is a polynomial of degree less than k and all the factors in the denominator are
di�erent (in the above sense) can be expanded out in Partial Fractions as
A1
x + b1
+
A2
x + b2
+ � � �+ Ak
x + bk
where the As are constants.
That sounds complicated. What it means is shown by the following examples
1
(x + 1)(x + 2)
=
A
x + 1
+
B
x + 2
x + 1
(x− 3)(2x + 3) =
A
x− 3 +
B
2x + 3
x2
(x− 1)(x + 1)(x− 3) =
A
x− 1 +
B
x + 1
+
C
x− 3
If we have produced this kind of expansion then the integration is easy. All we have to
know is that ∫
dx
ax + b
=
1
a
ln jax + bj
So, using the above examples:∫
dx
(x + 1)(x + 2)
= A
∫
dx
x + 1
+ B
∫
dx
x + 2
= A ln jx + 1j+ B ln jx + 2j
∫
x2 dx
(x− 1)(x + 1)(x− 3) = A
∫
dx
x− 1 + B
∫
dx
x + 1
+ C
∫
dx
x− 3
= A ln jx− 1j+ B ln jx + 1j+ C ln jx− 3j
The remaining problem is: how do we �nd the constants A, B etc.?
Consider the �rst example. We know, from the theorem, that
1
(x + 1)(x + 2)
=
A
x + 1
+
B
x + 2
=
A(x + 2) + B(x + 1)
(x + 1)(x + 2)
If this is going to work for all values of x then we must have
A(x + 2) + B(x + 1) = 1
for all values of x.
46 CHAPTER 3. METHODS OF INTEGRATION
Let me now show you two methods for �nding the values of A and B. The �rst is the
most general but the second is usually the easiest in these cases.
1) gather together powers of x and then compare coe�cients on both sides of the
equation: (A + B)x + (2A + B) = 1 so A + B = 0 and 2A + B = 1. These simultaneous
equations now solve to give A = 1 and B = −1.
2) A(x+2)+B(x+1) = 1 is true for all values of x, so it is true for any particular value
of x that we put into it. If we put x = −1 we will kill o� the B term and get (−1+2)A = 1,
so A = 1. If we put x = −2 we kill o� the A term and get (−2 + 1)B = 1, so B = −1.
So, either way,
1
(x + 1)(x + 2)
=
1
x + 1
− 1
x + 2
Example 3.23.
x2
(x− 1)(x + 1)(x− 3) =
A
x− 1 +
B
x + 1
+
C
x− 3
=
A(x + 1)(x− 3) + B(x− 1)(x− 3) + C(x− 1)(x + 1)
(x− 1)(x + 1)(x− 3)
So we must have
A(x + 1)(x− 3) + B(x− 1)(x− 3) + C(x− 1)(x + 1) = x2
for all values of x.
Put x = 1 to kill o� the B and C terms: (1 + 1)(1− 3)A = 1, A = −1
4
.
Put x = −1 to kill o� the A and C terms: 8B = 1, B = 1
8
.
Put x = 3 to kill o� the A and B terms: 8C = 9, C = 9
8
.
So
x2
(x− 1)(x + 1)(x− 3) = −
1
4
1
x− 1 +
1
8
1
x + 1
+
9
8
1
x− 3
Now we can do the integral∫
x2 dx
(x− 1)(x + 1)(x− 3) = −
1
4
∫
dx
x− 1 +
1
8
∫
dx
x + 1
+
9
8
∫
dx
x− 3
= −1
4
ln jx− 1j+ 1
8
ln jx + 1j+ 9
8
ln jx− 3j
If the function to be integrated has a numerator of degree not lower than that of the
denominator then you have to start by dividing the denominator into the numerator. For
example
x3 − 1
x(x + 2)
= (x− 2) + 4x− 1
x(x + 2)
The �rst term on the RHS is easy and you can do partial fractions on the second.
Question is: how do you do the division? You may well have met some way of doing this in school.
The basic method goes as follows. Suppose we are dividing x2 + x + 1 into 3x4 + x2 + x + 1. Look at the
3.6. INTEGRATION OF RATIONAL FUNCTIONS AND PARTIAL FRACTIONS 47
‘top’ terms of the two polynomials: x2 and 3x4. The �rst goes into the second 3x2 times. So multiply the
�rst polynomial by 3x2 and subtract it from the second:
(3x4 + x2 + x + 1)− 3x2(x2 + x + 1) = −3x3 − 2x2 + x + 1
Now start again. The top term x2 of the divisor goes into the top term −3x3 of our remainder −3x times.
So subtract −3x times the divisor from the remainder:
(−3x3 − 2x2 + x + 1)− (−3x)(x2 + x + 1) = x2 + 4x + 1
Keep going. The top term x2 of the divisor goes into the top term x2 of the remainder 1 time. So subtract
1 times the divisor from the remainder:
(x2 + 4x + 1)− 1:(x2 + x + 1) = 3x
We have now got down to the point where the remainder has lower degree than the divisor, so we stop.
The ‘multipliers’ that we have used make up the Quotient 3x2 − 3x + 1 and the �nal remainder is 3x. So
3x4 + x2 + x + 1
x2 + x + 1
= 3x2 − 3x + 1 + 3x
x2 + x + 1
Questions 7 (Hints and solutions start on page 68.)
Q 3.11. As some revision here are some integrals. Do the ones that can be done and explain