Chapter 3
INTEGRATION
Chapter Outline
3.1
Introduction
3.1.1
3.2
Differentiation in Reverse (Anti-Derivative)
Indefinite Integral
3.2.1
Properties of the Indefinite Integral
3.2.2
Integral of Polynomial Functions
3.2.3
Integral of Exponential Functions
3.2.4
Integral of Logarithmic Functions
3.2.5
Integral of Trigonometric Functions
3.3
Definite Integral
3.4
Techniques of Integration
3.5
3.4.1
Integration by Substitution
3.4.2
Integration by Parts
3.4.2
Integration by Partial Fractions
Applications of Definite Integration
3.5.1
Area of a Region Bounded by the Curve and Axis
3.5.2
Area Between Curves
3.5.3
Volume of the three-dimensional solids
Tutorials & Answers
1
3.1 INTRODUCTION
Integration is the reverse process of differentiation. It is sometimes called antidifferentiation. The topic of integration can be approached in several different ways.
Perhaps the simplest way of introducing it is to think of it as differentiation in reverse.
3.1.1 Differentiation in Reverse (Anti-Derivative)
Suppose we differentiate the function F ( x)  3x 2  7 x  2 . We obtain its
derivative as f ( x) 
dF
 6 x  7 . This process is illustrated in Figure 1.
dx
Differentiate
F ( x)  3x 2  7 x  2
f ( x)  6 x  7
Anti-Differentiate
Figure 1
In this case, we can say that the derivative of 3 x 2  7 x  2 is equal to 6x  7 . However,
there are many other functions which also have derivative 6x  7 . Some of these are
3x 2  7 x  3 , 3 x 2  7 x , 3x 2  7 x  11 and so on. The reason why all of these functions
have the same derivative is that the constant term disappears during differentiation. So,
all of these are anti-derivatives of 6x  7 . Given any anti-derivative of f ( x) , all others
can be obtained by simply adding a different constant. In other words, if F ( x) is an antiderivative of f ( x) , then so too is F ( x)  C for any constant C and this is actually
describe the definition of Indefinite Integration.
2
3.2 INDEFINITE INTEGRATION
We call the set of all anti-derivatives of a function as the indefinite integral of the
function. The indefinite integral of the function f ( x) is written as
 f ( x) dx  F ( x)  C
and read it as "the indefinite integral of f ( x) with respect to x ". The function f ( x) that
is being integrated is called the integrand, and the variable x is called the variable of
integration and the C is called the constant of integration.
3.2.1 Properties of the Indefinite Integral
Basically, there are three properties of anti-derivatives which been applied in
order to solve the integration for any kind of functions.
Theorem: Suppose that F ( x) and G ( x) are anti-derivatives of f ( x) and g ( x )
respectively, and C is a constant. Then:
a) A constant factor k can be taken out from an integral, that is
 k f ( x) dx  k  f ( x) dx  kF ( x)  C
,
where k is a constant
b) An anti-derivative of a sum is the sum of the anti-derivatives, that is
  f  x   g  x   dx   f  x  dx   g  x  dx  F ( x)  G ( x)  C
c) An anti-derivative of a difference is the difference of the anti-derivatives, that is
  f  x   g  x   dx   f  x  dx   g  x  dx  F ( x)  G ( x)  C
3.2.2 Integral of Polynomial Functions
Properties of the Integral of Polynomial Functions
x n 1
c
n 1
a)
n
 x dx 
b)
 kdx  kx  c ,
(General form)
where k is any number
3
Example 1
Determine the following integrals:
a)
x
7
b)  2 x 8 dx
dx
c)
1
x
6
dx
d)

3x dx
Solutions
a)
7
 x dx 
x 7 1
x8
c 
c
7 1
8
b)  2 x 8 dx  2 x 8 dx 
c)
2 x 81
 2 x 7
c 
c
 8 1
7
1
x 61
 x 5
6
dx

x
dx


c

c
 x6 
 6 1
5
1
d)
3x dx  3  x dx  3  x1 2

3
1
3
x2
x2
2 3 2
 3
c  3
c 
x c
1
3
3
1
2
2
Exercise 1
Solve the following integrals.
a)

4
x dx
1
b)

c)
x
x2
6
1
32
dx
4 54
x c
5
Answer:
3 23
x c
2
Answer: 
dx

Answer :
2
3
1
3
d)   3x dx
Answer: 9x  c
Example 2
Determine the following integrals:
a)
 2 x  5x
3

 4 x 2 dx
b)
 3x  5 dx
2
4
2
c
x
 2x  1 
d)   4 dx
 x 
2
2

c)   3x   dx
x

Solutions
a)
 2 x  5x
3

 4 x 2 dx   2 xdx   5 x 3 dx   4 x 2 dx
 2 xdx  5 x 3 dx  4 x 2 dx
 x2
 2
 2
  x4   x3 
  5   4   c
  4   3 
5
4
 x2  x4  x3  c
4
3
b)
 3x  5 dx   9 x
2
2

 30 x  25 dx
  9 x 2 dx   30 xdx   25dx
 9 x 2 dx  30  xdx  25 1dx
 x3 
 x2 
 9   30   25( x)  c
 3 
 2 
 3x 3  15 x 2  25 x  c
c)

2
2
4 

 2
 3x   dx    9 x  12  2 dx
x
x 


  9 x 2 dx   12dx  
4
dx
x2
 9  x 2 dx  12 1dx  4  x 2 dx
 x3 
 x 1 
 9    12 x  4 
c
 3
 1 
4
 3 x3  12 x   c
x
5
d)
2x
1
 2x  1 
dx   4 dx   4 dx
4
x
x
x 
 
 2  x 3 dx   x 4 dx
 x 2  x 3
 2
c

 2  3
1
1
  2  3 c
x 3x
Exercise 2
Solve:
a)
b)

4 x  32 dx

3x  23 dx
3
1
32 52
2
Answer:
x  16 x  18 x 2  c
5
x
3
Answer:
x2
4
1
81 103 162 73
x 
x  27 x 3  24 x 3  c
10
7
3.2.3 Integral of Exponential Functions
Formula of the Integral of Exponential Functions
a)  e x dx  e x  c
b)

e axb dx 
(Integral of Natural Exponential Function)
e axb
 c ; where a any number
a
c)  e u du  e u  c
or
 g ( x)(e
g ( x)
)dx  e g ( x )  c
(by using Substitution Rule)
Example 3
Solve the following integrals:
a)  e 2 x dx
d)
 e
2x
b)  e 2 x 3 dx

2
 e  2 x dx
e)
 e
2x

 e
f)
e
2
 3 dx
Solutions
a)  e 2 x dx 
c)
e2x
c
2
6
5x
x

 e 3 x dx
 e 3 x dx
b)

c)
 e
5x
d)
 e
2x
e 2 x 3 dx 
e 2 x 3
c
2

 e 3 x dx   e 5 x dx   e 3 x dx 
e 5 x  e 3 x 
e 5 x e 3 x
  c 
 

c
5  3 
5
3
 e  2 x dx   e 4 x  2  e 4 x dx

2
  e 4 x dx   2dx   e 4 x dx

 e 4 x
e4x
 2 x  
4
 4

  c

e4x
e 4 x

 2x 
c
4
4
e)
 e
2x

 3 dx 
 e

e
2
4x
4x

 6e 2 x  9 dx
dx 
 6e
2x

dx  9dx
e 4 x 6e 2 x

 9x  c
4
2
e4x

 3e 2 x  9 x  c
4

f)

e x  e 3 x dx   e 4 x dx 
e4x
c
4
Exercise 3
a)
b)
c)

e5x
dx
e2x
Answer :
e3 x
c
3

e 4 x  e8x
dx
e2x
Answer :
e 2 x e6 x

c
2
6

e 2x  e5x
dx
e9x
Answer : 
1
c
2e 2 x
7
3.2.4 Integral of Logarithmic Functions
Formula of Integral of Logarithmic Functions
a)
 xdx  ln x  c
b)

c)
 u du  ln u  c
1
a
dx  ln ax  b  c
ax  b
1
d)  (a x )dx 
or
g ( x)
 g ( x) dx  ln( g ( x))  c
(by using Substitution Rule)
ax
c
ln a
Example 4
Find:
a)
1
 x  3dx
b)  3 x dx
c)
Solutions
a)
1
 x  3dx  ln x  3  c
b)  3 x dx 
c)
3x
c
ln 3
7
1
1
 5  xdx  7 5  x dx  7 ln 5  x   1  c  7 ln 5  x  c
Exercise 4
Solve:
3
a)
 2 x  3dx
b)

5
dx
x
c)

34 x2 dx
Answer :
3
ln(2 x  3)  c
2
Answer : 5ln x  c
Answer :
34 x  2
c
4ln 3
8
7
 5  xdx
3.2.5 Integral of Trigonometric Functions
Formula of Integral of Trigonometric Functions
a)
 cos x dx = sin x  C
b)  sin x dx   cos x  C
c)  sec 2 x dx  tan x  C
d)  cos ec 2 x dx   cot x  C
1
e)  cos (ax  b) dx = sin (ax  b)  C
a
1
f)  sin (ax  b) dx   cos (ax  b)  C
a
g)  sec 2 (ax  b) dx 
1
tan (ax  b)  C
a
1
h)  cos ec 2 (ax  b) dx   cot (ax  b)  C
a
Example 5
Find:
a)  sin 3 x dx
b)  cos 5 x dx
c)  sec 2 4 x dx
d)  cos ec 2 2 x dx
e)  sin (3x  4) dx
f)  cos (2  5 x) dx
Solutions
a)  sin 3 x dx  
b)  cos 5 x dx 
cos 3 x
c
3
sin 5 x
c
5
c)  sec 2 4 x dx 
tan 4 x
c
4
d)  cos ec 2 2 x dx  
cot 2 x
c
2
9
e)  sin (3x  4) dx  
f)  cos (2  5 x) dx 
cos(3x  4)
c
3
sin( 2  5 x)
1
 c   sin( 2  5 x)  c
5
5
Exercise 5
a)  sec 2 (4  5 x) dx
1
Answer :  tan(4  5 x )  c
5
b)  cos ec 2 (2 x  5) dx
Answer : 
c)

cot(2 x  5)
c
2


Answer : cos   x   c
2


 sin  2  x dx
Many of the indefinite integrals are found by “reversing” derivative formulae. You will
see what we mean if you look at the basic integral formulae for some standard functions
in Table 1.
Table 1: Basic Derivatives and Integral Formulae
Corresponding Derivative Formula
d
Indefinite Integral
dx
 `1dx  x  C
d  x n 1 
n

  x , n  1
dx  n  1 

[ x]  1
x n dx =
x n1
 C , n  1
n 1
 x dx  ln x  C
 e dx  e  C
d
1
[ln x ] 
dx
x
d x
x
[e ]  e
dx
d kx
[e ]  kekx
dx
d x
[a ]  a x ln a
dx
d
[cos x]   sin x
dx
d
[sin x]  cos x
dx
d
[tan x]  sec2 x
dx
d
[cot x]   cos ec 2 x
dx
1
x

x
e kx dx 
e kx
C
k
ax
C
ln a

 sin x dx   cos x  C
 cos x dx = sin x  C
 sec x dx  tan x  C
 cos ec x dx   cot x  C
a x dx 
2
2
10
d
[sec x]  sec x tan x
dx
d
[cos ec x]   cosec x cot x
dx
d
[sinh x]  cosh x
dx
d
[cosh x]  sinh x
dx
d
[tanh x]  sec h 2 x
dx
d
1
[sin 1 x] 
dx
1 x2
 sec x tan x dx  sec x  C
 cos ec x cot xdx   cos ec x  C
 sinh x dx  cosh x  C
 cosh x dx = sinh x  C
 sec h x dx  tanh x  C
1
 1  x dx  sin x  C
d
1
[cos 1 x] 
dx
1 x2
d
1
[tan 1 x] 
dx
1 x2
d
1
[sinh 1 x] 
2
dx
x 1

d
[cosh 1 x] 
dx
2
1
2
1
1 x
1
1 x

1

x 1
1
2

d
[tanh 1 x] 
dx
1 x2
2
dx  cos 1 x  C
2
dx  tan 1 x  C
1
x 1
1
dx  sinh 1 x  C
2
dx  cosh1 x  C
x 1
1
dx  tanh 1 x  C
2
1 x
2
NOTE: Methods shown in 3.2.1 – 3.2.5 also can be applied in solving the
Definite Integral.
3.3 DEFINITE INTEGRATION
In this section, the concept of a “definite integrals” is introduced which will link
the concept of area to other important concepts such as length, volume, density,
probability and work.
Figure 2
11
Based on Figure 2, the curve f ( x) is nonnegative and continuous on an
interval [a, b] . The area of A which is under the graph of f ( x) over the interval [a, b]
can be represented by the definite integral
b
A   f ( x ) dx
a
where the number a is the lower limit and b is the upper limit of the integration.
Note that there is no constant in definite integral, therefore definite integral is
always in number. This is because the constant c is eliminated as shown below.
If
 f  x  dx  F  x   c , where c is a constant, then

b
a
f  x  dx   F  x  a   F b   c    F  a   c 
b
 F b  F  a 
Theorem (The Fundamental Theorem of Calculus):
If f ( x ) is continuous on [ a, b] , and F  x  is any anti-derivative of f ( x ) on [ a, b] , then
b
 f ( x) dx  F (b)  F (a)
a
Theorem: Basic Properties of Definite Integrals
If f(x) and g(x) are continuous functions on the interval [a, b], then
a)

a
b)

b

b
c)
d)
a
a
a
f ( x) dx  0 , if f(a) exists
a
f ( x) dx    f ( x) dx
b
b
kf ( x) dx  k  f ( x) dx
a
b
 k dx  k (b  a)
a
b
c
b
a
c
e)

f)
  f ( x)  g ( x) dx  
a
f ( x) dx   f ( x) dx   f ( x) dx , where a  c  b
b
b
a
a
b
f ( x) dx   g ( x) dx
a
12
Example 6
Solve the following integrals:
a)
d)

2

2
1
1
2

 3x   dx
x

 2x  1 
b)   4 dx
1
 x 
5
dx
x
e)
2
3


0
3
c)


sin   x dx
2

1
2
0


f)

0
e 4 x  e8 x
dx
e2x
4
(sec 2 x  2 sin 2 x) dx
Solutions
a)

2
1
2
2
2
4 

2
 3x   dx  1  9 x  12  2 dx
x
x 


2
2
  9 x 2 dx   12dx  
1
1
2
1
2
2
4
dx
x2
2
 9  x 2 dx  12  1dx  4  x 2 dx
1
1
1
2
4

  3 x 3  12( x)    (3(8)  12(2)  2)  (3  12  4)  11
x 1

b)
3 2x
3 1
 2x  1 
dx   4 dx   4 dx

4
1 x
1 x
1
x 
3
 
3
3
 2  x 3 dx   x 4 dx
1
1
3
3
 x 2 
x 3
 2


 2  1 3 1
3
1 
1  98
 1
 1 1 
   2  3         1   
3  81
 x 3 x  1  9 81  
c)
1
2
0

e 4 x  e8 x
dx 
e2x
1
2
 e
0
2x

 e 6 x dx
13
1
 e2x e6x  2

 

6  0
 2
 1 2 1  1 6 1    1 2 ( 0 ) 1 6 ( 0 ) 
  e 2  e 2    e
 e 
2
 2
6
6



1
1
1 1
e  e3  
2
6
2 6
1
1
2
 e  e3 
2
6
3

d)

2
1
21
5
2
dx  5  5 ln x1  (5 ln 2)  (5 ln 1)  3.4657  0  3.4657
1
x
x


 3

cos

x



3
2




e)
sin   x dx 
0
2

1




0

    

 cos     cos  0  
2 3  2

 cos

 cos
6
 0.866  0

2
 0.866

f)


0

  cos 2 x   4
4
(sec 2 x  2 sin 2 x) dx   tan x  2
 
2

  0


 tan x  cos 2 x 04

 

  tan  cos 2( )   (tan 0  cos 0)
4
4 

 1  cos

2
 200
 0 1
2
14
Exercise 6
a)
4

 
x
1
1 
dx
x
Answer :
e 2x  e5x
b) 
dx
1
e9x
2
c)


0

d)

0
Answer : 
1
sec 2 x dx
4
2
20
3
1 4 2
e  e 
2
Answer : 4
(2 x  cos x) dx
Answer :
2
4
1
3.4 TECHNIQUES OF INTEGRATION
In the previous sections, we learned how to integrate elementary functions using
integral formulas. Many integration problems which we encounter in the fields of science
and engineering are not expressed in such standard forms. Hence, we need some methods
to convert the given integrals to elementary forms before carrying out the integrations. In
this section we discuss some techniques of integration. These are

integration by substitution

integration by parts

integration by partial fractions.
We shall discuss each of these methods extensively by solving examples.
3.4.1 Integration by Substitution
This method is the first to be considered, whenever we try to obtain any integral.
The purpose of this method is to change the integrand into an expression of basic integral
forms. In principle, the process of integration by substitution can be done through five
steps as follows
Step 1:
Make appropriate choice of u, let u  g (x)
Step 2:
Obtain
du
 g ' ( x)
dx
15
Step 3:
Substitute
u  g (x) ,
du  g ' ( x)dx
After this stage, the whole integral must be in terms of u. This means
no more term in x can remain. If this step fails, we need to make
another appropriate choice of u.
Step 4:
Evaluate the integral obtained in terms of u.
Step 5:
Substitute u with g(x), so that the final answer will be in terms of x.
Some integral forms that can be evaluated using integration by substitution are

 f (ax  b)



 f ( x) f ' ( x)
dx
f ' ( x)
dx
f ( x)
dx
We shall illustrate each case above with examples.
i)
Function of a Linear Function of x
We are very required to integrate functions like those in standard list, but where x
is replaced by a linear function of x, e.g.
 (5x  4)
6
dx , which is very much
x
6
dx
except that x is replaced by (5 x  4) . If we put u to stand for (5 x  4) , the integral
becomes
u
6
dx and therefore we can complete the operation, we must change the
variables, thus:
u
Now
6

dx  u 6
dx
du
du
dx
du
dx 1
 5 , therefore

can be found from substitution u  (5 x  4) for
and
du
dx
du 5
the integral becomes:


u 6 dx  u 6
dx
1 6
1 u7
1
du  u 6  du 
u du  
c
du
5
5 7
5


Finally, we must express u in terms of the original variables, x, so that:
16

(5 x  4) 6 dx 
(5 x  4) 7
(5 x  4) 7
c 
c
57
35
Example 7
Evaluate the following integrals by using the substitution u  4x  1.
b)  (4 x  1) 2 dx
a)  (4 x  1) dx
Solutions
a) If u  4x  1, then
du
1
 4  du  dx .
dx
4
Substituting u and du give us
1
 (4 x  1) dx  4  u du
1
 u2  c
8
1
 (4 x  1) 2  c
8
If we solve directly:
4x 2
(
4
x

1
)
dx

 x  c  2x 2  x  c

2
Both answer are actually the same, because
1
1
(4 x  1) 2  c  (16 x 2  8 x  1)  c
8
8
1
c
8
 2 x2  x  C
 2 x2  x 
where C 
1
c
8
du
1
 4  du  dx .
dx
4
If we solve directly:
Substituting u and du give us
(4 x  1) dx   (16 x 2  8 x  1) dx

1 2
2
16 x3
 (4 x  1) dx  4  u du

 4 x2  x  C
3
1 3
Which
is
equivalent
to the previous because
 u c
12
1
1
1
(4 x  1)3  c  (64 x 3  48 x 2  12 x  1)  c
 (4 x  1) 3  c 12
12
12
16
1
 x3  4 x 2  x   c
3
12
16
1
 x3  4 x 2  x  C , where C 
c
3
12
b) If u  4x  1, then
17
Example 8
Evaluate the following integrals.
a)
c
 ax  b dx
5
b)  e 4 x dx
e)  cos( 2 x  3) dx

f)
tan( 3  2 x)dx
c)  (10 x  9) 2 dx
d)  sin 15 x dx
g)  35 x dx
h)
Solutions
a) If u  ax  b , then
c
 ax  b
1 1
dx  c   du
u a

c 1
du
au

c
c
ln u  c  ln ax  b  c
a
a
b) If u  4 x , then
e
4x
du
1
 a  du  dx
dx
a
du
1
 4  du  dx
dx
4
1
dx   e u  du
4
1 u
e du
4
1
1
 eu  c  e 4 x  c
4
4

c) If u  10x  9 , then
 (10 x  9)
5
2
du
1
 10  du  dx
dx
10
5
2
dx   u 
1
du
10
18
1
 4x  3
dx

5
1

u 2 du
10
5
1
2
1 u

c
10 5
1
2
7
1
 u2 c
35

7

d) If u  15x , then
1
(10 x  9) 2  c
35
du
1
 15  du  dx
dx
15
1
 sin 15 x dx   sin u  15 du
1
sin u du
15 
1
  cos u  c
15
1
  cos 15 x  c
15

e) If u  2x  3 , then
 cos(2 x  3)
du
1
 2  du  dx
dx
2
1
dx   cos u  du
2

f) If u  3  2x , then
1
1
1
cos u du  sin u  c  sin(2 x  4)  c

2
2
2
du
1
 2   du  dx
dx
2
 tan( 3  2 x) dx   tan u
1
  du
2
19

1
tan u du
2
1
  sec 2 u  c
2
1
  sec 2 (3  2 x)  c
2

g) If u  5x , then
3
5x
du
1
 5  du  dx
dx
5
1
dx   3u  du
5

1 u
3 du
5
1 3u
 
c
5 ln 3
35 x

c
5 ln 3

h) If u  4x  3 , then
1
 4x  3
du
1
 4  du  dx
dx
4
1 1
dx    du
u 4

1 1
1
ln( 4 x  3)
du  ln u  c 
c
4 u
4
4

Exercise 8
Find:
a)  e 3 x 5 dx
1
Answer :  e 3 x 5  c
3
b)  sec 2 (3x  1) dx
Answer :
1
tan(3x  1)  c
3
Answer :
1
sin(1  4 x )  c
4
c)

cos(1  4 x) dx
20
ii) Integral of the form

Consider the integral
f ' ( x)
dx
f ( x)
x
2
2x  3
dx . This is not one of our standard integrals,
 3x  5
so how shall we tackle it? This is an example of a type of integral which is very easy to
deal with but which depends largely on how keen your wits are. You will notice that if
we differentiate the denominator, we obtain the expression in the numerator. So, let u
stand for the denominator, i.e. u  x 2  3x  5 .

du
 2 x  3  du  (2 x  3)dx
dx
The given integral can then be written in terms of u:
x
2
2x  3
1
dx 
du  ln u  c
u
 3x  5

If we now put back what u stands for in terms of x, we get

2x  3
dx  ln( x 2  3x  5)  c
x  3x  5
2
Example 9
Evaluate the following indefinite integrals by using the substitution method.
a)
2x  3
 x 2  3x  2 dx
d)
 3x
x  cos x
dx
2
 6 sin x
b)
x
 x 2  4 dx
c)
12 x 2  16
 x 3  4 x dx
e)
 tan x dx
f)
 1  cos x
Solutions
a) If u  x 2  3x  2 , then
x
2
du
 2 x  3  du  (2 x  3)dx
dx
2x  3
1
dx    du  ln u  c  ln( x 2  3x  2)  c
u
 3x  2
b) If u  x 2  4 , then
du
1
 2 x  du  xdx
dx
2
21
sin x
dx
x
2
x
1 1
dx    du
u 2
4
1 1
du
2u
1
 ln u  c
2
1
 ln( x 2  4)  c
2

c) If u  x 3  4 x , then
du
 3x 2  4  du  (3x 2  4)dx
dx
12 x 2  16
4(3x 2  4)
 x 3  4x dx   x 3  4x dx
1
 4  du
u
 4 ln u  c
 4 ln( x 3  4 x)  c
d) If u  3x 2  6 sin x , then
du
 6 x  6 cos x
dx
du  6( x  cos) dx 
1
du  ( x  cos x)dx
6
x  cos x
1 1
dx    du
2
u 6
 6 sin x
 3x
1 1
du
6u
1
 ln u  c
6
1
 ln( 3x 2  6 sin x)  c
6

e) Given tan x 
sin x
. Therefore,
cos x
sin x
 tan x dx   cos x dx
22
If u  cos x , then
du
  sin x  du  sin x dx
dx
sin x
1
 tan x dx   cos x dx   u  du
1
    du
u
  ln u  c
  ln(cos x)  c
du
 sin x  du  sin x dx
dx
f) If u  1 cos x , then
sin x
 1  cos x
1
dx    du  ln u  c  ln( 1  cos x)  c
u
Exercise 9
a)
 cos x
b)
x
c)
sec 2 x
 tan x dx
iii)
sin x
2
Answer :  ln(cos x )  c
dx
x3
dx
 6x  2
Answer :
1
ln( x 2  6 x  2)  c
2
Answer : ln(tan x )  c
Integral of the Form
 f ( x) f ' ( x) dx
In very much the same way, we sometimes have integrals such as
 tan x sec
2
x dx
This, of course, is not a quotient but a product. Nevertheless we notice that one function
(sec 2 x) of the product is the derivative of the other function (tan x) .
If we put u  tan x , then du  sec 2 x dx and the integral can then be written

u du 

u2
c
2
 tan x sec 2 x dx 
tan 2 x
c
2
23
Example 10
Evaluate the following integrals.
a)
 4 x( 2 x
d)

2
 3) 6 dx
ln x
dx
x
b)
 sin x cos
e)
 x ln x
4
x dx
dx
c)
 e (e
f)
 2x( x  2)
x
x
 1) 3 dx
5
Solutions
a) If u  2 x 2  3 , then
 4 x( 2 x
2
du
 4 x  du  4 x dx
dx

 3) 6 dx  u 6 du

b) If u  cos x , then
 sin x cos
4
u7
1
 c  (2 x 3  3) 7  c
7
7
du
  sin x  du  sin x dx
dx

x dx  u 4  du

  u 4 du  
c) If u  e x  1 , then
 e (e
x
x
1
1
u5
 c   (cos x)5  c   cos5 x  c
5
5
5
du
 e x  du  e x dx
dx

 1) 3 dx  u 3 du
u4
c
4
1
 (e x  1) 4  c
4

d) If u  ln x , then
du 1
1
  du 
dx
dx x
x
24
dx

u2
1
ln x
1
dx  ln x  dx   u du   c  (ln x)2  c
2
2
x
x

du 1
1
  du 
dx
dx x
x
e) If u  ln x , then
 x ln x   ln x  x dx
dx
1


1
1
 du
u
 ln u  c
 ln(ln x)  c
f) If u  x  2 , then
 2x( x  2)
5
du
 1  du  dx . Also x  u  2
dx

  (2u  4)u du
  (2u  4u )du
dx  2(u  2)u 5 du
5
6
5
2u 7 4u 6

c
7
6
2u 7 2u 6


c
7
3
2u 6
2( x  2)6

(3u  7)  c 
 3( x  2)  7   c
21
21

Exercise 10
a)

b)
 2x
c)
2
xe x dx
x
3
cos x 4 dx
3x  2 dx
1 2
Answer :  e  x  c
2
Answer :
1
sin x 4  c
2
5
3
2
4
2
Answer :
(3x  2)  (3x  2) 2  c
45
27
25
3.4.2 Integration by Parts
Integration by parts is another important method to evaluate integrals. This
method is essentially an anti-derivative formulation of the formula for differentiating a
product of two functions or it is called as product rule. It is applied in the case when the
integration involve products of algebraic and transcendental functions, for examples



 x ln x dx
 xe dx
 e cos x dx .
x
x
Based on the formula of product rule
d
dv
du
(uv)  u
v
dx
dx
dx
when it is integrated with respect to x, it becomes

uv  u
and so,

dv
du
dx  v
dx ,
dx
dx
 u dx dx  uv   v dx
dv
du
dx
or simply
 u dv  uv   v du
This formula expresses an integral
 u dv
in term of a second integral,
 v du . Using
correct choices of u and dv, the second integral may be simpler to evaluate than the first.
Priority order on choosing the u:
- If one factor is a log function, that must be taken as ‘u’.
- If there is no log function but a power of x, that becomes ‘u’.
- If there is neither a log function nor power of x, then the exponential function is taken as
‘u’.
ln x
xn
26
e kx
Example 11
Evaluate the following integrals.
a)


b) ln x dx
x ln x dx
Solutions
dv  x dx
a) Let , u  ln x
du 1

dx x
1
du  dx
x

 dv   x dx
and
v
x ln x dx  (ln x)(
x2
)
2

x2
2
x2 1
 dx
2 x
x2
1
ln x   x dx
2
2
2
x
1  x2 
x2
x2
c 

ln x  
ln x 
c
2
2  2 
2
4

b) Let , u  ln x
dv  1 dx
and
du 1

dx x
1
du  dx
x
 dv   1 dx
vx
 ln x dx  (ln x)( x)   x  x
1
dx

 x ln x  1 dx
 x ln x  x  c
Example 12
Evaluate the following integrals.
a)

b)
xe x dx
x
2
cos x dx
Let , u  x
and
 x e dx
1
0
Solutions
a)
c)
dv  e x dx
27
2
x
 dv   e
du
1
dx
du  1dx

x
dx
v  ex

xe x dx  xe x  e x dx
 xe x  e x  c
b)
Let , u  x 2
dv  cos x dx
and
du
 2x
dx
du  2 x dx
x
2
 dv   cos x dx
v  sin x

cos x dx  x 2 sin x  sin x  2 x dx

 x 2 sin x  2 x sin x dx                (1)
Let , u  x
dv  sin x dx
and
du
1
dx
du  1 dx
 dv   sin x dx
v   cos x
 x sin x dx  x( cos x)    cos x  dx

  x cos x  cos x  dx
  x cos x  sin x  c                    (2)
Substitute (2) into 1:

2
 x 2 cos x dx  x sin x  2( x cos x  sin x  c)
 x 2 sin x  2 x cos x  2 sin x  2c
 x 2 sin x  2 x cos x  2 sin x  C ,
c)
Let , u  x 2
and
du
 2x
dx
du  2 x dx
 x e dx  x e   e
2
x
2
x
dv  e x dx
 dv   e
x
dx
v  ex
x
2 x dx
28
where C  2c

 x 2 e x  2 xe x dx                (1)
Let , u  x
dv  e x dx
and
 dv   e
du
1
dx
du  1dx

x
dx
v  ex

xe x dx  xe x  e x dx
 xe x  e x  c                  (2)
Substitute (2) into 1:

2 x
x
x
 x 2 e x dx  x e  2( xe  e  c)
 x 2 e x  2 xe x  2e x  2c
 e x ( x 2  2 x  2)  C ,

1



 
where C  2c
1
 x 2 e x dx  e x ( x 2  2 x  2) 0
0
 e1 (12  2(1)  2)  e 0 (0 2  2(0)  2)

 e1  2
 0.718
Example 13

Evaluate e x cos 2 x dx using integration by parts.
Solutions
Let , u  e x
dv  cos 2 x dx
and


du
dv  cos 2 x dx
 ex
dx
sin 2 x 1
 sin 2 x
du  e x dx v 
2
2
e
x

1
1
cos 2 x dx  e x  sin 2 x 
sin 2 x  e x dx
2
2


1 x
1
e sin 2 x 
e x sin 2 x dx              (1)
2
2
29
Let , u  e x
e
and
dv  sin 2 x dx
du
 ex
dx
 dv   sin 2 x dx
du  e x dx
v
cos 2 x
1
  cos 2 x
2
2

1
1
sin 2 x dx  e x   cos 2 x   cos 2 x  e x dx
2
2
x

1
1
  e x cos 2 x 
e x cos 2 x dx            (2)
2
2
Substitute (2) into (1):
e
e
e
x
cos 2 x dx 
1 x
1 1
1 x

e sin 2 x   e x cos 2 x 
e cos 2 x dx 
2
2 2
2

x
cos 2 x dx 
1 x
1
1
e sin 2 x  e x cos 2 x 
e x cos 2 x dx
2
4
4
x
cos 2 x dx 
1
1
1
e x cos 2 x dx  e x sin 2 x  e x cos 2 x
4
2
4




5
1
1
e x cos 2 x dx  e x sin 2 x  e x cos 2 x  c
4
2
4

 e x cos 2 x dx 

4 1 x
1

e sin 2 x  e x cos 2 x  c 
5  2
4

1 x
e 2 sin 2 x  cos 2 x   C ,
5
where C  4 c
15
Exercise 11
Evaluate the following integrals.
a)
x
2
sin 3x dx

b) e 5 x sin 3x dx
c)
x
3
2
3
ln x dx
Answer : 1 x 2 cos 3x  2 x sin 3x  2 cos 3x  C
3
9
27
Answer :  3 e5 x cos 3x  5 e5 x sin 3x  c
34
34
Answer : 81 ln 3  4 ln 2  65
4
16
3.4.3 Integration by Partial Fractions
Now, we will learn another method of integration which provides an approach to
integrating a general rational function. It is based on the idea of decomposing a rational
30
function into a sum of simple rational functions and then can be integrated by any
methods that we have studied in the previous sections.
Recall that a rational function is a quotient of two polynomials. Basically, it is
written as f ( x) 
P( x)
, where the both numerator P(x) and the denominator Q(x) are
Q( x)
polynomials. There are two types of rational function that we will consider:

Proper rational function - when the degree of the numerator is less than the degree
of the denominator.

Improper rational function – when the degree of the numerator is larger than the
degree of the denominator.
i) Type 1 : Proper Rational Function
The first step in integrating the proper rational function is by finding a
factorization of the denominator Q(x). It can be shown that any polynomial Q(x) can be
factored as a product of linear factors (of the form ax+b) and irreducible quadratic factors
(of the form ax 2  bx  c , where b 2  4ac  0 ). For instance, if Q( x)  x4  16 then
Q( x)  ( x 2  4)( x 2  4)  ( x  2)( x  2)( x 2  4) . This process is known as decomposition of the
partial fraction.
Linear Factor Rule:
For each factor of the form (ax  b)m , the partial fraction decomposition contains the
following sum of m partial fractions:
Am
A1
A2

 ...... 
2
m
ax  b  ax  b 
 ax  b 
where A1 , A2 ,..., Am are constants to be determined. If m  1 , only the first term in the
sum appears.
31
Example 14
Find
x
5x  1
dx
 x2
2
Solutions
Factorized Q(x)
x
5x  1
5x  1
dx 
dx
( x  1)( x  2)
 x2

2
Note: x 2  x  2  ( x  1)( x  2)
Applying the Linear Factor Rule to determine the shape of the partial fraction
5x  1
A
B
--------------------(1)


( x  1)( x  2) x  1 x  2
Multiply both sides by the denominator ( x  1)( x  2)
5 x  1  A( x  2)  B( x  1) ----------------------(2)
To find the value of A and B , we can apply this two method:
i) Substitute values of x that are chosen to make terms on the right drop out
Let ( x  1)  0; then choose x  1 to substitute in equation (2):
 5(1)  1  A(1  2)  B(1  1)
6  3 A  A  2
Let ( x  2)  0; then choose x  2 to substitute in equation (2):
 5(2)  1  A(2  2)  B(2  1)
 9  3B  B  3
ii) Multiply out the right side of (2) and equate corresponding coefficients on the two
sides
5 x  1  Ax  2 A  Bx  B
 ( A  B) x  (2 A  B)
A B  5
2A  B  1
which implies A  2 and B  3
32
Then, taking the value A  2 and B  3 from any methods as show before to substitute
into (1)
5x  1
2
3


( x  1)( x  2) x  1 x  2
Thus the integral can now be written as below:

5x  1
dx 
( x  1)( x  2)

2
dx 
x 1

3
dx
x2
 2 ln x  1  3 ln x  2
Example 15
Find
x2
  x  1 x  1 dx
2
Solutions
x2
A
B
C
 ( x  1)( x  1)2 dx   ( x  1) dx   ( x  1) dx   ( x  1)2 dx
x2
A
B
C
------------(1)



2
x  1 x  1 ( x  1) 2
( x  1)( x  1)
x 2  A( x  1) 2  B( x  1)( x  1)  C ( x  1) ---------(2)
Let ( x  1)  0; i.e. substitute x  1 in equation (2):
 (1) 2  A(1  1) 2  B(1  1)( 1  1)  C (1  1)
1  4 A  A 
1
4
Let ( x  1)  0; i.e. substitute x  1 in equation (2):
 (1) 2  A(1  1) 2  B(1  1)(1  1)  C (1  1)
1  2C  C 
1
2
When the crafty substitution has come to an end, we can find the remaining constants (in
this case, just B) by equating coefficients. Choose the highest power involved, i.e. x2 in
this example:
33
From equ. (2): x 2  A( x  1) 2  B( x  1)( x  1)  C ( x  1)
x 2  A( x 2  2 x  1)  B( x 2  1)  Cx  C
x 2  Ax 2  2 Ax  A  Bx 2  B  Cx  C
x 2  ( A  B) x 2  (2 A  C ) x  ( A  B  C )
Compare [x2]: 1  A  B
1
Substitute A 
1
3
 B  B 
4
4
1
3
1
, B  and C  in equ. (1):
4
2
4
x2
1 1
3 1
1
1
 

 
2
4 x  1 4 x  1 2 ( x  1) 2
( x  1)( x  1)

x2
dx 
( x  1)( x  1) 2

1
3
1
dx 
dx 
dx
4( x  1)
4( x  1)
2( x  1) 2

1
4


1
3
1 ( x  1) 1
ln x  1  ln x  1  
c
4
4
2
1

1
3
1
ln x  1  ln x  1 
c
4
4
2( x  1)

1
3
dx 
x 1
4



1
1
1
dx 
dx
x 1
2 ( x  1) 2
Quadratic Factor Rule:
For each factor of the form (ax 2  bx  c)m , the partial fraction decomposition contains
the following sum of m partial fractions:
Am x  Bm
A1 x  B1
A2 x  B2

 ...... 
2
2
m
2
ax  bx  c  ax  bx  c 
 ax2  bx  c 
where A1 , A2 ,..., Am , B1 , B2 ,..., Bm are constants to be determined. If m  1 , only the first
term in the sum appears.
34
Example 16
Find

x2 1
dx
( x  2) 3
Solution

x2 1
A
B
C
dx 
dx 
dx 
dx
3
2
( x  2)
( x  2)
( x  2)
( x  2) 3



x2 1
A
B
C
-------------------(1)



3
2
x  2 ( x  2)
( x  2)
( x  2) 3
x 2  1  A( x  2) 2  B( x  2)  C -------------------------(2)
Let ( x  2)  0; i.e. substitute x  2 in equation (2):
 (2) 2  1  A(2  2) 2  B(2  2)  C
5  C  C  5
When the crafty substitution has come to an end, we can find the remaining constants (in
this case, A and B) by equating coefficients.
From equ. (2): x 2  1  A( x  2) 2  B( x  2)  C
x 2  1  A( x 2  4 x  4)  Bx  2B  C
x 2  1  Ax 2  4 Ax  4 A  Bx  2 B  C
x 2  1  Ax 2  (4 A  B) x  (4 A  2B  C )
Compare the coefficient of,
:
1  A  A  1
[constant]:
1  4 A  2B  C
[x2]
1  4(1)  2 B  5
2B  8  B  4
Substitute A  1 , B  4 and C  5 in equ. (1):
x2 1
1
(4)
5



3
2
x  2 ( x  2)
( x  2)
( x  2) 3

x2 1
1
4
5
dx 
dx 
dx 
dx
3
2
( x  2)
( x  2)
( x  2)
( x  2) 3



35




1
dx  4 ( x  2) 2 dx  5 ( x  2) 3 dx
( x  2)
 ln x  2  4 
 ln x  2 
( x  2) 1
( x  2) 2
 5
c
(1)
(2)
4
5

c
( x  2) 2( x  2) 2
Example 17
Evaluate

2
dx
x  2x
3
Solution
x

3
2
dx 
 2x
 x( x
2
dx 
x( x  2)
2

2
dx
 2)
2
A
dx 
x

Bx  C
dx
x2  2
2
A Bx  C
  2
------------------------------------(1)
x ( x  2) x x  2
2
2  A( x 2  2)  ( Bx  C )( x) ------------------------------------(2)
Let x  0; i.e. substitute x  0 in equation (2):
 2  A[(0) 2  2)]  [ B(0)  C ](0)
2  2 A  A  1
When the crafty substitution has come to an end, we can find the remaining constants (in
this case, B and C) by equating coefficients.
From equ. (2): 2  A( x 2  2)  ( Bx  C )( x)
2  Ax 2  2 A  Bx 2  Cx
2  ( A  B) x 2  Cx  2 A
Compare the coefficient of,
[x2]:
0  A B
0  1  B  B  1
[x]:
C0
36
Substitute A  1 , B  1 and C  0 in equ. (1):
Use substitution method
2
1
 1x
  2
2
x( x  2) x x  2
x
du
du
 2x 
 xdx
dx
2
x
1 du 1 1
1
dx 


du  ln u  c
2
x 2
u 2
2 u
2
x
1
2
 2
dx  ln x  2  c
x 2
2



1
 ln x  ln x 2  2  c
2
Example 18
1  x  2x 2  x3
 x5  2x3  x
Solution
1  x  2 x 2  x 3 1  x  2 x 2  x 3 A Bx  C Dx  E

  2

2
x x 1
x 5  2x 3  x
x( x 2  1) 2
x2 1

x
dx ;
2
Let u  x 2  2

Find
2
  Bx  C( x
1  x  2x 2  x 3  A x 2  1
2

2


 1)( x)  Dx  E ( x)
= ( A  B) x 4  Cx 3  (2 A  B  D) x 2  (C  E ) x  A
x4 : A  B  0
x 3 : C  1
x2 : 2A  B  D  2
x1 : C  E  1
x0 : A  1
 A  1, B  1, C  1, D  1, E  0
1  x  2x 2  x 3 1
x
1
x
  2
 2

5
3
2
x x 1 x 1 x 1 2
x  2x  x


Then integrate that problem
1  x  2x 2  x 3
1
x
1
x
 x 5  2x 3  x dx   xdx   x 2  1dx   x 2  1 dx   x 2  1 2 dx

37

1  x  2x 2  x 3
1
1
2
1
 x 5  2 x 3  x dx  ln x  2 ln x  1  tan ( x)  2( x 2  1)  c
Conclusion of Partial Fraction Decomposition:Factor in
Term in partial
Denominator Q(x)
fraction decomposition
ax+b
A
A1
A2

 ......  m
ax  b ax  b
ax  b
 ax  b 
m
Am
A1
A2

 ...... 
2
m
ax  b  ax  b 
 ax  b 
Ax  B
ax  bx  c
ax 2  bx  c
2
 ax
2
 bx  c 
m
Am x  Bm
A1 x  B1
A2 x  B2

 ...... 
2
2
m
ax  bx  c  ax 2  bx  c 
 ax2  bx  c 
Exercise 12
x 3  4 x 2  3x  4
dx
x 2  4x  3
a)

b)
x
2
3x  2
dx
 2 x  24
Answer :
x2
 2ln x  3  2ln x  1  c
2
Answer : ln x  4  2ln x  6  c
x2 1
c)  3
dx
x  6 x 2  11x  6
Answer : ln x  1  5ln x  2  5ln x  3  c
 2 x 2  14 x  49
d) 
dx
x3  7x2
Answer : 3ln x 
e)
ii)

6x  7
dx
( x  2) 3
Answer : 
7
 5ln x  7  c
x
6
5

C
x  2 ( x  2)2
Type 2 : Improper Rational Function
In order to solve the improper rational function by using integration of partial
fractions, we have to take the preliminary step of dividing denominator Q(x) into
numerator P(x) by long division until a remainder R(x) is obtained such that
38
f ( x) 
P( x)
R( x)
 S ( x) 
Q( x)
Q( x)
(1)
where P(x) and Q(x) are polynomials.
Example 19
Find

3x 2  x  1
dx
x 1
Solution
By long division:
divisor
3x  2
x  1 3x  x  1
quotient
2
3x 2  3x
 2x  1
 2x  2
3
remainder
take the quotient, remainder and divisor, then plug into the formula (1)
3x 2  x  1
3
 (3x  2) 
x 1
x 1
the integration can be solve as follows
3x 2  x  1
3
  (3 x  2) dx  
dx
x 1
x 1
(3 x  2) 2
(3 x  2) 2

 3ln( x  1)  c 
 3ln( x  1)  c
23
6
or
3x 2

 2 x  3ln x  1  c
2
Example 20
Find
3x 4  3x3  5 x 2  x  1
dx

x2  x  2
39
Solution
Use long division
3x 2
1
x  x  2 3x  3x  5 x  x  1
2
4
3
2
3x 4  3x3  6 x 2
x2  x  1
x2  x  2
1
3x 4  3x3  5 x 2  x  1
1
2
dx

(3
x

1)
dx



 x 2  x  2 dx
x2  x  2
1
  (3 x 2  1) dx  
dx
( x  1)( x  2)
A
B
  (3 x 2  1) dx  

dx
( x  1) ( x  2)
Then solve A and B by using the methods that have been discussed before,
3x 4  3x3  5 x 2  x  1
1 x 1

dx  x3  x  ln
C
2
3 x2
x  x2
Exercise 13
a)
x3  x
 x  1 dx
Answer :
x3 x 2
  2 x  2ln x  1  c
3 2
b)

x3
dx
x 1
Answer :
1 3 1 2
x  x  x  ln x  1  c
3
2
c)
 x  3dx
x
Answer : x  3ln x 2  3  c
3.5 APPLICATIONS OF DEFINITE INTEGRATION
This is the last section for this chapter. Under this section, we will show that the
area between two curves can be obtained by integration the function of the curves which
40
based on the concept of definite integral. Other than that, we will see that the integration
method also can be used to find the volumes of certain three-dimensional solids.
3.5.1 Area of a Region Bounded by the Curve and Axis
Before we consider the problem of finding the area between two curves, it will be
helpful to review the basic principle that underlies the calculation of area as a definite
integral. Recall that as in the section 3.3, if the curve f ( x) is nonnegative and continuous
on an interval [a, b] . The area of A which is under the graph of f ( x) over the interval
[a, b] can be represented by the definite integral
b
A   f ( x ) dx
a
Example 21
The graph of y  2 x is shown below. Find the area of the shaded region A.
1
4
2
1
Solution
From the graph we find that region A is above the x-axis. Since the shaded region is
bounded x  1 and x  4 , then
4
4
Area of region, A   y dx   2 x dx
1
1
4
 32 
4x 

 3 

 1
4(8) 4(1) 28



unit 2
3
3
3
In the Example 21, the function given, f ( x)  0 for all x in a, b , which indicates that
the region is above the x-axis and the area is positive. On the contrary, if f ( x)  0 in the
41
interval b, c , it indicates that region is below the x-axis and the value of the definite
integral f is negative (see Figure 3). Since area is always positive, then we take

c
A   f ( x)dx for all curves below x-axis.
b
y
y  f (x)
Area
+ve
0
a
b
Area
-ve
c
x
Figure 3
Example 22
Find the area of the shaded region which bounded by f ( x)  x 2  2 x .
Solution
From the graph, it is obvious that f ( x)  0 for 1  x  2 and f ( x)  0 for 2  x  3 .
Hence,
2
 x3

 4  2 
Area, A    ( x  2 x) dx     x 2            2 unit 2
1
 3  3 
3
1
2
2
42
3
 x3

 27
 8
 4

Area, B  ( x  2 x)dx   x 2     9    4  unit 2
2
 3  3
3
2  3

3
2
Total area  2 
4 10
 unit 2
3 3
In finding the area of the region which bounded by curves, it is important for us to sketch
the graph of the f ( x) , since we might make a mistake by taking the integration straight
from x  0 to x  3 and the result will be
3
 x3

 27

Area  ( x  2 x)dx    x 2     9  0  0
0

3
0  3

3
2
Therefore, it is very important to sketch the graph so that we can see whether the area is
positive or negative.
The same concept is applied in dealing with the negative or positive area for
f ( y )  0 which indicates that the region is on the right hand side of y-axis and the area is
positive. Whereas for f ( y )  0 , it means that the region is on the left hand side of y-axis
and the area is negative (See Figure 4).
y
x  f ( y)
Area
+ve
Area
-ve
x
0
Figure 4
43
Example 23
Find the area of the region enclosed by y  2 x and y-axis between y  1 and y  2 .
Solution
Note that the region is on the right hand side of the y-axis, hence the area is positive.

2
Area, A  xdy 
1

2
1
y2
dy
4
2
 y3 
8
1
7
  

 unit 2
 12  1 12 12 12
Example 24
Below is the graph of y  x 3 . Calculate the area of the shaded region.
44
Solution
Since there are two regions, one is located on the right hand side of the y-axis and the
other on the left, therefore
1
0
 43   43 
1
0
3y  3y 
3 6 3
3
 

   0    0     unit 2
Area  y dy  y dy  
 4   4 
0
1
4 4 2
4
 
 1

 0 

1
3

1
3

1
1
3
Since both regions have equal area we can also obtain the total area as 2 y dy .
0
3.5.2 Area Between Curves
We will now consider the following extension of the area problem.
Area between Curves:
Suppose that f and g are continuous function on an interval [a,b] and if f ( x)  g ( x) for
a  x  b , then the area of the region bounded above by y  f ( x) , below by y  g ( x) ,
on the left by the line x  a and on right by the line x  b is given by
A

b
f ( x)dx 
a

b
g ( x)dx 
a
  f ( x)  g ( x)dx
b
a
In this section we are going to look at finding the area between two curves. There are
actually two cases that we are going to be looking at.
Case 1: To determine the area between y  f (x) and y  g (x) on the interval a, b .
We are also going to assume that f ( x)  g ( x) .
y
A
x
45
A = [area under y = f(x)] - [area under y = g(x)] or
b
b
=

=
 [ f ( x)  g ( x)]dx
a
f ( x)dx   g ( x)dx or
a
b
 upper

b
 lower

  function dx    function dx
a
a
b
a
Case 2: To determine the area between x  f ( y ) and x  g ( y ) on the interval c, d 
with that.
y
A
x
A=
=

d
c
d
f ( y)dy   g ( y)dy or
c

d
c
d  left
 right



dy   
dy
c
 function 
 function 
d
 [ f ( y)  g ( y)]dy
c
Example 25
Find the area of the region enclosed by the parabolas y  x 2 and y  2 x  x 2 .
Solution
Step 1:
Find the points intersections
x 2  2x  x 2
2x 2  2x  0
2 x( x  1)  0 so x = 0, 1
points intersections (0, 0) and (1, 1)
46
Step 2:
Sketch the curves
y  x2
y
y  x2
x
y  2x  x 2
Step 3:
From Step 2 we can find areas between that curves.
1
 x2 x3 
1
(2 x  x )dx  x dx  (2 x  2 x )dx  2    unit 2
0
0
0
3 0 3
2

1
2

1
2

1
2
Example 26
Determine the area of the region enclosed by y  x 2 and y  x .
Solution
Step 1:
Find the points intersections
x2  x
(x 2 )2  x
x4  x  0
x( x 3  1)  0
x  0 or x 3  1  0 . So x = 0, 1 and points intersections (0, 0) and (1, 1)
Step 2:
Sketch the curves
47
Step 3:
From Step 1 we can find areas between that curves.
1
 23

3x
x3 
1
2
2

( x )dx  x dx  ( x  x )dx 

 unit 2


0
0
0
2
3
3

 0


1
1

1
Example 27
Determine the area of the region enclosed by x 
1 2
y  3 and y  x  1 .
2
Solution
Step 1:
Find the points intersections
1 2
y  3  y 1
2
1 2
y  y40
2
y 2  2 y  8  ( y  4)( y  2)  0 so y = -2, 4
points intersections (-1, -2) and (5, 4)
Step 2:
Sketch the curves
Step 3
A=

d
c
2
d  left
4
4  y

 right



dy   
dy   ( y  1)dy   
 3 dy
c
2
2
 function 
 function 
 2

4
 y2

 y3 y2

 y  4 dy  

 4 y   18unit 2
= 
2
2
 2

 6
 2

4
48
Exercise 14
a) Find the area of the region bounded by the graphs y 
x2
and y  x as shown in the
3
following figure.
Answer :
3
unit 2
2
b) Calculate the area of the region between the curve y  9  x 2 and y  x 2  1 from
x  0 to x  3 .
Answer :
46
unit 2
3
3.5.3 Volumes of the three-dimensional solids
In this section, we will show that the method of definite integral can also be used
to find the volume of a solid generated by revolving the area of a region about an axis
through 360 or 2 radian. Usually, the x-axis, y-axis or any line which is parallel to
these axes is taken to be the axis of revolution. Figure 5 and Figure 6 show how each
solid is formed when the area of the region is rotated about x and y-axis.
y
0
a
b
x
Figure 5 (a): Area of region rotated about x-axis
49
Figure 5 (b): Solid generated from the rotation
y
a
b
x
Figure 6 (a): Area of region rotated about y-axis
Figure 6 (b): Solid generated from the rotation
50
i)
Volume of solid generated by the region bounded by the curve and the axis
Theorem:
If the function y  f (x) is continuous and f ( x)  0 for all x in the close interval a, b ,
then the volume, V, obtained by revolving 360 the region bounded by y  f (x) , the xaxis, the lines x  a and x  b about the x-axis is

b
V   { f ( x)}2 dx
a
In the same way, when the region is rotated about the y-axis, the volume is obtained as
stated in the following theorem.
Theorem:
If the function x  g ( y ) is continuous and g ( y )  0 for all y in the close interval c, d  ,
then the volume, V, obtained by revolving 360 the region bounded by x  g ( y ) , the yaxis, the lines y  c and y  d about the y-axis is

d
V   {g ( y )}2 dy
c
Example 28
Find the volume of the solid obtained when the region bounded by the curve y  3x 2 , xaxis and the line x  1 is revolved about the x-axis through 360 .
51
Solution
Since the area is revolved about the x-axis, the disc has width of x and radius of
y  3x 2 .
V
   f ( x) dx
1
2
0
    3 x 2  dx
1
2
0
    9 x 4  dx
1
0
1
 9 x5 
 

 5 0
9
  unit 3
5
Example 29
Find the volume of the solid obtained when the region bounded by the curve
y  3x 2 , ( x  0) , y-axis and the line y  1 is revolved about the y-axis through 360 .
Solution
52
y
. Using Theorem 3.7,
3
The disc has width of y and radius of x 
Volume, V 
   f ( y) dy
1
2
0
2
 y
 dy
  

0
3



1
1
 y2 
 y
   dy    
0  3
 6 0

1
1
  unit 3
6
ii)
Volume of solid generated by the region bounded by the two curves
Theorem:
Let f(x) and g(x) be continuous functions and f ( x)  0 , g ( x)  0 with f ( x)  g ( x) on
the interval a, b . If R is the region bounded by the curves y  f (x) and y  g (x) , from
x  a and x  b , then the volume, V, obtained by revolving R about the x-axis is

b

b
V   ( f ( x)) 2 dx   ( g ( x)) 2 dx
a

a
b
  [( f ( x)) 2  ( g ( x)) 2 ]dx
a
Theorem:
Let f(y) and g(y) be continuous functions and f ( y )  0 , g ( y )  0 with f ( y )  g ( y ) on
the interval c, d  . If R is the region bounded by the curves x  f ( y ) and x  g ( y ) , from
x  c and x  d , then the volume, V, obtained by revolving R about the y-axis is

d

d
V   ( f ( y )) 2 dy   ( g ( y )) 2 dy
c

c
d
  [( f ( y )) 2  ( g ( y )) 2 ]dy
c
53
Example 30
Calculate the volume of the solid formed when the area bounded by the curves y  x 2
and y  x is rotated about
a) x-axis
b) y-axis
Solution
a)
Rotating about the x-axis:
Two solids are generated when the two curves are rotated about the x-axis. The
bigger one is formed by y  x and the smaller one by y  x 2 and the
intersections are at x  0 and x  1 (see figure below.)
Since y  x  y  x 2 , the volume is calculated using Theorem 3.6. Therefore,
V 
 (
1

x ) 2  ( x 2 ) 2 dx
0
METHOD NOTE:
Do not calculate the volume by
taking the radius of the enclosed
region as ( x  x 2 ) that is
1
 x 2 x5 
   
 2 5 0
1 1 3
       unit 3
 2 5  10
V 
(
1
x  x 2 ) 2 dx
0
b)
Rotating about the y-axis:
When rotating about the y-axis, y  x 2  y  x and the intersections are at
y  0 and y  1 .
54
Writing x in terms of y, we have y  x 2  x  y and y  x  x  y 2 .
Therefore,
V 
 (
1

y ) 2  ( y 2 ) 2 dy
0
METHOD NOTE:
Do not calculate the volume by
taking the radius of the enclosed
region as ( y  y 2 ) that is

1
  ( y  y 4 )dy
0
1
 y2 y5 
  
5 0
2
V 
(
1
0
1 1 3
       unit 3
 2 5  10
55
y  y 2 ) 2 dy
Tutorials
1. Determine the following integrals:
a)
 2e dx
f)  4  2xdx
b)
 12dx
g)
 2e
c)
 9x 3 dx
h)
5
d)
 5x 1
i) 
3
dx
2x  3
e)

x
1
4
dx
sin  6 x  1
3
dx
j) 
3x2
1 x
dx
dx
sec2  2  5 x 
5
2. Determine the following integrals:
a) I    x3  x 2  x  1dx
b) I    4 x3  9 x 2  8 x  2  dx given that I 
11
1
when x 
16
2
3. Determine
a)
 4e
b)
 6e
c)
 3x  2 dx
d)
 3sec 1  4x  dx
e)
  6e
2 x 3
1 3 x
dx
dx
5
2

3 x 5

4

 52 x 1 dx
3x  2

4. Evaluate each of the following definite integrals:
2
a)
 12dx
1
56
dx
8
b)
 9x dx
1
2
0
4
c)
  5x  1
4
dx
2

d)

sin  6 x  1
3
0
dx
5. Determine the following integrals:
a)  ( x 2  x) 2 (2 x  1)dx
b)
 x( x
c)  ( x  1)( x 2  2 x  4) 3 dx
d)
2 x 2
 x e dx
e)  ecos x sin xdx
f)

g)  ( x  3) sin( x 2  6 x)dx
h)
ln y 2
 y dy
 1)3 dx
3
1
ex
i)  2 dx
x
2
x
2 x7  1
 4 x  1
8
1  ln x
dx
x
j) 
6. Find:
2
a)
2
3
 x( x  1) dx
1
b)
x e
2 x3  2
dx
0
0
7. Determine (by parts)
a)
x
2
cos xdx
b)
 x cos 2 xdx
c)
x
2
ln xdx
d)
x
e)
 x e dx
3 x
3
x  1dx
f)  ln x 3 dx
0
57
3
dx
8. Evaluate (partial fractions)
a)
3x 2
 ( x 1)( x2  x  1) dx
d)
 x (1  x )
dx
2
2
b)
2 x2  x  1
 ( x  1)( x2  1) dx
e)
 9x
2
c)
4 x 2  7 x  13
 ( x  2)( x2  1) dx
x 1
dx
 18 x  17
9. Find the area of the region bounded by
a)
f ( x)  x 2 , x-axis between x  2 and x  2
b)
y  sin( x) , x-axis between x   and x  
c)
y  x 3 , y-axis between y  0 and y  4
d)
y
x  2 , y-axis between y  2 and y  4
10. Find the area of the region bounded by the following curves:
2
, y  3 x
x
a)
y
b)
y  4  x 2 , y  x 2  2x
11. A and B are the points of intersections between the curves
y  4x  x 2
and
y
4
1.
x
Find the coordinates of A and B. Hence find the area bounded by two curves.
12. Evaluate the shaded area:
58
13. Find the area of the region bounded by the graphs y 2  8  x and y 2  x as shown
in the following figure.
14. Find the volumes when the region bounded by the given curves, the x-axis and the
given lines is rotated about the x-axis
a)
y
x  2, x  0
b)
y  x2 , x  5
15. Calculate the volumes when the region bounded by the given curves, the y-axis and
the given lines is rotated about the y-axis.
a)
y  x  2 , y  0, y  2
b)
y
x, y4
16. The region R is bounded by the curves y  x 2 ( x  0) , y  4 x 2 ( x  0) and the line
y  1.
a)
Sketch and shade the region R
b)
Calculate the area R
c)
Find the volume of the solid obtained when the region R is rotated about
y-axis.
59
ANSWERS
b) 12x  c
1. a) 2e x  c
3
27 4
x c
c)
4
e) 
d)
cos (6 x  1)
c
18
(5 x  1)5
c
25
3
(4  2 x) 2
f) 
c
3
g)
2e3 x  2
c
3
h) 
i)
3ln(2 x  3)
c
2
51 x
c
ln 5
j) 
tan(2  5 x)
c
25
2. a) I 
x 4 x3 x 2
 
xc
4
3
2
b) I  x 4  3x3  4 x 2  2 x  1
3. a) 2e2 x 3  c
c)
b) 2e13 x  c
5ln(3 x  3)
c
3
d)
3 tan (1  4 x)
c
4
4
52 x 1
e) 2e3 x 5  ln(3x  2) 
c
3
2ln 5
4. a) 36
b) 135.76
c) 96682
5. a)
d) 0
1 2
( x  x )3  c
3
1
c)  ( x 2  2 x  4)2  c
4
e) ecos x  c
b)
1 2
( x  1) 4  c
8
d)
1 x2  2
e
c
3
1
f)  ( x8  4 x  1) 2  c
8
1
g)  cos( x 2  6 x)  c
2
h)
1
x
i) e  c
j)
6. a) 78
2
1
ln y 2   c

4
3
2
 l ln x  2  c
3
b) 4.232
60
7. a) x 2 sin x  2 x cos x  2sin x  c
c)
x3
x3
ln x   c
3
9
2
4
x( x  1) 2  ( x  1) 2  c
3
15
b) ln  x  1 x 2  1   c


1
c) 3ln  x  2   ln  x 2  1  5 tan 1 x  c
2
d) 
1
 tan 1 x  c
x
1
ln  9 x 2  18 x  17   c
8
b) 4 unit 2
16
unit 2
3
c) 4.762 unit
d)
2
10. a) 0.114 unit 2
11. (a) A(1,3) and B(4,0);
12.
d)
f) 3ln 27  9
8. a) ln  x  1  x 2  x  1  c
9. a)
1
1
x sin 2 x  cos 2 x  c
2
4
3
e) x3 e x  3x 2 e x  6 xe x  6e x  c
e)
b)
8
unit 2
3
b) 9 unit 2
6.4548 unit 2
2
unit 2
3
13.
64
unit 2
3
14. a)
40
 unit 3
3
b) 625 unit 3
15. a)
56
 unit 3
3
b) 204.8 unit 3
16. a)
1
unit 2
3
b)
61
3
 unit 3
8
5