BQT 133-Business Mathematics
Teaching Module
CHAPTER 5 : INTEGRAL CALCULUS
5.1
INTRODUCTION
Integration is the reverse process of differentiation. For that reason, it is also
known as anti-derivative which means that if given derivative of a function
we can work backwards to find the function from which it is derived.
For example: If
d
F ( x)  f ( x) , then F(x) is the anti-derivative of f(x) and the
dx
process of finding F(x) is called integration. It is written as
 f ( x)dx  F ( x)  c , where the symbol  f ( x)dx is the integral of f(x), f(x) is
called the integrand and c is the constant of integration.
Definition 5.1 : Anti Derivatives
Given a function f(x) an antiderivative of f(x) is any function F(x) such that
F ( x)  f ( x)
Beside F(x), c is also part of the product of the integration because of the
explanation given below:
d 3
( x  1)  3 x 2
dx
d
(ii) ( x 3  1)  3x 2
dx
(i)
From (i) and (ii) we can conclude that the functions F ( x)  x 3  1 and
F ( x)  x 3  1 are both anti-derivatives of f ( x)  3x 2 . In that case, we can also
conclude that the function F ( x)  x 3 plus any constant (positive or negative),
can also be the anti-derivatives of the function f ( x)  3x 2 . Therefore, if we
replace this constant with the letter c, then, we can say that F ( x)  x 3  c is
the anti-derivative of f ( x)  3x 2 . Hence, the anti-derivative is written as
2
3
 3x dx  x  c and this integral is known as indefinite integral because of
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the indefinite value of c. The value of c could be determined if the
corresponding values of x and y are given.
Definition 5.2 : Indefinite Integral
If F(x) is any anti-derivative of f(x) then the most general anti-derivative of
f(x) is called an indefinite integral and denoted,
 f ( x)dx  F ( x)  c , c is any constant
In this definition  is called the integral symbol, f(x) is called the
integrand, x is
called the integration variable and the “c” is called the constant of
integration.
Derivative Formula
Equivalent Integral Formula
 
 2xdx  x  c
 e dx  e  c
d 2
x  2x
dx
d x
e  ex
dx
2
 
x
x
The result of this discussion and information on differentiation (see Chapter
2) is summarized in Table 3.1 which shows the relationship between
derivative and indefinite integral formulas for some elementary functions.
Table 5.1: Derivative and Integral Formulas of Elementary Functions
Like differentiation, integration has many practical applications. The
obvious are, for examples, in kinematics, estimating populations growth,
measuring the arc length, surface area, volume of solid and center of
gravity.
5.2
STANDARD INTEGRAL
f (x)
xn
1
a
ex
 f ( x)  dx
x n 1
 C , n  1
n 1
xC
ax  C
e x +C
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ax
ax
C
ln a
1
x
ln x  C
From the above table , we can use this basic function to solve the integral of
polynomial function, integral of exponential functions, integral of
logarithmic functions and integral of trigonometric functions.
5.3
INDEFINITE INTEGRALS
In this section we shall prove two basic properties of indefinite integrals.
The basic properties of indefinite integrals are stated in the following
theorem.
Theorem 5.1: Basic Properties of Indefinite Integrals
1.
The constant factor k can be taken out from an integral, that is
 kdx  k  f ( x)dx ,
2.
where k is a constant
The integral of a sum or differences equals the sum or difference of
the integrals, that is   f ( x)  g ( x) dx   f ( x)dx   g ( x)dx
5.3.1 Integral of Polynomial Functions
Properties of the Integral of Polynomial Functions
x n 1
c
n 1
1.
n
 x dx 
2.
 kdx  kx  c ,
(General form)
where k is any number
Example 5.1
Determine the following integrals:
(a)  x 7 dx
(b)  2 x 8 dx
(c)
1
x
6
dx
(d)

3x dx
Solutions:
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(a)  x 7 dx 
BQT 133
x 7 1
x8
c 
c
7 1
8
2 x 81
 2 x 7
c 
c
 8 1
7
1
x 61
 x 5
(c)  6 dx   x 6 dx 
c 
c
 6 1
5
x
(b)  2 x 8 dx  2 x 8 dx 
(d)

3x dx  3  x dx  3  x
12
1
1
2
3
2
3
x
x
2 3 2
 3
c  3
c 
x c
1
3
3
1
2
2
Exercise 5.1:
(a)

4
(b)
x dx

1
6
x2
1
(c)  3 2 dx
x
dx

2
3
(d)   3x dx
Example 5.2
Determine the following integrals:
(a)  (2 x  5 x 3  4 x 2 )dx
(b)  (3x  5) 2 dx
2x  1
(d)   4 dx
2
2
(c)   3x   dx
x

 x

Solutions:
(a)  (2 x  5 x 3  4 x 2 )dx



 2 xdx  5 x dx  4 x dx
 2 xdx  5 x 3 dx  4 x 2 dx
3
2
 x2
 2
 2
  x4   x3 
  5   4   c
  4   3 
5
4
 x2  x4  x3  c
4
3
(b)  (3x  5) 2 dx
  (9 x 2  30 x  25)dx
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  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

 3x   dx
x

4 

   9 x 2  12  2 dx
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
3

1
 


4
 3 x 3  12 x 
x
(d)
 2x  1 
dx
x4 
2x
1
  4 dx   4 dx
x
x
 
 2 x 3 dx   x  4 dx
 x  2  x 3
 
 2
 2  3
1
1
 2  3
x
3x
Exercise 5.2:
(a)

(4 x  3) 2
x
dx
(b)

(3x  2) 3
3
x2
dx
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5.3.2 Integral of Exponential Functions
Formula of the Integral of Exponential Functions
1.
e
2.

3.
e
x
dx  e x  c
e axb dx 
u
(Integral of Natural Exponential Function)
e axb
 c ; where a any number
a
du  e u  c
 g ( x)(e
or
g ( x)
)dx  e g ( x )  c
(by using Substitution Rule)
Example 5.3
Solve the following integrals:
(a)  e 2 x dx
(b)  e 2 x 3 dx
(d)
 e
2x

2
 e  2 x dx
(e)
 e
2x

(c)
 e
(f)
e
2
 3 dx
5x
x

 e 3 x dx
 e 3 x dx
Solutions:
(a)  e 2 x dx 
e2x
c
2
(b)  e 2 x 3 dx 
(c)
 e
(d)
 e
5x
2x
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
2
 e  2 x dx


  e 4 x  2  e 4 x dx
  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
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(e)
 e
2x
BQT 133

2
 3 dx 

 e
e
4x
4x
dx 
4x

 6e 2 x  9 dx
 6e
2x

dx  9dx
2x
e
6e

 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 5.3
(a)

e5x
dx
e2x
(b)

e 4 x  e8x
dx
e2x
(c)

e 2x  e5x
dx
e9x
5.3.3 Integral of Logarithmic Functions
Formula of Integral of Logarithmic Functions
1.
2.
3.
 xdx  ln x  c
a
 ax  bdx  ln ax  b  c
1
1
 u du  ln u  c
or
g ( x)
 g ( x) dx  ln( g ( x))  c
(by using Substitution Rule)
4.
x
 (a )dx 
ax
c
ln a
Example 5.4
Find:
(a)
1
 x  3dx
(b)  3 x dx
(c)
7
 5  xdx
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Solutions:
(a)
1
 x  3dx  ln x  3  c
3x
(b)  3 dx 
c
ln 3
x
(c)
7
1
1
 5  xdx  7 5  x dx  7 ln 5  x   1  c  7 ln 5  x  c
Exercise 5.4
(a)
3
 2 x  3dx
(b)

(c)  34 x2 dx
5
dx
x
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 5.2
Table 5.2: Basic Derivatives and Integral Formulae
Corresponding Derivative Formula
Indefinite Integral
d
 `1dx  x  C
[ x]  1
dx
d  x n 1 
n

  x , n  1
dx  n  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
[e ]  e x
dx
d kx
[e ]  kekx
dx
d x
[a ]  a x ln a
dx
1
x
dx 
e kx
C
k
a x dx 
ax
C
ln a
e

x
kx
5.4 DEFINITE INTEGRALS
We have learnt about the indefinite integral of f which is denoted by
 f ( x)dx . In this section we will discuss about the definite integral of f for
continuous function on closed interval a, b, denoted by

b
a
f ( x) dx . The
definition of the definite integral is summarized in a theorem called
Fundamental Theorem of Calculus.
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Definition 5.2 : Definite Integral
A definite integral is an integral

b
a
f ( x) dx
With upper and lower limits it mean f(x) continuous on closed interval [a,b]
and F(x) is the anti-derivative (indefinite integral)of f(x) on [a,b] then
b
b
(Fundamental Theorems of
a f ( x) dx  F ( x)a  F (b)  F (a)
Calculus)
From this definition, it is clear that the function f must be continuous on the
interval a  x  b so that it is differentiable on the interval. 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
 f ( x) dx  F ( x)
b
b
a
a
 F (b)  c  F (a)  c
 F (b)  F (a)
Theorem 5.2: Basic Properties of Definite Integrals
If f(x) and g(x) are continuous functions on the interval [a, b], then


a
(c)

b
(d)
(e)
 k dx  k (b  a)
 f ( x) dx   f ( x) dx  
(f)
  f ( x)  g ( x) dx  
(a)
(b)
a
b
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
a
b
c
b
a
a
c
b
b
a
a
f ( x) dx , where a  c  b
b
f ( x) dx   g ( x) dx
a
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BQT 133
Example 5.7
Solve the following integrals:
2x  1
(b) 1  4 dx
 x 
2
(a) 1  3 x  dx
x

25
(d) 1 dx
x
2
3
(c)
1
2
0

e 4 x  e8 x
dx
e2x
Solutions:
(a)

2
1
2
2

 3 x   dx
x

2
4 
   9 x 2  12  2 dx
1
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)
 2x  1 
dx
1
x4 
3 2x
3 1
  4 dx   4 dx
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
3 x  1  9 81  
 x
(c)
1
2
0

e 4 x  e8 x
dx
e2x

1
2
 e
0
2x

 e 6 x dx
10
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BQT 133
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
5.5 TECHNIQUES OF INTEGRATION
In Section 5.3 and 5.4 we learned how to integrate elementary functions
using integral formulas. Many integration problem 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 is
integration by substitution
We shall discuss each of these methods extensively by solving examples.
5.5.1 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
Step 3:
Substitute
du
 g ' ( x)
dx
11
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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:
of x.
Substitute u with g(x), so that the final answer will be in terms
Some integral forms that can be evaluated using integration by substitution,
are
(i)
 f (ax  b)
(ii)

(iii)
 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
 5 , therefore
can be found from substitution u  (5 x  4) for
du
dx
dx 1

and the integral becomes:
du 5
12
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
BQT 133

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:

(5 x  4) 7
(5 x  4) 7
(5 x  4) dx 
c 
c
57
35
6
Example 5.8
Evaluate the following integrals by using the substitution u  4x  1.
(a)  (4 x  1) dx
(b)  (4 x  1) 2 dx
Solutions:
du
1
 4  du  dx . Substituting u and du give us
dx
4
1
 (4 x  1) dx  4  u du
1
 u2  c
8
1
 (4 x  1) 2  c
8
(a) If u  4x  1, then
Remarks
By integrating directly will get
 (4 x  1) dx 
4x 2
 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
 2x 2  x   c
8
 2x 2  x  C,
(b) If u  4x  1, then
 (4 x  1)
where C 
1
c
8
du
1
 4  du  dx . Substituting u and du give us
dx
4
2
1 2
u du
4
1
 u3  c
12
1
 (4 x  1) 3  c
12
dx 
Remarks
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© Universiti Malaysia Perlis 2009
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BQT 133
By integrating directly will get
2
 (4 x  1) dx   (16 x  8x  1) dx 
16 x 3
 4x 2  x  C
3
Which is equivalent to the previous because
1
1
(4 x  1) 3  c  (64 x 3  48 x 2  12 x  1)  c
12
12
16 3
1
x  4x 2  x   c
3
12
16
 x 3  4x 2  x  C,
3

where C 
1
c
12
Example 5.9
Evaluate the following integrals.
c
 ax  b dx
1
(e) 
dx
4x  3
(b)  e 4 x dx
(a)
5
(c)  (10 x  9) 2 dx
(d)  35 x dx
Solutions:
du
1
 a  du  dx
dx
a
c
1 1
 ax  b dx  c  u  a du
c 1
  du
a u
c
c
 ln u  c  ln ax  b  c
a
a
(a) If u  ax  b , then
(b) If u  4 x , then
e
4x
du
1
 4  du  dx
dx
4
1
dx   e u  du
4
1 u

e du
4
1
 eu  c
4
1 4x
 e c
4

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© Universiti Malaysia Perlis 2009
81898397
BQT 133
du
1
 10  du  dx
dx
10
5
5
1
 (10 x  9) 2 dx   u 2  10 du
5
1

u 2 du
10
(c) If u  10x  9 , then

5
1
2
1 u

c
10 5
1
2
7
1
 u2 c
35

7
1
 (10 x  9) 2  c
35
(d) 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

(e) 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 5.9:
(a)  e 3 x 5 dx
(b)  sec 2 (3x  1) dx
(c)  cos(1  4 x) dx
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© Universiti Malaysia Perlis 2009
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BQT 133
(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
 3x  5
integrals, 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
 2x  3
dx
 du  (2 x  3)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
x
2
2x  3
dx  ln( x 2  3x  5)  c
 3x  5
Example 5.10
Evaluate the following indefinite integrals by using the substitution method.
2x  3
(a)  2
dx
x  3x  2
x
(b)  2
dx
x 4
12 x 2  16
(c)  3
dx
x  4x
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
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© Universiti Malaysia Perlis 2009
81898397
x
2
BQT 133
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
du
 3x 2  4  du  (3x 2  4)dx
dx
12 x 2  16
4(3x 2  4)
dx

 x 3  4x
 x 3  4x dx
1
 4  du
u
 4 ln u  c
(c) If u  x 3  4 x , then
 4 ln( x 3  4 x)  c
Exercise 5.10:
x
(a)
(iii)
2
x3
dx
 6x  2
Integral of the form
 f ( x) f ' ( x) dx
Example 5.11
Evaluate the following integrals.
(b)  e x (e x  1) 3 dx (c)
(a)  4 x(2 x 2  3) 6 dx
(d)

ln x
dx
x
(e)  2 x( x  2) 5 dx
 x ln x
dx
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
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© Universiti Malaysia Perlis 2009
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BQT 133

du
 e x  du  e x dx
dx
(b) If u  e x  1 , then
 e (e
x
x
u7
1
 c  (2 x 2  3)7  c
7
7

 1) 3 dx  u 3 du
u4
c
4
1
 (e x  1) 4  c
4

du 1
1
  du 
dx
dx x
x
ln x
1
dx  ln x  dx
x
x
(c) If u  ln x , then


  u du
u2
c
2
1
 (ln x) 2  c
2

du 1
1
  du 
dx
dx x
x
dx
1 1

 dx
x ln x
ln x x
1

 du
u
 ln u  c
 ln(ln x)  c
(d) If u  ln x , then



(e) 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
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© Universiti Malaysia Perlis 2009
81898397
BQT 133
2u 7 4u 6


c
7
6
2u 7 2u 6


c
7
3
2u 6
2( x  2) 6
3( x  2)  7  c

(3u  7)  c 
21
21
Exercise 5.11:
Evaluate the following integrals.
2
(a)  xe x dx
(b)  x 3x  2 dx
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© Universiti Malaysia Perlis 2009
81898397
5.6
BQT 133
APPLICATION OF INTEGRATION
5.6.1 Consumer’s Surplus and Supplier’s Surplus
Suppose that there are eight people in a store wanting to buy a can of
chili and that the price of the chili is not marked. Each person pays as much
as he or she is willing to pay. One person pays RM 6. Another pays RM5.50.
The third pays RM5, the forth RM4.50, the fifth RM4, the sixth RM3, the
seventh RM2 and the eighth RM1. The store collects
6+5.50+5+4.50+4+3+2+1 = RM31. If the store had marked a price of RM1,
then all eight would have paid RM1 and the store would have only collected
RM8, a reduction of RM23. In other words, sales worth RM31 to the
consumers would have been purchased for RM8. The difference is called the
consumer surplus. It is the difference between what consumers are willing to
pay for a product or service and what they actually do pay. The consumer
surplus represents the total savings to consumers who are willing to pay
more than price for the product. The supplier’s surplus measures the
difference between the amount of money a supplier is willing to accept at a
given price for a product and the amount the supplier actually does receive.
Definition 5.3
The consumer’s surplus at a price level of p is
q
Consumer surplus (CS)   D  x  dx  pq
0
Where D  x  is demand function and pq is the revenue.
Definition 5.4
The supplier’s surplus at a price level of p is
q
Supplier surplus (SS)  pq   S  x  dx
0
Where S  x  is supply function and pq is the revenue.
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© Universiti Malaysia Perlis 2009
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BQT 133
Example 5.12
Find the consumer’s surplus for the given function and quantity
(a) D( x)  0.01x 2  x  20, q  15 .Answer:RM135
20 x  40
(b) D( x)  2
, q  12 .Answer:RM19.70
x  4x  5
Example 5.13
Find the supplier’s surplus for S ( x)  0.05x 2  x  10, q  20 . Answer
RM466.67.
Example 5.14
Find the equilibrium point, consumer surplus and supplier surplus for the
given supply and demand function.
D  x    x  107.5 and S  x   0.02 x 2  2 x  20
Answers
Equilibrium point=25
CS=RM312.50
SS=RM833.33
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© Universiti Malaysia Perlis 2009