chapter9-integration-111012121054-phpapp01.pdf

QQM1023 Managerial Mathematics

Chapter 9: Integration 229

9.1 DEFINITION OF INTEGRATION

9.1.1 INTEGRATION IS ANTI DERIVATIVE

In Chapter 7 and 8 we have learned the techniques how to get the

differentiation from a certain function. A function F(x) an anti-derivative

of a function f(x) if F(x) = f(x). The anti-derivate of f is

+= ,)()( CxFdxxf where C is constant. If )(xf

dxdy = then = dxxfdy )(

= dxxfy )( The table below shows some example of functions, each paired with

one of its anti-derivatives.

Function, f(x) Anti-derivative, F(x)

1 x

3 3x + C (C is a constant)

4x3 x4

xx 23 + 244

xx +

9.2 INDEFINITE INTEGRATION & INTEGRATION RULES

( ) ( ) CxFdxxf +=

integral sign integrand constant of integration

read; indefinite integral of f(x)

dxxfbx

ax==

)(



( ) ++= + cxndxx nn 111

k dx = kx + C

9.2.1 RULES OF INTEGRATION #1 : CONSTANT RULES

k is a constant

a) =dx4 b) =dxe

9.2.2 RULES OF INTEGRATION #2 : POWER RULES

Alternatively, the process of integration of a function can be simplified

by using formula.

Where, 1n

a) += Cxdxx 22

b) =dxx2 c) =dxx 21 d) dxx 4 =

Divide the term by new index

Add an arbitrary constant C

Example 1

Example 2



[ ] = dxxgdxxfdxxgxf )()()()(

= dxxfkdxxfk )()(. 9.2.3 RULES OF INTEGRATION #3 : CONSTANT MULTIPLE RULE

,with k 0

a) = dxxdxx 33 44 b) = dxex 2 = CxC

x +=+

444

4

9.2.4 RULES OF INTEGRATION #4 : SUM AND DIFFERENCE

a) ( ) + dxxx 172 Solution :

( ) +=+ dxxdxdxxdxxx 1717 22 = Cx

xx +

+

27

3

23

= Cxxx ++ 2

3

27

3

Example 3

Example 4



( ) ( )( ) ( ) Cbaxnadxbax nn +++=+ + 111

b) ( )dxxx 32 43

c) ++ dxxx 31 2

3

d) ( ) dxx 52

9.2.5 RULES OF INTEGRATION #5 : INTEGRATION FOR

( ) ( )nbaxxfdxdy +==

a) ( ) dxx + 236 ( ) ( )( ) ( ) Cxdxx +++=+ + 122 36123 136

= ( )

( ) CxCx ++=+

+ 363

1336 1

Example 5



cxa

dxax

+= ln11

( ) ( ) ++=+ cbaxadxbax ln11

b) ( ) dxx 654

c) ( ) dxx + 3/125

d) ( ) dxx 463

9.2.6 RULES OF INTEGRATION #6 : INTEGRAL OF FUNCTION THAT WILL PRODUCE ln: i)

a) Cxdxx+= ln4141

b) = dxx2

ii)

Example 6



+=

+=+=

0;ln

)

0;1)

)

aCa

adxaiii

aCea

dxeii

Cedxei

xx

axax

xx

a) ( ) Cxdx

xdx

x

++=+=+

21ln23

2113

213

b) + dxx 121

=

c) + dxx 23 4 =

9.2.7 RULES OF INTEGRATION #7 : INTEGRATION OF EXPONENTIAL FUNCTIONS

Cdxc

Cedxeb

Cedxea

xx

xx

xx

+=

+=+=

5ln55)

21)

)

22

Example 7

Example 8



d) dxe x5

e) ( ) dxe x2/1

f) dxe x21

g) dxx7



a)

k dx = kx + C

b)

( ) ++= + cxndxx nn 111

c)

= dxxfkdxxfk )()(.

d)

[ ] = dxxgdxxfdxxgxf )()()()(

e)

( ) ( )( ) ( ) Cbaxnadxbax nn +++=+ + 111

f)

cxa

dxax

+= ln11

g)

( ) ( ) ++=+ cbaxadxbax ln11

h)

+=

+=+=

0;ln

)

0;1)

)

aCa

adxaiii

aCea

dxeii

Cedxei

xx

axax

xx

Rules of Integration

; a 0



9.3 INTEGRATION BY SUBSTITUTION

In this section, we are going to discuss one technique called

integration by substitution that usually uses to transform

complicated integration into more easy form.

)('.))(( xgxgf = duuf )( = CuF +)( = CxgF +))((

a) Solve ( ) dxxx 72 2312 Solution :

Step 1:

Substitute ( )nkk dcxbxax ++++ ...1 = nu)( for n = 2, 3, 4, - dcxbxaxu kk ++++= ...1 For the example above: u = 3x2 2

Step 2:

From dcxbxaxu kk ++++= ...1 , get dxdu

.

Hence, u = 3x2 2, then xdxdu 6=

Step 1: Substitute u = g(x), du = g(x) to obtain the integral .)( duuf Step 2: Integrate with respect to u. Step 3: Replace u by g(x) in the result.

Example 9



Step 3:

Change ........=dxdu

to dxdu =.....

and substitute into question.

xdxdu 6= then dx

xdu =6

. (dx in term of du)

= dxxx 72 )23(12 xduux6

.)(12 7 = duu72 ( ) cu +

+=+17

1712

cucu +=+

= 8841

812

( ) cx += 82 2341 #

Solve:

a) ( ) dxee xx + 233 1 b) dxxx 12 c)

( ) dxx

x + 21ln d) ( ) + 22 1x x

EXERCISE



( ) ( )[ ]bab

acxFdxxf += .

( ) ( )aFbF =

9.4 DEFINITE INTEGRAL

Definite integral can be easily recognized by numbers assigned

to the upper and lower parts of the integral sign, such as

ba

dxxf )( , where a and b are known as the lower and upper limits

of the integral respectively.

Generally, if ( ) ( ) cxFdxxf += . then

Evaluate dxxx

+6

1

432

Step 1: Integrate the terms with respect to x

=

+

542 54

6

1

xx

Step 2: Substitute x with 6 and 1

= 9.22025)1(

4)1(2

56

4)6(2 5454 =

+

+

Example 10



Evaluate the following definite integrals:

a) ( ) +21

2 .1 dxx

b) ( )dxx .1210

c) ( )221

2 2 +x

EXERCISE :



1A = ( )b

adxxf .

9.5 AREA UNDER A CURVE AND BETWEEN TWO CURVES

9.5.1 AREA UNDER A CURVE

One of definite integral applications is area under a curve.

i. Consider this graph,

If 1A is the area under a curve ( )xfy = between ax = and bx = then:

Find the area enclosed by:

a) 2xy = , 0=x , 2=x and x-axis

Example 11 :



2A = ( )d

cdyyf .

b) 21x

y = , 3,1 == xx and x-axis

c) ( ) 32 = xxf , 1=x , 2=x and x-axis

ii. Consider graph below:

if 2A is area under a curve ( )yfx = between cy = and dy = then



A3 = ( ) ( )[ ]dxxgxfb

a

a) Find the area enclosed by xy 42 = , lines 4,1 == yy and y-axis.

b) Find the first quarter enclosed by x

y 2= , y-axis and lines 4,2 == yy

9.5.2 AREA ENCLOSED BY TWO CURVES

If function f and g is continuous and ( ) )(xgxf > within [ ]ba, , then area enclosed by ( )xfy = and ( )xg for

bxa



a) Find the area enclose by the curves 12 = xy and 168 = xy

b) Find the area enclosed by the curves xy 22 = and yx 22 =

Example 13:



integrate

integrate

integrate integrate

integrate integrate

9.6 APLICATION IN ECONOMIC AND BUSINESS

9.6.1 APLICATION IN MARGINAL COST, MARGINAL REVENUES ETC.

Normally in order to get the marginal cost, we will differentiate the cost

function, but how to get the cost function if we are given the marginal

cost?

Marginal cost function Total cost function

Marginal Average cost function Average Cost function

Marginal revenue function Total Revenue function

Marginal Average revenue Average revenue function

Marginal profit function Profit function

Marginal Average profit function Average profit function

Suppose the marginal cost to produce product A is:

100202 += xxCT the fixed cost for the production is RM 9000 where x is the quantity. Find

the total cost function.

Example 14 :



( ) 000

0.. xydxxDSC

x=

9.6.2 CONSUMERS AND PRODUCERS SURPLUS

In this topic, we use the definite integral to solve the problem in finding

the consumers surplus and producers surplus.

If the demand function is ( )xDy = and the supply is ( )xSy = , then the intersection of two points is called equilibrium points, ( )00 , yx .

The amount that can be saved by the consumer when the market price is lower than the price demanded is called as

CONSUMERS SURPLUS (C.S).

Demand / Supply Supply = S(x) Consumers Surplus y0 Equilibrium Producers Surplus Demand = D(x) x0



( )dxxSyxSPx

= 00

00 ..

Amount that is gained by the producer when the market price is higher is called as PRODUCERS SURPLUS (P.S).

Demand function for Embun Boutique is 642 += xxy while the supply function is 2xy = where x is the quantity and y is the price. Find the consumers surplus (C.S) and producers

surplus (P.S).

Solutions:

Step 1:

Get the equilibrium point ( )00, yx .

)1,1(int11

lim31,

0)3)(22(0642

64

2

2

22

===

==

=++=+=+

pomEquilibriu

yinatedexorx

thereforexx

xxxxx

NOTE :The equilibrium point is only considered on the first

quarter of the plane.

Example 15 :



Step 2:

( ) .................... 000

0== xydxxDSC

x

[ ]

322

106231

123

)1)(1(64

623

1

0

1

0

2

RM

dxxx

xxx

=

+=

=

+=

+

( ) ................... 00

00 == dxxSyxSPx

32

311

33

1

)1)(1(

1

0

1

0

2

RM

x

dxx

=

=

=

=



1. The demand function for a product is

,05.0100)( qqfp == where p is the price per unit (in RM) for q units.

The supply function is .1.010)( qqgp +==

Determine consumers surplus and producers surplus under market

equilibrium.

Answer: CS = RM 9000 and PS = RM 18,000

2. Given the function;

qqf 05.0100)( = and qqg 1.010)( += where p is the price per unit (in RM) for q units of product .

a) Determine which of the function is a;

i) Demand function

ii) Supply function

b) Find the market equilibrium.

Answer : (600,70)

EXERCISE:



c) Determine the consumers surplus as well as the producers

surplus.

Answer : CS = RM9000

PS = RM18000

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