(chapter 8) Integration

QQM1023 Managerial Mathematics x =b

∫ f ( x) dx

x=a

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

∫ f ( x)dx = F ( x) + C , where C is constant. dy = f (x) dx

If

then

∫ dy = ∫ f ( x)dx y=

∫ f ( x)dx

The table below shows some example of functions, each paired with one of its anti-derivatives.

9.2

Function, f(x)

Anti-derivative, F(x)

1

x

3

3x + C (C is a constant)

4x3

x4

x3 + 2x

x4 + x2 4

INDEFINITE INTEGRATION & INTEGRATION RULES

∫ f ′(x )dx = F (x ) + C integral sign

integrand

constant of integration

read; indefinite integral of f’(x)

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9.2.1 RULES OF INTEGRATION #1 : CONSTANT RULES

∫ k dx = kx + C k is a constant

Example 1 a)

∫ 4 dx =

b)

∫ e dx =

9.2.2 RULES OF INTEGRATION #2 : POWER RULES

Alternatively, the process of integration of a function can be simplified by using formula.

n x ∫ dx =

1 x n +1 + c (n + 1) Where,

n ≠ −1

Example 2 a)

x2 ∫ x dx = 2 + C

Divide the term by new index

Add an arbitrary constant C

b)

∫ x dx = 2

c)

∫x

d)

∫x

1 2

dx =

−4

dx =

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9.2.3 RULES OF INTEGRATION #3 : CONSTANT MULTIPLE RULE

∫ k. f ( x) dx = k ∫ f ( x) dx ,with k ≠ 0

Example 3 a)

∫ 4x =

3

dx = 4 ∫ x 3 dx

b)

∫ ex

2

dx =

 x4  4  + C = x 4 + C 4

9.2.4 RULES OF INTEGRATION #4 : SUM AND DIFFERENCE

∫ [ f ( x) ± g ( x)] dx = ∫ f ( x)dx

±

∫ g ( x)dx

Example 4 a)

∫ (x

2

)

+ 7 x − 1 dx

Solution :

∫ (x

2

)

+ 7 x − 1 dx = ∫ x 2 dx + ∫ 7 xdx − ∫ 1dx  x2 x3 + 7  = 3  2

  − x + C 

x3 7 2 + x − x + C = 3 2 Chapter 9: Integration

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b)

∫ (3x

c)

 1 2 + x + 3 ∫  x 3  dx

d)

∫ (2 x − 5) dx

2

)

− 4 x 3 dx

9.2.5 RULES OF INTEGRATION #5 : INTEGRATION FOR

dy = f ′( x ) = (ax + b )n dx n ( ) ax + b dx = ∫

1 (ax + b )n +1 + C (a )(n + 1)

Example 5 a)

∫ (− 6 + 3x )

−2

dx

−2 ∫ (− 6 + 3x ) dx =

=

Chapter 9: Integration

1 (− 6 + 3x )−2+1 + C (3)(− 2 + 1)

(− 6 + 3x )−1 + C = − −3

1 +C 3(− 6 + 3 x )

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b)

6

∫ (4 x − 5)

c)

∫ (5 x + 2)

d)

∫ (3 − 6 x )

dx

1/ 3

−4

dx

dx

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

1 1 dx = ln x + c ∫ ax a Example 6

1

1

a)

∫ 4 x dx = 4 ln x + C

b)

2 ∫ x dx = ii)

1 1 dx = ln (ax + b ) + c ∫ (ax + b ) a

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Example 7

3

1

∫ 1 + 2 x dx = 3∫ 1 + 2 x dx a)

=

3 ln (1 + 2 x ) + C 2

b)

1 ∫ 2 x + 1 dx =

c)

4 ∫ 3x + 2 dx =

9.2.7 RULES OF INTEGRATION #7 : INTEGRATION OF EXPONENTIAL FUNCTIONS

i)

∫e

x

dx = e x + C

1 ax e + C ;a ≠ 0 a ax x iii ) ∫ a dx = + C ;a ≠ 0 ln a ii )

ax e ∫ dx =

Example 8

a)

∫ e dx = e x

x

+C

1 2x e +C 2 5x x c) ∫ 5 dx = +C ln 5

b)

2x ∫ e dx =

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d)

5x e ∫ dx

e)

∫e

f)

g)

(1 / 2 )x dx

1

∫ e 2 x dx

∫7

x

dx

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Rules of Integration ∫ k dx = kx + C a)

n ∫ x dx = b)

1 x n +1 + c (n + 1)

∫ k. f ( x) dx = k ∫ f ( x) dx c)

d)

∫ [ f ( x) ± g ( x)] dx = ∫ f ( x)dx 1

∫ (ax + b ) dx = (a )(n + 1) (ax + b ) n

e)

f)

g)

±

∫ g ( x)dx

n +1

+C

1 1 dx = ln x + c ∫ ax a 1 1 dx = ln (ax + b ) + c ∫ (ax + b ) a i)

∫e

ii )

∫

e ax dx =

iii )

∫

ax

h)

Chapter 9: Integration

x

dx = e x + C 1 ax e + C ; a; a≠ 0≠ 0 a ax dx = + C ;a ≠ 0 ln a 236

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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.

∫ f ( g ( x)).g ' ( x)

=

∫ f (u )du

=

F (u ) + C

=

F ( g ( x )) + C

Step 1: Substitute u = g(x), du = g’(x) to obtain the integral

∫ f (u )du.

Step 2: Integrate with respect to u. Step 3: Replace u by g(x) in the result.

Example 9 a) Solve

∫ (

)7

12 x 3 x 2 − 2 dx

Solution : Step 1: Substitute

(ax k + bx k −1 + ... + cx + d )n = (u) n

for n = 2, 3, 4, … -

u = ax k + bx k −1 + ... + cx + d

For the example above:

u = 3x2 – 2

Step 2: From u = ax k + bx k −1 + ... + cx + d , get

Hence, u = 3x2 – 2, then

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du . dx

du = 6x dx 237

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Step 3: Change

du du = ........ to = dx dx .....

du du = 6 x then = dx . 6x dx

and substitute into question.

(dx in term of du)

2 7 7 ∫12 x(3x − 2) dx = ∫12 x (u ) .

du 6x

∫

= 2u 7 du

 1  u 7 +1  + c = 2  (7 + 1)  1 1  = 2 u8  + c = u8 + c 4 8  =

(

)

8 1 2 3x − 2 + c # 4

EXERCISE Solve: a)

b)

)

(

∫

2 e3 x e3 x + 1 dx

∫

x x 2 −1 dx  

c)

∫

d)

∫

(ln x + 1)2 dx x x

(x + 1) 2

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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 b

∫ f ( x)dx , where a and b are known as the lower and upper limits a

of the integral respectively. Generally, if

∫ f ′( x ).dx = F ( x ) + c then b

∫

f ′( x ).dx = [F ( x ) + c ]ba

a

= F (b ) − F (a )

Example 10 Evaluate

6

∫ 2x

3

+ x 4 dx

−1

Step 1: Integrate the terms with respect to x

 2 x 4 x5  =  +  4 5 

6

−1

Step 2: Substitute x with 6 and –1

 2(6) 4 6 5   2(−1) 4 (−1) 5  + + − =   = 2202.9 5  4 5   4

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EXERCISE : Evaluate the following definite integrals:

∫ (x 2

a)

2

)

+ 1 .dx

1

1 b)

∫ (2 x − 1).dx

0

2

c)

∫ (x

2

2

+2

)

1

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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,

A1 is the area under a curve y = f ( x ) between x = a

If

x=b

and

then:

b

A1 = ∫ f ( x ).dx a

Example 11 : Find the area enclosed by: a)

y = x 2 , x = 0 , x = 2 and x-axis

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1

b)

y=

c)

f ( x ) = x 2 − 3 , x = 1 , x = 2 and x-axis

ii.

x

2

,

x = 1, x = 3 and x-axis

Consider graph below:

if

A2 is area under a curve x = f ( y ) between

y = c and

y = d then

d

A2 = ∫ f ( y ).dy c

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Example 12 : a)

Find the area enclosed by

y 2 = 4 x , lines y = 1, y = 4

and y-axis.

b)

Find the first quarter enclosed by

y=

2 , y-axis and lines x

y = 2, y = 4

9.5.2 AREA ENCLOSED BY TWO CURVES If function

f

and

g

is continuous and

f ( x ) > g ( x) within

[a, b], then area enclosed by y = f ( x ) and g ( x ) for a < x < b is:

b

A3 = Chapter 9: Integration

∫ [ f (x ) − g (x )]dx

a

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Example 13: a) Find the area enclose by the curves

y = x2 −1

and

y2 = 2x

and

y = 8 x − 16

b) Find the area enclosed by the curves

x2 = 2 y

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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?

integrate

Marginal cost function Marginal Average cost function Marginal revenue function Marginal Average revenue

Total cost function

integrate

integrate integrate

integrate integrate Marginal Average profit function

Average Cost function Total Revenue function

Average revenue function

Marginal profit function

Profit function Average profit function

Example 14 : Suppose the marginal cost to produce product A is:

TC ′ = x 2 − 20 x + 100 the fixed cost for the production is RM 9000 where x is the quantity. Find the total cost function.

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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

y = D( x ) and the supply is y = S ( x ), then

the intersection of two points is called equilibrium points,

( x0 , y 0 ) .

Demand / Supply

Supply = S(x)

Consumers’ Surplus Equilibrium

y0

Producers’ Surplus

Demand = D(x) x0 ƒ

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).

C.S =

x0

∫ D( x ).dx − y0 x0

0

Chapter 9: Integration

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

x0

P.S = x0 y0 − ∫ S ( x ).dx 0

Example 15 : Demand function for Embun Boutique is while the supply function is

y = −x2 − 4x + 6

y = x 2 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

( x0 , y 0 ) .

− x 2 − 4x + 6 = x 2 − 2x 2 − 4x + 6 = 0 (−2 x + 2)( x + 3) = 0 therefore, x = 1 or x = −3 → e lim inated y = 12 =1 ∴ Equilibrium po int = (1,1)

NOTE

:The equilibrium point is only considered on the first quarter of the plane.

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Step 2:

C.S =

x0

∫ D( x ).dx − y0 x0 = ..................

0 1

= ∫ − x 2 − 4 x + 6 dx − (1)(1) 0

1

  x − 2 x 2 + 6 x   3 3

= −

−1 0

 1  = − − 2 + 6 − [0] − 1  3  2 = RM 2 3

x0

P.S = x0 y0 − ∫ S ( x ).dx = ................. 0

1

= (1)(1) − ∫ x 2 dx 0

1

3 = 1−  x   3  = 1−

0

1 3

= RM

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2 3

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EXERCISE: 1.

The demand function for a product is

p = f (q ) = 100 − 0.05q, where p is the price per unit (in RM) for q units.

The supply function is

p = g ( q ) = 10 + 0.1q.

Determine consumers’ surplus and producers’ surplus under market equilibrium. Answer: CS = RM 9000

and PS = RM 18,000

2. Given the function;

f ( q ) = 100 − 0.05q

and

g ( q ) = 10 + 0.1q

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)

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c) Determine the consumers’ surplus as well as the producers’ surplus.

Answer : CS = RM9000 PS = RM18000

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