BA201 Engineering Mathematic UNIT 4 Integration

TOPIC 4

INTEGRATION 1

PREPARED BY SITI NORSHAFINAZ BINTI MD NOH

4.1 DEFINITION OF INTEGRATION

1) The process of obtaining

from y ( a function of x ) is known as differentiation

and the reverse process of obtaining y from

is called integration.

Differentiation process

Integration process

2) Integration of y respect to x , is denoted by ∫ ( ) .

3) Integration is divided into two parts which are indefinite integral and definite

integral and first we will approach to indefinite integral as the basic of integration.

4.2 INDEFINATE INTEGRALS

1) We shall use the following notation for the family of integration.

∫ ( )

2) If integration of y with respect to x, is denoted by ∫ ( )

𝑦

𝑦

TOPIC 4 : INTEGRATION

TOPIC 4

INTEGRATION 2


3) If 𝑦 ( ), then

( ) and ∫ ( ) 𝑦 is known as integration of y

respect to x where c is an arbitrary constant.

Formulas in integration :

i) a dx ax c ; where a is a constant.

ii) 1

1

nn ax

ax dx cn

; where n is an integer,

iii) ( ) ( ) ( ) ( )p x q x dx p x dx q x dx

Tips!

Note that a function is integrate respect to the function given and it is not compulsory to be

integrated respect to x only. (see example given)

Example 4.1

Find :

a) 3

5

x dx b) 3 r dr

c) 6 dk d) xdx

Solution:

a) 23 3

5 5 2

xx dx c

23

10x c

In indefinite integral, c must be stated. It is

known that as an arbitrary constant where

the value is yet to be determined.

TOPIC 4

INTEGRATION 3


Can only integrate terms by terms in

addition or subtraction. If the function

given is in multiplication, the

function must be expanded.

Tips! Operation

b) 4

3 4

rr dr c

c) 6 6dk k c

d) 1

2xdx x dx

32

32

32

2

3

xc

x c

4.3 Determining the integration of algebraic expressions involving operations.

Example 4.2

Find :

a) 2

6x dx b) 3 2 2 5x x dx

c)

4

4 5 2x xdx

x

d)

3

32

5x dx

x

Solution :

a) 2

26 12 36x dx x x dx

3 2

32

12 363 2

6 363

x xx c

xx x c

The given function is integrate respect to k,

therefore the integration will be in terms of k.

Make sure that x is always in index form.

TOPIC 4

INTEGRATION 4


b) 23 2 2 5 6 15 4 10x x dx x x x dx

2

3 2

3 2

6 11 10

6 11 103 2

112 10

2

x x dx

x xx c

x x x c

c) 2

4 4

4 5 2 20 8 5 2x x x x xdx dx

x x

2

4 4 4

4 3 2

3 2 1

3 2

20 3 2

20 3 2

20 3 23 2 1

20 3 2

3 2

x xdx

x x x

x x x dx

x x xc

cx x x

d) 3

32

5x dx

x

. (Let’s try this question in class)

Can only integrate terms by

terms in addition or

subtraction. If the function

given is in division, the

function must be simplified.

Tips! Operation

TOPIC 4

INTEGRATION 5


4.4 Definite Integrals of Algebraic Expressions

1. If x=a to x=b for ( )f x dx , then it is written as ( )

b

a

f x dx and it is called definite

integral.

( ) ( ) ( )

b

a

f x dx F b F a

2. For the definite integral ( )

b

a

f x dx , the constant a is known as the lower limit of

integration while b is known as the upper limit of the integration.

Properties Of Definite Integrals.

a) ( ) ( )

b b

a a

kf x dx k f x dx

b) ( ) ( ) ( ) ( )

b b b

a a a

f x g x dx f x dx g x dx

c) ( ) ( )

b a

a b

f x dx f x dx

d) ( ) ( ) ( )

b c c

a b a

f x dx f x dx f x dx

TOPIC 4

INTEGRATION 6


Evaluate definite integrals using the properties of definite integrals. Example 4.3 1) Evaluate each of the following definite integrals below.

a) 3

1

(3 5)x dx b) 0

1

( 1) 3x x dx

c) 2 2 3

1

4 5x x xdx

x

Solution

a)

33 2

1 1

3(3 5) 5

2

xx dx x

2 2

333 3 3 1

5(3) 5 12 2

= 22

b) 0 0

2

1 1

( 1) 3 3 3x x dx x x x dx

= 3

5

0

2

1

03 2

1

03

2

1

3

2

2 3

2 33 2

33

10 1 3 1

3

x x dx

x xx

xx x

TOPIC 4

INTEGRATION 7


c) 2 22 3

2

1 1

4 54 5

x x xdx x x dx

x

22 3

1

2 3 2 3

4 52 3

2 2 1 14(2) 5 4(1) 5

2 3 2 3

x xx

= 6

85

2) Given that 4

2

( ) 6g x dx and 8

4

( ) 10g x dx , find the values of

a) 4 4

2 8

( ) ( )g x dx g x dx b) 4

2

( ) 2g x x dx

c) 2 8

4 4

5 ( ) ( )g x dx g x dx d) k if 8

4

( ) 9kx g x dx

Solution :

a) 4 4

2 8

( ) ( ) 6 ( 10)g x dx g x dx

= 16

b) 4 4 4

2 2 2

( ) 2 ( ) 2g x x dx g x dx x dx

42

2

22

26

2

6 (4) 2

6 (16 4)

6 12

6

x

TOPIC 4

INTEGRATION 8


c) 2 8 4 8

4 4 2 4

5 ( ) ( ) 5 ( ) ( )g x dx g x dx g x dx g x dx

( 5)(6) 10

30 10

40

d) 8

4

( ) 9kx g x dx

8 8

4 4

( ) 9kx dx g x dx

82

4

10 92

kx

2 2

8 410 9

2 2

k k

32 8 1

24 1

1

24

k k

k

k

TOPIC 4

INTEGRATION 9


4.5 Integration for other functions

4.5.1 Integration of trigonometric functions

There are an overwhelming number of combinations of trigonometric functions which appear

in integrals. This section examines integration of trigonometric functions.

Formula integration of trigonometric functions

a) cos( )

sin( )

( )

ax bax b dx c

dax b

dx

sin( )ax b

ca

b) sin( )

cos( )

( )

ax bax b dx c

dax b

dx

sin( )ax b

ca

c) 2 tan( )sec ( )

( )

ax bax b dx c

dax b

dx

tan( )ax b

ca

TOPIC 4

INTEGRATION 10


Example 4.4

Find :

a) ∫ ( ) b) ∫ (

)

) ∫ (

)

Solution

a) ∫ ( ) ( )

) ∫ (

)

(

)

c)

cx

c

x

dxx

22tan

21

2tan

2

sec2

4.5.2 Integration of exponential functions

Formula of integration of exponential functions :

ca

e

c

axdx

d

edxe

ax

axax

TOPIC 4

INTEGRATION 11


Example 4.5

Find :

a) dxe x

35 b) dxeee xxx 32

c) dxe

eeex

xxx

35 24

Solution

a)

ce

dxex

x

3

3535

b) dxeee xxx 32 dxee xx 35

cee xx

35

35

c) dxe

eeex

xxx

35 24 dxee xx )24( 24

cee

x xx

24

44

cee

xxx

2

2

44

24

4.5.3 Integration of reciprocal functions

The integral of dxx

1 has an infinite discontinuity between x and 1 and has does not exist.

Thus, ln x is an antiderivative of x

1, therefore the integration of reciprocal functions is defined

by

ca

(ax)

c

axdx

d

axdx

ax

ln

)(

)ln(

1

TOPIC 4

INTEGRATION 12


Example 4.6

Find

a) dxx

5 b) dx

x

54

3

Solution

a) cxdxx

ln5 5

b) cx)(dxx

54ln3

54

3

4.6 Integration through substitution method

Integration of an expression in the form of nbax

Integration of expression in the form of nbax can be determined by using substitution

baxu where a and b are constants.

STEPS

If baxu a

duudxbax n

n

adx

du c

na

u n

)1(

1

dxa

du c

na

bax n

)1(

)( 1

Integration of nbax can also be done by applying the following formula

cna

baxdxbax

nn

)1(

1

TOPIC 4

INTEGRATION 13


Example 4.7

Solve the integral below :

a) dxx3

24 b)

dxx

426

2

c) dxxx4252 d)

1

0

32 2 dxxx

Solution

a) By using substitution method,

xu 24 dxx3

24

2

3 duu

2dx

du; dx

du

2 duu 3

2

1

cu

42

1 4

cx 4

248

1

By using formula,

dxx3

24

cx

)2(4

244

cx

8

244

b) By using substitution method,

xu 26

dxx

426

2du

u

4

1

2dx

du; dxdu 2 duu

4

cu

3

3

TOPIC 4

INTEGRATION 14


cx

3263

1

By using formula,

cx

cx

dxxdxx

3

3

4

4

263

1

)3(2

262

26226

2

a) By using substitution method,

25 xu duudxxx 44252

xdx

du2 c

u

5

5

xdxdu 2

cx

5

552

b) By using substitution method,

22 xu

1

0

32 2 dxxx

2

3 duu

xdx

du2 ; xdx

du

2 duu 3

2

1

8

65

16818

1

2218

1

28

1

42

1

44

1

0

42

4

x

u

TOPIC 4

INTEGRATION 15


TIPS!

4.7 Using substitution method involving problems in integration

4.7.1 Trigonometric basic identities and Double-angle formulae

In certain cases, trigonometric integral needs to apply identity trigonometric in order to solve

the solution by using formula integration of trigonometric functions. You can see clearly in our

next examples.

Example 4.8

Find the integral below m:

a) dxx sin 2

b) dxx tan2

Solution

a) dxx sin2

By using double-angle formulae,

xx 2sin212cos

2

2cos1sin 2 x

x

dxx sin2

dxx

2

2cos1

dxx 2cos12

1

cx

x

2

2sin

2

1

b) dxx tan2

By using trigonometry identity;

xx 22 tan1sec

xx 22 tan1sec

dxx tan2 dxx 1sec2

cxx tan

It is recommended to use substitution method in

multiplication and division which cannot be

simplified to a single function.

TOPIC 4

INTEGRATION 16


4.7.2 Exponential function and Reciprocal functions

The best way to solve integration of exponential functions and reciprocal function in

multiplication or division operations is by using substitution method.

Example 4.9

Find :

a) dxee xx 242 3 b) dxxx

x

53

593

2

c) dxxe x )6sin(6cos

Solution

a) dxee xx 242 3

Assume 32 xeu

xedx

du 22

dxedu x2

2

dxee xx 242 3

2

4 duu

duu 4

2

1

ce

cu

x

10

3

52

1

2

5

a) dxxx

x

53

593

2

Assume xxu 53 3

59 2 xdx

du

TOPIC 4

INTEGRATION 17


cxx

cu

u

dudx

xx

x

53ln

)ln(

53

59

3

3

2

a) dxxe x )6sin(6cos

Assume )6cos( xu

)6sin(6 xdx

du

dxxdu

)6sin(6

dxxe x )6sin(6cos

6

dueu

ce

ce

due

x

u

u

)6cos(

6

1

6

1

6

1

TOPIC 4

INTEGRATION 18


Activity 4(a)

1. Integrate the functions below

a. 2

1 b. 2x

Feedback for Activity 4(a)

1. a.

x + 6 and 3 +

x or any

2

1x + C

b. x2 and 6 + x2 or any x2 + C

TOPIC 4

INTEGRATION 19


Activity 4(b)

1. Evaluate

a. kdk5

b. dxx27.0

2. Evaluate

a. dxx 54

b. dzz

4

5 2

Feedback for Activity 4(b)

1. a. ck

2

5 2

b. cx 223.0

2. a. cx

4

1

b. cz

12

5 3

TOPIC 4

INTEGRATION 20


Activity 4(c)

Evaluate :

1. dx3

2. dxx )14(

3. dxxx )123( 2

4. dtt

tt )9

2(2

5.

dx

x

x2

1

Feedback for Activity 4(c)

1. 3x + c

2. 2x2- x + c

3. x3 + x2 –x + c

4. ct

tt

9

23

4 2

2

3

5. cxxx 2

1

2

3

2

5

23

1

5

2

TOPIC 4

INTEGRATION 21


Activity 4(d)

Integrate x.

1. a. b.

2. a. b. c.

3. a b.

Feedback for Activity 4(d)

1. a.

b.

2. a.

b.

cxdxkosxdxkosx sin333

cxdxxsekdxxsek

tan4

1

4

1

4

22

dxx 2

1dx

x 3

dxkosx3

dxxsek

4

2

dxx sin3

1

dxe x

2

dxe x

4

cxdxx

dxx

ln2

11

2

1

2

1

cxdxx

dxx

ln31

33

TOPIC 4

INTEGRATION 22


c. ckosx

ckosxxdxxdx

33

1sin

3

1sin

3

1

3. a. b.

ce

dxex

x 4

44 c

edxe

xx

2

22

TOPIC 4

INTEGRATION 23


Activity 4(e)

Evaluate the given integral

1. a.

b.

2. a.

b.

c.

3. a

b.

dtt 3

1

dzz

13

dxkosx3

dxxsek

2

2

dkk sin2

1

dxex

2

1

dxe x

4

TOPIC 4

INTEGRATION 24


Feedback for Activity 4(e)

1. a.

b. cz ln13

2. a. - 3sin x + c

b. cx tan2

1

c. - ckosk 2

1

3. a. ce x

4

4

b. ce

x

2

1

2

ct ln3

1

TOPIC 4

INTEGRATION 25


Activity 4(f)

1. Evaluate the following integrals.

a. dxxx ]4[ 23

b. dtt

t ]1

3[3

3

c. dxx

]32

[2

2. Evaluate

a. dkkk ]44[ 2

b. dzz 2)32(

c. dxx

x

2

542

Feedback for Activity 4(f)

1. a. cxx

34

3

4

4

b. ct

t 2

4

2

1

4

3

c. cxx

32

2. a. ckkk

423

23

b. czzz 963

4 23 c. cxx

42

TOPIC 4

INTEGRATION 26


Activity 4(g)

1. Using simple substitution, evaluate.

a. dxx 3)24( b. dzz 6)83(

Feedback for Activity 4(g)

1. a. cx

16

)24( 4

b. cz

56

)83( 7

TOPIC 4

INTEGRATION 27


Activity 4(h)

1. Evaluate:

a. dxx 4)32( b. dzz 3)63(

c. dtt 5)75( d. dxx 3)84(6

e. dxx 3)27( f. dt

t 2)31(

g. dxx 3)54(

1 h. dx

x

4)53(2

3

2. Evaluate

a. dkkk 732 )1(

b. dzzzz )33()3( 233

c. dppp

p

3 3

2

3

1

TOPIC 4

INTEGRATION 28


Feedback for Activity 4(h)

1. a. cx

10

)32( 5

b. cz

12

)63( 4

c. ct

42

)75( 6

d. cx

8

)84(3 4

e. cx

2)27(14

1 f. c

t

)31(3

g. cx

2)54(8

1 h. c

x

3)53(6

1

2. a. ck 83124

1

b. css 323

3

1 c. cpp 3

23 3

2

1

TOPIC 4

INTEGRATION 29


Activity 4(i)

Evaluate the given indefinite integrals

a.. dxx4sin b. dxxx 2cos

c. dxx7tan2 d. dxx

x

2sin

cos2

Feedback for Activity 4(i)

a. - Cx 4cos4

1 b. Cx 2sin

2

1

c. Cxx 7tan7

1 d. -cot x – csc x + C

TOPIC 4

INTEGRATION 30


Activity 4(j)

Evaluate the given integrals

1. x

dx

3 2.

dxx

x

12 3. dx

x

xln

4. dxe x10 5. dxex x322 6. dt

e

et

t

1

Feedback for Activity 4(j)

1. Cx ln3

1 2. Cx )1(ln

2

1 2

3. Cx 2)(ln2

1 4. Ce x 10

10

1

5. - Ce x 32

6

1 6. -e-t + t + C

TOPIC 4

INTEGRATION 31


Activity 4(k)

1. Evaluate a.

b.

c.

2. Integrate

a.

b.

c.

3. Integrate

a.

b.

c.

dxx3

1

dt

t 32

1

dk

k

k

33

dte t 23

dxe

eex

xx

2

3

dxeee xxx )( 32

dxx

kos3

dxx

3sin 2

xkosxdxsin2

TOPIC 4

INTEGRATION 32


Feedback for Activity 4(k)

1. a.

b.

c.

2. a.

b.

c.

3. a.

b.

c.

cx ln3

1

ct )32ln(2

1

ckk

)3ln(3

1 3

ce t 23

3

1

cee xx 3

3

1

cee xx

43

43

cx

3sin3

cx

x 3

2sin

4

3

2

1

cxkos

2

2

TOPIC 4

INTEGRATION 33


Activity 4(l)

Evaluate each of the following:

1 a. 3

2

2 )5( dxxx

b. dxx )33

2(

3

0

2. a. 4

2 )21)(31( dttt

b. 1

1 )32( dkk

3. Find dxx

xx

51

2 3

4

4. An arrow was following a straight line with velocity, v = 6t 4 + 3t2 at time t. What is the

distance of the arrow that moves from t = 1till t =10 ?

TOPIC 4

INTEGRATION 34


Feedback for Activity 4(l)

1 a. b.

2 a. b.

3

4. 120 997.8 meter

2

2

25

3

2

2

35

3

3

2

5

3

)5(

2323

3

2

23

3

2

2

xx

dxxx

6

)0(36

)0(2)3(3

6

)3(2

323

2

)33

2(

22

3

0

2

3

0

xx

dxx

115

)2(22

22)4(2

2

44

22

)61(

)21)(31(

32

32

4

2

32

4

2

2

4

2

tt

t

dttt

dttt

6

)1(3)1()1(31

3

)32(

22

1

1

2

1

1

kk

dkk

1

)2(

5

2

)2(

)1(

5

2

)1(

5

2

)5(

5

22

1

2

2

1

2

2

1

2 3

4

x

x

dxxx

dxx

xx

TOPIC 4

INTEGRATION 35


Activity 4(m)

1. If 2

7 )(

1

2 dxxf and

2

3 )(

2

1 dxxf , evaluate :

a. 2

2 )( dxxf

b.

2

1

1

2 )(2 )( dxxfdxxf

c.

1

2

1

2 )(2 )( dxxfdxxf

2. Evaluate each of the following if 1 )(3

2 dxxf and 4 )(

3

1 dxxg

a. dxxf )1)(3(3

2

b. } )( )({23

2

3

1 dxxfdxxg

TOPIC 4

INTEGRATION 36


Feedback for Activity 4(m)

1. a.

52

3

2

7

)( )(

)(

2

1

1

2

2

2

dxxfdxxf

dxxf

b.

2

16

2

13

2

32

2

7

)(22

7

)(2 )(

2

1

2

1

1

2

dxxf

dxxfdxxf

c.

2

16

2

13

2

32

2

7

)(22

7

)(2 )(

2

1

1

2

1

2

dxxf

dxxfdxxf

2. a. b.

2

)23(1

1

1)(3

)1)(3(

3

2

3

2

3

2

3

2

x

dxdxxf

dxxf

10

28

)1(2)4(2

)(2 )(2

) )( )((2

3

2

3

1

3

2

3

1

dxxfdxxg

dxxfdxxg

TOPIC 4

INTEGRATION 37


Activity 4(n)

1. Evaluate :

a.

1

0

3)12(2 dxx where u = 2x + 1

b.

dz1z2

z43

2

22

where u = 2z2 +1

2. Evaluate these definite integrals by using a suitable substitution.

a.

2

1

43 dt)1t5(t

b. dk1k

k3

02

2. Management cost for a building increased from time to time. If the rate of the

management cost increased and given by 20t

1500

dt

dx

where x is cost in thousand ringgit

and t is time in years, evaluate the management cost for 5 years.

TOPIC 4

INTEGRATION 38


Feedback for Activity 4(n)

1. a. 20

b. 171

10

2. a. 1638

1

b. 1

3. RM 15 835.92 million

TOPIC 4

INTEGRATION 39


SELF ASSESSMENT 4 (a)

1. Evaluate the following integral

a. dz7

b. dtt32

c. dxx 4

10

2. Evaluate

a. dxxxx 96 2

b. dzx2

52

3. Rewrite each expression and using derivative of a sum, to solve the following

integrals

a. (3x - 2)2

b. 5

2 )1(

x

xx

c. 2

)1)(1(

k

kk

4. Evaluate

a. dss334

b. dzx 2)76(

c.

dkk

k)6(

TOPIC 4

INTEGRATION 40


SOLUTION : SELF ASSESSMENT 4 (a)

Have you tried ?

1. a. 7z +c

b. ct

5

22 5

c. cx

33

10

2. a. cxxx 323

3

2

2

92

b. cxxx

25103

4 23

3. a. cxxx 463 23

b. cxx

1

2

12

c. ck

k 1

4. a. css 4

4

34

b.

c. √ -

TOPIC 4

INTEGRATION 41


SELF ASSESSMENT 4 (b)

Evaluate

1. a.

b.

2. An epidemic struck Klang. The affected population is rising at the rate of

per day where t is the number of days after the outbreak of the epidemic. Find

the epidemic function S(t) for new cases each day.

73 43 ttdt

dS

,)31( 4 dxx

,52 dtt

TOPIC 4

INTEGRATION 42


SOLUTION: SELF ASSESSMENT 4 (b)

1. a b.

2.

Let u = (t4 + 7) 2

1

32

14 47

2

1tt

dt

du

2

1

4

3

)7(

2

t

t

= duut

2

2

3 = cu

t3

2

1 = ct

tct

t 2

343

2

14 7

2

1]7[

2

1

dxx 4)31(

cx

cx

15

)31(

)14(3

)31(

5

14

c

t

ct

dtt

dtt

3

52

)12

1(2

52

52

52

2

3

12

1

2

1

dtttS 73 43

3

2

1

443

2

)7(73

t

dutttS

TOPIC 4

INTEGRATION 43


SELF ASSESSMENT 4 (c)

1. Evaluate :

a.

3

1

2 dx)5x4x( b.

200

100

2 dt)120t2t03.0(

2. The rate of sale for a shop is given as J (t) = -3t2 + 300t where t is the number of days

after the sale. Determine the total of sale J(t) for the first week after the campaign.

3. The rate of toll collection at a highway is given as this model

23 t240t20dt

dk

Where t is time in hours and k is the sum of money that has been collected. This model is

used from time 1200 (t = 0) till time 1000. What is the total of collection that has been made

in this period?

TOPIC 4

INTEGRATION 44


SOLUTION: SELF ASSESSMENT 4 (c)

1. a. 3

40

b. 52 000

2. RM 7007

3. RM30 000

Documents

BA201 Engineering Mathematic UNIT 4 Integration