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BITS Pilani, K K Birla Goa Campus BITS Pilani K K Birla Goa Campus Engineering Mathematics – I (MATH ZC 161) Dr. Amit Setia (Assistant Professor) Department of Mathematics

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Page 1: DocumentL9

BITS Pilani, K K Birla Goa Campus

BITS PilaniK K Birla Goa Campus

Engineering Mathematics – I

(MATH ZC 161)

Dr. Amit Setia (Assistant Professor)Department of Mathematics

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BITS PilaniK K Birla Goa Campus

BITS Pilani, K K Birla Goa Campus

Chapter 1 & chapter 3Pratap Singh, Review of Elementary Calculus, WILPD- Notes

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Summary of lecture 8

• L’Hopital’s Rule

• Rationalization method to find limit

• How to check continuity of a function

• Derivative

BITS Pilani, K K Birla Goa Campus

• Derivative

• Rules of derivatives

• Higher order derivatives

• Derivative by Chain Rule

• Derivative by Substitution31/08/2012 3Engineering Mathematics – I (MATH ZC 161)

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• Integration

• Fundamental Integration Formulae

• Some properties of Integration

• Various methods for integration

BITS Pilani, K K Birla Goa Campus

• Various methods for integration

31/08/2012 4Engineering Mathematics – I (MATH ZC 161)

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Outline of lecture 9

• Integration by parts

• Integration of some standard forms

• Definite Integral

• Properties of definite integral

BITS Pilani, K K Birla Goa Campus

• Properties of definite integral

• Some examples based on the properties of

definite integral

31/08/2012 5Engineering Mathematics – I (MATH ZC 161)

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Don’t forget

You need to

of "Pratap Singh, Review of Elementary Calculus,

remember all the formulae given in Chapter 2

BITS Pilani, K K Birla Goa Campus

of "Pratap Singh, Review of Elementary Calculus,

WILPD Notes" uploaded on Taxila in a folder

as MATH_ZC161_Lecture_Notes

24/08/2012 6Engineering Mathematics – I (MATH ZC 161)

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• If u and v are functions of x then the formula

for integration by parts is

duuvdx u vdx vdx dx

= −∫ ∫ ∫ ∫

Article: 4.10

Integration by Parts

BITS Pilani, K K Birla Goa Campus

u is called I functions and v is called II function.

duuvdx u vdx vdx dx

dx = −

∫ ∫ ∫ ∫

31/08/2012 7Engineering Mathematics – I (MATH ZC 161)

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• Rule 1:

If both u and v are integrable functions,

then take that function as the first function

which can be finished by repeated

BITS Pilani, K K Birla Goa Campus

which can be finished by repeated

differentiation.

31/08/2012 8Engineering Mathematics – I (MATH ZC 161)

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Example

BITS Pilani, K K Birla Goa Campus

31/08/2012 9Engineering Mathematics – I (MATH ZC 161)

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• Rule 2:

If the integral contains logarithmic or inverse

trigonometric function,

then take logarithmic or inverse trigometric

BITS Pilani, K K Birla Goa Campus

then take logarithmic or inverse trigometric

function as the first function and other

function as the II function.

If there is no other function then take 1 as the

second function.

31/08/2012 10Engineering Mathematics – I (MATH ZC 161)

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Example

BITS Pilani, K K Birla Goa Campus

31/08/2012 11Engineering Mathematics – I (MATH ZC 161)

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BITS Pilani, K K Birla Goa Campus

31/08/2012 12Engineering Mathematics – I (MATH ZC 161)

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Example

( )

( ) ( ) ( )( )( )

2

2 2

log 1

log 1 log 1

Integrating by parts, we get

III

x dx

dx dx x dx dx

dx = −

∫ ∫ ∫

BITS Pilani, K K Birla Goa Campus

31/08/2012 13Engineering Mathematics – I (MATH ZC 161)

( ) ( )( )( ) ( ) ( )

( ) ( ) ( )

( ) ( )

2

2

2

2loglog

log 2 log 1

log 2 log

III

dx

xx x x dx

x

x x x dx

x x x x x C

= −

= −

= − − +

∫ ∫ ∫

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• Rule 3:

If both functions are integrable and none can

be finished by repeated differentiation, then

any function can be taken as the I function

BITS Pilani, K K Birla Goa Campus

any function can be taken as the I function

and other as the II function. Repeat the rule of

integration by parts.

31/08/2012 14Engineering Mathematics – I (MATH ZC 161)

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3

2

cosec

cosec cosec

Integrating by parts, we get

III

x dx

x x dx=

Example

BITS Pilani, K K Birla Goa Campus

( ) ( )( )( ) ( )( )

( )

2 2

2

cosec ccosec

cosec

cosec c

c o

ot cosec cot

sec

cot co

osec

cosec co tt

x dx x dx

x x

x dx

x dx

x x x x dx

dx

dx

x x

= −

= −

=

∫ ∫∫

∫31/08/2012 15Engineering Mathematics – I (MATH ZC 161)

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( )( )( )

2

3

3

cosec cot cosec cosec 1

cosec cot cosec cosec

cosec cot cosec cosec

x x x x dx

x x x x dx

x x x dx x dx

= − − −

= − − −

= − − +

∫ ∫

BITS Pilani, K K Birla Goa Campus

3

3

3

2 cosec cosec cot cosec

2 cosec cosec cot log cosec cot

1cosec cosec co

2

x dx x x x dx

x dx x x x x C

x dx x

⇒ = − +

⇒ = − + − +

⇒ = −

∫ ∫

∫ ∫

∫1

t log cosec cot2 2

Cx x x+ − +

31/08/2012 16Engineering Mathematics – I (MATH ZC 161)

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• Rule 4:

That function is taken as the first function

which comes first in the word ILATE

where I = inverse circular function,

BITS Pilani, K K Birla Goa Campus

where I = inverse circular function,

L = Logarithmic function,

A = Algebraic function,

T = Trigonometric function,

E = Exponential function.

31/08/2012 17Engineering Mathematics – I (MATH ZC 161)

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Example

BITS Pilani, K K Birla Goa Campus

31/08/2012 18Engineering Mathematics – I (MATH ZC 161)

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putting sin cosx a dx a dθ θ θ= ⇒ =

Article: 4.11

Some standard forms

BITS Pilani, K K Birla Goa Campus

31/08/2012 19Engineering Mathematics – I (MATH ZC 161)

putting sin cosx a dx a dθ θ θ= ⇒ =

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BITS Pilani, K K Birla Goa Campus

31/08/2012 20Engineering Mathematics – I (MATH ZC 161)

Please try !

use sec sec tanHint: x a dx aθ θ θ= ⇒ =

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BITS Pilani, K K Birla Goa Campus

31/08/2012 21Engineering Mathematics – I (MATH ZC 161)

2

Please try !

use tanHint: secx a dx a dθ θ θ= ⇒ =

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Please try !

do integration by parts,Hint:

BITS Pilani, K K Birla Goa Campus

31/08/2012 22Engineering Mathematics – I (MATH ZC 161)

2 2

do integration by parts,

take as I function and 1 as II f

H

u

in

nc

t:

tiona x−

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Similarly,

BITS Pilani, K K Birla Goa Campus

31/08/2012 23Engineering Mathematics – I (MATH ZC 161)

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The integrals of the above form can be evaluated

Article: 4.12

BITS Pilani, K K Birla Goa Campus

31/08/2012 24Engineering Mathematics – I (MATH ZC 161)

The integrals of the above form can be evaluated

by writing in any one of the 6 forms just described.

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Example

BITS Pilani, K K Birla Goa Campus

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Example

BITS Pilani, K K Birla Goa Campus

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Example

BITS Pilani, K K Birla Goa Campus

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Article: 4.13

BITS Pilani, K K Birla Goa Campus

31/08/2012 30Engineering Mathematics – I (MATH ZC 161)

To solve the integral of above form, we write

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Example

( )1

BITS Pilani, K K Birla Goa Campus

31/08/2012 31Engineering Mathematics – I (MATH ZC 161)

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( )

BITS Pilani, K K Birla Goa Campus

31/08/2012 32Engineering Mathematics – I (MATH ZC 161)

( )1 becomes

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( )2

Putting in first integral 2 5

2&

in 2 integral 1

2

nd

x x t

x dx t

x u

dx du

d

+ =

+ +

=

⇒ + =

=

BITS Pilani, K K Birla Goa Campus

31/08/2012 33Engineering Mathematics – I (MATH ZC 161)

dx du⇒ =

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( )( )2Evaluate: 3 45x x x dx+ − −∫

Problem

BITS Pilani, K K Birla Goa Campus

31/08/2012 34Engineering Mathematics – I (MATH ZC 161)

Please try !

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Article: 4.14

BITS Pilani, K K Birla Goa Campus

31/08/2012 35Engineering Mathematics – I (MATH ZC 161)

22

1 2tan

To solve the integral of above fo

sec2 2 2 1

rm, we take

x x dtt dx dt dx

t⇒ ⇒

= = = +

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BITS Pilani, K K Birla Goa Campus

31/08/2012 36Engineering Mathematics – I (MATH ZC 161)

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Example

BITS Pilani, K K Birla Goa Campus

31/08/2012 37Engineering Mathematics – I (MATH ZC 161)

2

2tan

2 1|

x dtt dx

t = =⇒ +

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BITS Pilani, K K Birla Goa Campus

31/08/2012 38Engineering Mathematics – I (MATH ZC 161)

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Article: 4.15

BITS Pilani, K K Birla Goa Campus

31/08/2012 39Engineering Mathematics – I (MATH ZC 161)

( )1 2

To solve the integral of above form, we take

sin cos Denominator Denominatord

a x b x K Kdx + = +

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( )

To solve the integral of above form, we take

2sin 3cos 3sin 4cos 3si

2sin 3cos

3sin 4cos

n 4cosd

x x K x x K x

x xd

xx

x

x + = + + +

+

+∫

Example

BITS Pilani, K K Birla Goa Campus

31/08/2012 40Engineering Mathematics – I (MATH ZC 161)

( )1 22sin 3cos 3sin 4cos 3sin 4cosd

x x K x x K xdx

x + = + + +

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1 2

1 2

Equating coefficient of sin and cos , we get

3 4 2

4 3 3

18 1,

x x

K K

K K

K K

∴− =

+ =

⇒ = =

BITS Pilani, K K Birla Goa Campus

31/08/2012 41Engineering Mathematics – I (MATH ZC 161)

1 2

18 1,

25 25K K⇒ = =

18 1log | 3sin 4cos |

25 25x x x C= + + +

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Article: 4.16

BITS Pilani, K K Birla Goa Campus

31/08/2012 42Engineering Mathematics – I (MATH ZC 161)

( )sin cos Denominator Denominator

where p,q,r

To solve the integral of above form, we

can be evaluated by comparing coefficients of

cos , sin and const

ta

ant terms

ke

da x b x

x

dx

x

C p q r + + = + +

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Given integral

BITS Pilani, K K Birla Goa Campus

31/08/2012 43Engineering Mathematics – I (MATH ZC 161)

The last integral part can be evaluated by using method

described in article 4.14

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( ) [ ]( ) ( )

If is a continuous function on ,

Fundamental Theorem of

, and

,

Calculus

f x a b

f x dx F x C= +∫

Article: 4.17 Definite Integral

BITS Pilani, K K Birla Goa Campus

31/08/2012 44Engineering Mathematics – I (MATH ZC 161)

( ) ( )

( ) ( ) ( )

,

then b

a

f x dx F x C

f x dx F b F a

= +

= −

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Example

/22

0 0sin cos cos

I IIx x dx x x x dx

π π = + ∫ ∫

BITS Pilani, K K Birla Goa Campus

31/08/2012 45Engineering Mathematics – I (MATH ZC 161)

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BITS Pilani, K K Birla Goa Campus

31/08/2012 46Engineering Mathematics – I (MATH ZC 161)

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( )To evaluate

if we are taking a substitution ( ),

then find the values of t corresponding to

and .

b

af x dx

x g t

x a x b

=

= =

Remark

BITS Pilani, K K Birla Goa Campus

31/08/2012 47Engineering Mathematics – I (MATH ZC 161)

and .

If and

when x = a and x = b respectively, then

x a x b

t A t B

= == =

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Example

BITS Pilani, K K Birla Goa Campus

31/08/2012 48Engineering Mathematics – I (MATH ZC 161)

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BITS Pilani, K K Birla Goa Campus

31/08/2012 49Engineering Mathematics – I (MATH ZC 161)

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Properties of definite integral

BITS Pilani, K K Birla Goa Campus

31/08/2012 50Engineering Mathematics – I (MATH ZC 161)

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BITS Pilani, K K Birla Goa Campus

31/08/2012 51Engineering Mathematics – I (MATH ZC 161)

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BITS Pilani, K K Birla Goa Campus

31/08/2012 52Engineering Mathematics – I (MATH ZC 161)

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Example

BITS Pilani, K K Birla Goa Campus

31/08/2012 53Engineering Mathematics – I (MATH ZC 161)

( )of property 7 , as∵

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Example

BITS Pilani, K K Birla Goa Campus

31/08/2012 54Engineering Mathematics – I (MATH ZC 161)

( )of property 4∵

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BITS Pilani, K K Birla Goa Campus

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If sin and cos are replaced

by tan and cot or

by cosec and sec respectively

x x

x x

x x

Remark

BITS Pilani, K K Birla Goa Campus

31/08/2012 56Engineering Mathematics – I (MATH ZC 161)

in the integral I,

then similarly we can evaluate the integral.

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Example

BITS Pilani, K K Birla Goa Campus

31/08/2012 57Engineering Mathematics – I (MATH ZC 161)

( )| of property 3∵

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BITS Pilani, K K Birla Goa Campus

31/08/2012 58Engineering Mathematics – I (MATH ZC 161)

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Thanks

BITS Pilani, K K Birla Goa Campus

Thanks