Download - DocumentL9
![Page 1: DocumentL9](https://reader034.vdocuments.net/reader034/viewer/2022042515/545d3e91b1af9f1b5d8b48e3/html5/thumbnails/1.jpg)
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
![Page 2: DocumentL9](https://reader034.vdocuments.net/reader034/viewer/2022042515/545d3e91b1af9f1b5d8b48e3/html5/thumbnails/2.jpg)
BITS PilaniK K Birla Goa Campus
BITS Pilani, K K Birla Goa Campus
Chapter 1 & chapter 3Pratap Singh, Review of Elementary Calculus, WILPD- Notes
![Page 3: DocumentL9](https://reader034.vdocuments.net/reader034/viewer/2022042515/545d3e91b1af9f1b5d8b48e3/html5/thumbnails/3.jpg)
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)
![Page 4: DocumentL9](https://reader034.vdocuments.net/reader034/viewer/2022042515/545d3e91b1af9f1b5d8b48e3/html5/thumbnails/4.jpg)
• 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)
![Page 5: DocumentL9](https://reader034.vdocuments.net/reader034/viewer/2022042515/545d3e91b1af9f1b5d8b48e3/html5/thumbnails/5.jpg)
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)
![Page 6: DocumentL9](https://reader034.vdocuments.net/reader034/viewer/2022042515/545d3e91b1af9f1b5d8b48e3/html5/thumbnails/6.jpg)
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)
![Page 7: DocumentL9](https://reader034.vdocuments.net/reader034/viewer/2022042515/545d3e91b1af9f1b5d8b48e3/html5/thumbnails/7.jpg)
• 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)
![Page 8: DocumentL9](https://reader034.vdocuments.net/reader034/viewer/2022042515/545d3e91b1af9f1b5d8b48e3/html5/thumbnails/8.jpg)
• 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)
![Page 9: DocumentL9](https://reader034.vdocuments.net/reader034/viewer/2022042515/545d3e91b1af9f1b5d8b48e3/html5/thumbnails/9.jpg)
Example
BITS Pilani, K K Birla Goa Campus
31/08/2012 9Engineering Mathematics – I (MATH ZC 161)
![Page 10: DocumentL9](https://reader034.vdocuments.net/reader034/viewer/2022042515/545d3e91b1af9f1b5d8b48e3/html5/thumbnails/10.jpg)
• 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)
![Page 11: DocumentL9](https://reader034.vdocuments.net/reader034/viewer/2022042515/545d3e91b1af9f1b5d8b48e3/html5/thumbnails/11.jpg)
Example
BITS Pilani, K K Birla Goa Campus
31/08/2012 11Engineering Mathematics – I (MATH ZC 161)
![Page 12: DocumentL9](https://reader034.vdocuments.net/reader034/viewer/2022042515/545d3e91b1af9f1b5d8b48e3/html5/thumbnails/12.jpg)
BITS Pilani, K K Birla Goa Campus
31/08/2012 12Engineering Mathematics – I (MATH ZC 161)
![Page 13: DocumentL9](https://reader034.vdocuments.net/reader034/viewer/2022042515/545d3e91b1af9f1b5d8b48e3/html5/thumbnails/13.jpg)
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
= −
= −
= − − +
∫ ∫ ∫
∫
∫
![Page 14: DocumentL9](https://reader034.vdocuments.net/reader034/viewer/2022042515/545d3e91b1af9f1b5d8b48e3/html5/thumbnails/14.jpg)
• 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)
![Page 15: DocumentL9](https://reader034.vdocuments.net/reader034/viewer/2022042515/545d3e91b1af9f1b5d8b48e3/html5/thumbnails/15.jpg)
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)
![Page 16: DocumentL9](https://reader034.vdocuments.net/reader034/viewer/2022042515/545d3e91b1af9f1b5d8b48e3/html5/thumbnails/16.jpg)
( )( )( )
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)
![Page 17: DocumentL9](https://reader034.vdocuments.net/reader034/viewer/2022042515/545d3e91b1af9f1b5d8b48e3/html5/thumbnails/17.jpg)
• 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)
![Page 18: DocumentL9](https://reader034.vdocuments.net/reader034/viewer/2022042515/545d3e91b1af9f1b5d8b48e3/html5/thumbnails/18.jpg)
Example
BITS Pilani, K K Birla Goa Campus
31/08/2012 18Engineering Mathematics – I (MATH ZC 161)
![Page 19: DocumentL9](https://reader034.vdocuments.net/reader034/viewer/2022042515/545d3e91b1af9f1b5d8b48e3/html5/thumbnails/19.jpg)
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= ⇒ =
![Page 20: DocumentL9](https://reader034.vdocuments.net/reader034/viewer/2022042515/545d3e91b1af9f1b5d8b48e3/html5/thumbnails/20.jpg)
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= ⇒ =
![Page 21: DocumentL9](https://reader034.vdocuments.net/reader034/viewer/2022042515/545d3e91b1af9f1b5d8b48e3/html5/thumbnails/21.jpg)
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= ⇒ =
![Page 22: DocumentL9](https://reader034.vdocuments.net/reader034/viewer/2022042515/545d3e91b1af9f1b5d8b48e3/html5/thumbnails/22.jpg)
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−
![Page 23: DocumentL9](https://reader034.vdocuments.net/reader034/viewer/2022042515/545d3e91b1af9f1b5d8b48e3/html5/thumbnails/23.jpg)
Similarly,
BITS Pilani, K K Birla Goa Campus
31/08/2012 23Engineering Mathematics – I (MATH ZC 161)
![Page 24: DocumentL9](https://reader034.vdocuments.net/reader034/viewer/2022042515/545d3e91b1af9f1b5d8b48e3/html5/thumbnails/24.jpg)
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.
![Page 25: DocumentL9](https://reader034.vdocuments.net/reader034/viewer/2022042515/545d3e91b1af9f1b5d8b48e3/html5/thumbnails/25.jpg)
Example
BITS Pilani, K K Birla Goa Campus
31/08/2012 25Engineering Mathematics – I (MATH ZC 161)
![Page 26: DocumentL9](https://reader034.vdocuments.net/reader034/viewer/2022042515/545d3e91b1af9f1b5d8b48e3/html5/thumbnails/26.jpg)
Example
BITS Pilani, K K Birla Goa Campus
31/08/2012 26Engineering Mathematics – I (MATH ZC 161)
![Page 27: DocumentL9](https://reader034.vdocuments.net/reader034/viewer/2022042515/545d3e91b1af9f1b5d8b48e3/html5/thumbnails/27.jpg)
BITS Pilani, K K Birla Goa Campus
31/08/2012 27Engineering Mathematics – I (MATH ZC 161)
![Page 28: DocumentL9](https://reader034.vdocuments.net/reader034/viewer/2022042515/545d3e91b1af9f1b5d8b48e3/html5/thumbnails/28.jpg)
Example
BITS Pilani, K K Birla Goa Campus
31/08/2012 28Engineering Mathematics – I (MATH ZC 161)
![Page 29: DocumentL9](https://reader034.vdocuments.net/reader034/viewer/2022042515/545d3e91b1af9f1b5d8b48e3/html5/thumbnails/29.jpg)
BITS Pilani, K K Birla Goa Campus
31/08/2012 29Engineering Mathematics – I (MATH ZC 161)
![Page 30: DocumentL9](https://reader034.vdocuments.net/reader034/viewer/2022042515/545d3e91b1af9f1b5d8b48e3/html5/thumbnails/30.jpg)
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
![Page 31: DocumentL9](https://reader034.vdocuments.net/reader034/viewer/2022042515/545d3e91b1af9f1b5d8b48e3/html5/thumbnails/31.jpg)
Example
( )1
BITS Pilani, K K Birla Goa Campus
31/08/2012 31Engineering Mathematics – I (MATH ZC 161)
![Page 32: DocumentL9](https://reader034.vdocuments.net/reader034/viewer/2022042515/545d3e91b1af9f1b5d8b48e3/html5/thumbnails/32.jpg)
( )
BITS Pilani, K K Birla Goa Campus
31/08/2012 32Engineering Mathematics – I (MATH ZC 161)
( )1 becomes
![Page 33: DocumentL9](https://reader034.vdocuments.net/reader034/viewer/2022042515/545d3e91b1af9f1b5d8b48e3/html5/thumbnails/33.jpg)
( )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⇒ =
![Page 34: DocumentL9](https://reader034.vdocuments.net/reader034/viewer/2022042515/545d3e91b1af9f1b5d8b48e3/html5/thumbnails/34.jpg)
( )( )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 !
![Page 35: DocumentL9](https://reader034.vdocuments.net/reader034/viewer/2022042515/545d3e91b1af9f1b5d8b48e3/html5/thumbnails/35.jpg)
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⇒ ⇒
= = = +
![Page 36: DocumentL9](https://reader034.vdocuments.net/reader034/viewer/2022042515/545d3e91b1af9f1b5d8b48e3/html5/thumbnails/36.jpg)
BITS Pilani, K K Birla Goa Campus
31/08/2012 36Engineering Mathematics – I (MATH ZC 161)
![Page 37: DocumentL9](https://reader034.vdocuments.net/reader034/viewer/2022042515/545d3e91b1af9f1b5d8b48e3/html5/thumbnails/37.jpg)
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 = =⇒ +
∵
![Page 38: DocumentL9](https://reader034.vdocuments.net/reader034/viewer/2022042515/545d3e91b1af9f1b5d8b48e3/html5/thumbnails/38.jpg)
BITS Pilani, K K Birla Goa Campus
31/08/2012 38Engineering Mathematics – I (MATH ZC 161)
![Page 39: DocumentL9](https://reader034.vdocuments.net/reader034/viewer/2022042515/545d3e91b1af9f1b5d8b48e3/html5/thumbnails/39.jpg)
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 + = +
![Page 40: DocumentL9](https://reader034.vdocuments.net/reader034/viewer/2022042515/545d3e91b1af9f1b5d8b48e3/html5/thumbnails/40.jpg)
( )
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 + = + + +
![Page 41: DocumentL9](https://reader034.vdocuments.net/reader034/viewer/2022042515/545d3e91b1af9f1b5d8b48e3/html5/thumbnails/41.jpg)
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= + + +
![Page 42: DocumentL9](https://reader034.vdocuments.net/reader034/viewer/2022042515/545d3e91b1af9f1b5d8b48e3/html5/thumbnails/42.jpg)
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 + + = + +
![Page 43: DocumentL9](https://reader034.vdocuments.net/reader034/viewer/2022042515/545d3e91b1af9f1b5d8b48e3/html5/thumbnails/43.jpg)
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
![Page 44: DocumentL9](https://reader034.vdocuments.net/reader034/viewer/2022042515/545d3e91b1af9f1b5d8b48e3/html5/thumbnails/44.jpg)
( ) [ ]( ) ( )
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
= +
= −
∫
∫
![Page 45: DocumentL9](https://reader034.vdocuments.net/reader034/viewer/2022042515/545d3e91b1af9f1b5d8b48e3/html5/thumbnails/45.jpg)
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)
![Page 46: DocumentL9](https://reader034.vdocuments.net/reader034/viewer/2022042515/545d3e91b1af9f1b5d8b48e3/html5/thumbnails/46.jpg)
BITS Pilani, K K Birla Goa Campus
31/08/2012 46Engineering Mathematics – I (MATH ZC 161)
![Page 47: DocumentL9](https://reader034.vdocuments.net/reader034/viewer/2022042515/545d3e91b1af9f1b5d8b48e3/html5/thumbnails/47.jpg)
( )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
= == =
![Page 48: DocumentL9](https://reader034.vdocuments.net/reader034/viewer/2022042515/545d3e91b1af9f1b5d8b48e3/html5/thumbnails/48.jpg)
Example
BITS Pilani, K K Birla Goa Campus
31/08/2012 48Engineering Mathematics – I (MATH ZC 161)
![Page 49: DocumentL9](https://reader034.vdocuments.net/reader034/viewer/2022042515/545d3e91b1af9f1b5d8b48e3/html5/thumbnails/49.jpg)
BITS Pilani, K K Birla Goa Campus
31/08/2012 49Engineering Mathematics – I (MATH ZC 161)
![Page 50: DocumentL9](https://reader034.vdocuments.net/reader034/viewer/2022042515/545d3e91b1af9f1b5d8b48e3/html5/thumbnails/50.jpg)
Properties of definite integral
BITS Pilani, K K Birla Goa Campus
31/08/2012 50Engineering Mathematics – I (MATH ZC 161)
![Page 51: DocumentL9](https://reader034.vdocuments.net/reader034/viewer/2022042515/545d3e91b1af9f1b5d8b48e3/html5/thumbnails/51.jpg)
BITS Pilani, K K Birla Goa Campus
31/08/2012 51Engineering Mathematics – I (MATH ZC 161)
![Page 52: DocumentL9](https://reader034.vdocuments.net/reader034/viewer/2022042515/545d3e91b1af9f1b5d8b48e3/html5/thumbnails/52.jpg)
BITS Pilani, K K Birla Goa Campus
31/08/2012 52Engineering Mathematics – I (MATH ZC 161)
![Page 53: DocumentL9](https://reader034.vdocuments.net/reader034/viewer/2022042515/545d3e91b1af9f1b5d8b48e3/html5/thumbnails/53.jpg)
Example
BITS Pilani, K K Birla Goa Campus
31/08/2012 53Engineering Mathematics – I (MATH ZC 161)
( )of property 7 , as∵
![Page 54: DocumentL9](https://reader034.vdocuments.net/reader034/viewer/2022042515/545d3e91b1af9f1b5d8b48e3/html5/thumbnails/54.jpg)
Example
BITS Pilani, K K Birla Goa Campus
31/08/2012 54Engineering Mathematics – I (MATH ZC 161)
( )of property 4∵
![Page 55: DocumentL9](https://reader034.vdocuments.net/reader034/viewer/2022042515/545d3e91b1af9f1b5d8b48e3/html5/thumbnails/55.jpg)
BITS Pilani, K K Birla Goa Campus
31/08/2012 55Engineering Mathematics – I (MATH ZC 161)
![Page 56: DocumentL9](https://reader034.vdocuments.net/reader034/viewer/2022042515/545d3e91b1af9f1b5d8b48e3/html5/thumbnails/56.jpg)
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.
![Page 57: DocumentL9](https://reader034.vdocuments.net/reader034/viewer/2022042515/545d3e91b1af9f1b5d8b48e3/html5/thumbnails/57.jpg)
Example
BITS Pilani, K K Birla Goa Campus
31/08/2012 57Engineering Mathematics – I (MATH ZC 161)
( )| of property 3∵
![Page 58: DocumentL9](https://reader034.vdocuments.net/reader034/viewer/2022042515/545d3e91b1af9f1b5d8b48e3/html5/thumbnails/58.jpg)
BITS Pilani, K K Birla Goa Campus
31/08/2012 58Engineering Mathematics – I (MATH ZC 161)
![Page 59: DocumentL9](https://reader034.vdocuments.net/reader034/viewer/2022042515/545d3e91b1af9f1b5d8b48e3/html5/thumbnails/59.jpg)
Thanks
BITS Pilani, K K Birla Goa Campus
Thanks