Question

@ganeshie8

Answer 1

Sketch the area of integration and evaluate:

\[\int\limits_{0}^{\infty} \int\limits_{x}^{\infty} e^{-y}dydx i\]

Answer 2

@rishavraj
mai har ek type ka example solve karna chahathi hu

Answer 3

wht is tht i????
and kaun s book se padh rahi hai tu???

Answer 4

arrey humhe ek sheet milthi hai clg se
so usse problems uta rahi hu

Answer 5

okkk bt wht is tht "i" ??iota??

Answer 6

ooops i is not there

Answer 7

\[\int\limits_{x}^{\infty}e^{-y} dy = -[e^{- \infty} + e^{-x}]\]

Answer 8

well is it the normal integration?
sketch karna hai
woh kaise karthe hai
i mean x y limits leke na?

Answer 9

graph ka nahi pata....lol and yeah normal integration k tarah I guess....
answer is 1 ??

Answer 10

ans nahi pata baba

Answer 11

lol ok...ans 1 hoga :P

Answer 12

@rvc

Answer 13

haan wait next type
\[\int\limits_{0}^{\pi/2}\int\limits_{0}^{1-\sin\theta}r^2~\cos\theta~dr~d\theta\]

Answer 14

okkkk M'am

Answer 15

wow u back :P

Answer 16

so first of all solve the 'r' part

Answer 17

since it is \(\theta\) limits

Answer 18

yeah because dekh..... after u are done with r ... u need to assign the limits in it .....

Answer 19

yes now next time

Answer 20

type* xD

Answer 21

hold on..

Answer 22

now can u proceed???? @rvc

Answer 23

wait
r limits na
so diff wrt r first

btw @baru join in

Answer 24

hmmmm ???? :/

Answer 25

it should be r^3/3 first na

Answer 26

lol I skipped many steps.....

Answer 27

try substituting u=(sin -1 )

Answer 28

yup plug sin(theta) = 1 or sin(theta)-1 = u

Answer 29

*sin(theta) = u

Answer 30

hmm
sunn
\[\int\limits_{0}^{\pi/4} \int\limits_{0}^{\sqrt(\cos2\theta)}\frac{r}{(1+r^2)^2}dr~d\theta \]

Answer 31

sub \(u=1+r^2\\du=2rdr\)

Answer 32

yes just did that xD

Answer 33

okay lets see the next type :
Change the order of integration :
\[\int\limits_{0}^{a}\int\limits_{\sqrt{a^2-y^2}}^{y+a} \phi(x,y)dxdy\]

Answer 34

dinner time