Question

need help with this double integral question :)

http://gyazo.com/a3e6e7133a02d92949f4a14f0cb3961e

Answer 1

@amistre64

Answer 2

What are you stuck with?

Answer 3

\[\int\limits_{\pi}^{2\pi}\int\limits_{0}^{1}xsin(xy)dxdy\]

Answer 4

since your limits are constants; you can prolly more easily swap it for dydx

Answer 5

which proff did ... ironically :)

Answer 6

IRONICALLY?!?! LOOOOOOOOOOL

Answer 7

\[\int\limits_{\pi}^{2\pi}\int\limits_{0}^{1}xsin(xy)~dydx\]
\[\int\limits_{\pi}^{2\pi}\left(\int\limits_{0}^{1}xsin(xy)~dy\right)dx\]

Answer 8

doh ... loathesome integrals!!

Answer 9

@amistre64 doesn't the zero to one have to stick with the dx differential though?

Answer 10

\[\int_{x=0}^{x=1}\left(\int_{y=\pi}^{y=2\pi}xsin(xy)~dy\right)dx\]

Answer 11

it tends to help me out if i label the integrals ....

Answer 12

Isn't the general integral of that xsin(xy)/y|

Answer 13

\[\int_{x=0}^{x=1}\left(\int_{y=\pi}^{y=2\pi}xsin(xy)~dy\right)dx\]
\[\int_{x=0}^{x=1}-cos(2\pi~x)+cos(\pi~x)~dx\]

Answer 14

I thought you didn't have to do parts integration since your X term out front evidently doesn't have a Y in it?

Answer 15

you still have to integrate your limits respectively

x = [0,1] has to address your dx

y = [pi, 2pi] has to address your dy

Answer 16

the dx, dy parts tells you what you focus on while leaving the other term a constant

Answer 17

LOLOL I forgot about that

Answer 18

I didn't realize you were evaluating limits in your 2nd step of that response

Answer 19

let x = k

int k sin(ky) dy -> -cos(ky)

-cos(x 2pi) - -cos(x pi)

Answer 20

int cos(x pi) - cos(x 2pi) dx

sin(x pi)/pi - sin(x 2pi)/2pi

(sin(pi)/pi - sin(2pi)/2pi) - (sin(0 pi)/pi - sin(0 2pi)/2pi)

(0 - 0) - (0 - 0)

so with any luck :)

Answer 21

thanks @amistre64