Question

Evaluate the Double Intergal

Answer 1

I got:

\[\frac{ 7 }{ 9 } + \frac{ 2 }{ \pi }\]

Answer 2

Just want to see if that's right.

Answer 3

*checking*

Answer 4

Typo sorry.

Answer 5

\[\int\limits\limits_{0}^{1}\int\limits\limits_{-2}^{-1}x^2y^2+\cos(\pi x)+\sin(\pi y)dxdy\]

Answer 6

I got this \[\frac{ 2 }{ 3 }-\frac{ 2 }{ pi }\]

Answer 7

One second

Answer 8

No Nope. I get 7/9 + 2/pi again. How did you get that?

Answer 9

Can I do double intergals on wolfram to check?

Answer 10

7/9 + 2/pi is correct.

Answer 11

Woot!

Answer 12

sorry my mistake

Answer 13

It's fine :) .

Answer 14

can I see how u did it?

Answer 15

Too much work to type up :/ .

Answer 16

I just wanna compare with mine no need to write completely

Answer 17

\[\int\limits\limits\limits_{0}^{1}\int\limits\limits\limits_{-2}^{-1}x^2y^2+\cos(\pi x)+\sin(\pi y)dxdy\]

\[\int\limits\limits\limits_{0}^{1}[\int\limits\limits\limits_{-2}^{-1}x^2y^2+\cos(\pi x)+\sin(\pi y)dx]dy\]

\[\int\limits\limits\limits_{0}^{1}\int\limits\limits\limits_{-2}^{-1}\frac{ x^3y^2 }{3 }+\frac{ 1 }{ \pi }\sin(\pi x)+xsin(\pi y)\]

\[\int\limits\limits\limits_{0}^{1}(\frac{ 7y^2 }{ 3 }+\sin(\pi y)dy\]

Answer 18

I am sure the rest you can solve.

Answer 19

thanks.

Answer 20

No problem.

Answer 21

No wonder I wasn't getting it, my eyesight is horrible and thought sin(pi*y) was sin(xy)