Question

Double Integrals y'all!

∫3 to 1 ∫1 to 0 (2xy)/((x^2)+1) dx dy

Answer 1

\[\int_{1}^{3}\int_{0}^{1}\frac{2xy}{x^2+1}dxdy\]What's so tough about this one?

Answer 2

just do it in turns...

Answer 3

first x, later y

Answer 4

I'm studying for finals and going through my entire textbook and after having 4 other finals prior to today, you could say my brain is fried. sorry for being a college student and having off days

Answer 5

We know how you feel. All of us have gone through that.

Answer 6

well kind of a harsh response before then.

Answer 7

not harsh just saying it is easy to do, if you are in the right state of mind or if you get a quick tip on how to start it off

Answer 8

let's not argue, just answer his/her question people :D

Answer 9

I'm not here to argue. I just want to make sure I am doing things right, this is supposed to be study help yeah?

Answer 10

http://www.youtube.com/watch?feature=player_detailpage&v=DYsv6L-VcsQ#t=190s

Answer 11

this will explain better than we can

Answer 12

limit of 0-1 is for x
so integrate
sub in 1 for x
integrate again sub in y values
sub in 3
- sub in values of 1

enjoy :D

Answer 13

\[\int\limits_{1}^{3}\int\limits_{0}^{1}(2xy /x ^{2}+1)dxdy=\int\limits_{1}^{3}y[Ln(x ^{2}+1]_{0}^{1}dy=Ln2/2(y ^{2})|_{1}^{3}=(Ln(2)/2)(9-1)=4Ln(2)\]

Answer 14

lol, didn't fit....... till the end

Answer 15

4Ln(2)

Answer 16

Thanks everyone. I appreciate not allowing me to completely brainfart today