# Thread: [SOLVED] finding the double integral of an e^x problem

1. ## [SOLVED] finding the double integral of an e^x problem

I have the double integral of....

∫∫xye^((x^2)y) dxdy

the inner integral goes from 0 to 1
the outer integral goes from 0 to 2

Do we need to u-sub for this, because I can't see what we need to look for if that's the case.

if it IS u-sub, this is what I may have:
u = x^2
du = 2xdx
xdx = (1/2)du

2. Yes, your substitution is correct.

$\displaystyle \int_0^2 \int_0^1 x \, y \, e^{x^2 y} \, dx \, dy$

Let $\displaystyle u=x^2 \implies du = 2x \, dx$

$\displaystyle \frac{1}{2} \int_0^2 \int_0^1 y \, e^{u y} \, du \, dy$

$\displaystyle = \frac{1}{2} \int_0^2 \left[ e^{u y} \right]_0^1 \, dy$

$\displaystyle = \frac{1}{2} \int_0^2 ( e^y - 1 ) \, dy$

3. Thanks! I got the same answer, but it's good to have a second opinion!

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# integration 0 to 1 .integration 0 to x . xye^(-x)^2 dydx with full description

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