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Compute $\int_{y=0}^{1} \int_{x=y}^{1} \frac{x^2}{y^2} e^{\frac{-x^2}{y}}dxdy.$

Here's my idea.

Switch the order of integration by Fubini's Theorem. Then compute $\int_{y=0}^{1} \frac{x^2}{y^2} e^{\frac{-x^2}{y}}dy$ with $u = \frac{-1}{y}$, $du = \frac{1}{y^2}dy$ to have $\int_{-\infty}^{-1} x^2e^{ux^2}du = e^{-x^2}$ but now I get stuck with $\int_{x=y}^{1}e^{-x^2}dx$ because this can't be computed.

Does anyone have suggestions?

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    I know *I* had to make a sketch to figure out how to ex$p$ress the region in the other order. +1 to André's comment.2012-04-28

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You have to be careful when switching the order of integration. The region over which you are integrating when you do $\int_{y=0}^1\int_{x=y}^1 f(x,y)\,dxdy$ is the triangle between $y=0$, $x=1$, and $x=y$ (check).

If you want to switch the order of integration, then this is not simply $\int_{x=y}^1\int_{y=0}^1f(x,y)\,dydx;$ this would not make sense, since the lower limit of the outside integral depends on $y$, but $y$ is free.

Instead, you need to think about how to describe that same region using $x$ first and $y$ second. It should not be hard to verify that you cover the same region as before if you use $\int_{x=0}^1\int_{y=0}^x f(x,y)\,dydx.$

Try that.

(Your "inside integral" then is incorrect above, since the limits are wrong; the outside integral of course leads to problems because your limits of integration don't make sense)

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    It works, thanks. $1 - e^{-1}$. Had trouble remembering rules of multivariable calculus since it has been a while.2019-03-06