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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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Perhaps the problem came from not making a sketch of the region of integration. Despite all too many years of double integrals, I still always make a sketch. –  André Nicolas Apr 28 '12 at 4:07
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I know I had to make a sketch to figure out how to express the region in the other order. +1 to André's comment. –  Arturo Magidin Apr 28 '12 at 4:12
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1 Answer

up vote 2 down vote accepted

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. –  user16647 Apr 28 '12 at 4:11
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