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I'm not sure if I am doing something wrong or if the question has a typo...

$$\int\limits_{0}^{3}\!\!\int\limits_{x^{2}}^{9}xe^{y^{2}}\mathsf{d}y\ \mathsf{d}x$$

I notice that $0\leq x\leq 3$ and $x^{2}\leq y\leq 9$. Given these bounds, when I reverse the order I'm coming up with $0\leq y\leq 9$ and $\sqrt{y}\leq x\leq 3$...

$$\int\limits_{0}^{9}\!\!\int\limits_{\sqrt{y}}^{3}xe^{y^{2}}\mathsf{d}x\ \mathsf{d}y$$

So now I can easily integrate with respect to x, but I get stuck on the next step...


I can integrate $\displaystyle\frac{1}{2}\int\limits_{0}^{9}ye^{y^{2}}\mathsf{d}y$ with no problem, but I don't see how I can easily integrate $\displaystyle\frac{1}{2}\int\limits_{0}^{9}9e^{y^{2}}\mathsf{d}y$.

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Basically, as everyone more or less said, your limits are off. Particularly, your limits define everything outside the region that you should be integrating within the rectangle formed by $0<x<3$ and $0<y<9$. – Ataraxia Jul 31 '13 at 22:41
up vote 5 down vote accepted

As I always yell to my students, DRAW pictures of the region. Your inner limits are incorrect.

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Oh, ok.. I see - I did draw the region, but I picked out the wrong part as the region in question. – agent154 Jul 31 '13 at 22:39
Well, promise you won't ever do that again! :) – Ted Shifrin Jul 31 '13 at 22:48

You made a mistake with the limits when you reversed the order. It should be $0 \le y \le 9$, $0 \le x \le \sqrt{y}$. (Draw a picture!)

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Really, the approach is to look at a plot of the integration region. You may reverse the order of integration by using inverse functions, more or less; the integral is equal to

$$\int_0^9 dy\,e^{y^2} \, \int_0^{\sqrt{y}} dx \, x = \frac12 \int_0^9 dy\,e^{y^2} y$$

which you said you can do.

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