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Compute the integral: \begin{equation} \iint\limits_{\substack{\displaystyle 0 \leqslant x \leqslant 1\\\displaystyle 0 \leqslant y \leqslant 1}}\exp\left\{\min(x^2, y^2)\right\}\mathrm{d}x\mathrm{d}y \end{equation} $D$ is the rectangle with vertices $(0,0)$, $(0,1)$, $(1,0)$ and $(1,1)$ and $\min(x^2,y^2)$ is the minimum of the numbers $x^2$ and $y^2$.

I dont have a clue about how I can integrate this function. I thought of using Taylor series but I am not sure if this would work either. Can anyone help me out?

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First: you should try to use Latex. Just check the FAQ :) Second: Do you know Fubini's theorem? then you can integrate first over $x$ and then over $y$. You just have to paremetrise the set $[0,1]^2$ in terms of $x^2$ being larger than $y^2$ and the other reround – Quickbeam2k1 Feb 23 '13 at 11:12
Thank you. Ill definately take a look into how I can use that :). – dreamer Feb 23 '13 at 12:12

Note that the derivative of $x\mapsto e^{x^2}$ is $x\mapsto 2xe^{x^2}$, hence by symmetry along the line $x=y$ $$\begin{align} \int_0^1\int_0^1e^{\min\{x^2,y^2\}}\,\mathrm dy\,\mathrm dx &= 2\int_0^1\int_x^1e^{x^2}\,\mathrm dy\,\mathrm dx\\ &=2\int_0^1(1-x)e^{x^2}\,\mathrm dx\\ &=2\int_0^1e^{x^2}\,\mathrm dx-\int_0^12xe^{x^2}\,\mathrm dx\\ &=2\int_0^1e^{x^2}\,\mathrm dx-e+1. \end{align}$$ Unfortunately, the remaining integral is non-elementary.

A similar integal with max instead of min is much easier: $$\begin{align} \int_0^1\int_0^1e^{\max\{x^2,y^2\}}\,\mathrm dy\,\mathrm dx &= 2\int_0^1\int_0^xe^{x^2}\,\mathrm dy\,\mathrm dx\\ &=2\int_0^1 x e^{x^2}\,\mathrm dx\\ &=e-1. \end{align}$$

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Thank you. I still dont really get the step though where you go over to integrating e^(x^2) with limits 0-1 and x-1. What exactly do you do when you say that you sue symmetry? – dreamer Feb 23 '13 at 12:02
And wouldnt it be possible to evaluate the second integral using Taylor series? – dreamer Feb 23 '13 at 12:09
Oh and last but not least I also dont understand why you integrate the function e^(x^2) only and not for instance e^(y^2). – dreamer Feb 23 '13 at 12:11

Hint: Split $D$ into two triangles, according to $x\ge y$ or $y\ge x$, and calculate both integrals on these triangles.

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I think I get your idea, however Im still not entirely sure how to do this in practice. Which function would I then integrate? Could you please show me how this works? – dreamer Feb 23 '13 at 12:03
It's the same as Hagen's answer. One triangle will have those $(x,y)$ points for which $0\le x\le 1$ and $x\le y\le 1$. – Berci Feb 23 '13 at 17:06

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