What is the Integral of e to the Maximum Power?

[Ackbach]

Here is this week's POTW:

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Evaluate the integral
$$\int_0^1 \int_0^1 e^{\max(x^2, \, y^2)} \, dy \, dx.$$

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Ackbach]

An honorable mention goes to kiwi for a valiant attempt, but not quite correct. Here is my solution:

Spoiler

\begin{align*}
\int_0^1 \int_0^1 e^{\max(x^2, \, y^2)} \, dy \, dx
&=\int_0^1\left[\underbrace{\int_0^x e^{\max(x^2, \, y^2)} \, dy}_{
y<x}
+\underbrace{\int_x^1 e^{\max(x^2, \, y^2)} \, dy}_{y>x} \right] dx \\
&=\int_0^1\left[\int_0^x e^{x^2} \, dy+\int_x^1 e^{y^2} \, dy \right] dx
\\
&=\int_0^1 x \, e^{x^2} \, dx+\int_0^1\int_x^1 e^{y^2} \, dy \, dx \\
&=\int_0^1 e^u \, \frac{du}{2}+\int_0^1 \int_0^y e^{y^2} \, dx \, dy \\
&=\frac{e^u}{2} \Bigg|_0^1+\int_0^1 y \, e^{y^2} \, dy \\
&=\left(\frac{e}{2}-\frac12\right)+\left(\frac{e}{2}-\frac12\right) \\
&=e-1.
\end{align*}