\int_{0}^{1}\int_{0}^{1} min(x,y) dxdy
= \int_{0}^{1}\int_{y}^{1} y dxdy
= \int_{0}^{1} y(1-y)dy
= 1/6
What's wrong with it?
Thank you.
On the square you can write the function
$\displaystyle \text{min}(x,y)=\begin{cases} x, \qaud \text{ if } y \ge x \\ y, \qaud \text{ if } y \le x \end{cases}$
So you can break this into two different integrals over each triangle. Or you can do one of the integrals and multiply it by 2. Why does this work?