Thread: Simple integration

\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.

Can you write this in LaTex? I can't seem to follow what you have there.

Originally Posted by xiongs8421

\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?