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



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?