$\displaystyle Centroid = \frac {\int_0^h y\frac {b}{h}(h-y)dy}{\int_0^h \frac {b}{h}(h-y)dy}$

I have the answer which is $\displaystyle \frac {\frac {1}{6}bh^2}{\frac {1}{2}bh}$ which produces $\displaystyle \frac {h}{3}$ as you may know that is a triangle.

However I cant get the same answer, infact I appear to be way off. Can anyone suggest how to integrate this problem? maybe even just the top would be great help.

Thanks to those who can help.