Answer
$\displaystyle \frac{1}{2}\pi-\frac{4}{3}$
Work Step by Step
$A=\displaystyle \int_{0}^{\pi}(\sin^{2}x-\sin^{3}x)dx$
$=\displaystyle \int_{0}^{\pi}[\frac{1}{2}(1-\cos 2x)-\sin x(1-\cos^{2}x)]dx$
$=\displaystyle \int_{0}^{\pi}\left(\frac{1}{2}-\frac{1}{2}\cos 2x\right)dx-\int_{0}^{\pi}\sin x(1-\cos^{2}x)]dx$
$=\displaystyle \left(\frac{1}{2}\pi-0\right)-(0-0)+\int_{1}^{-1}\left(1-u^{2}\right)du\quad $
$\left[\begin{array}{c}{u=\cos x}\\{du=-\sin xdx}\end{array}\right]$
$=\displaystyle \left(\frac{1}{2}\pi-0\right)-(0-0)+2\left[\frac{1}{3}u^{3}-u\right]_{0}^{1}$
$=\displaystyle \frac{1}{2}\pi+2\left(\frac{1}{3}-1\right)$
$=\displaystyle \frac{1}{2}\pi-\frac{4}{3}$
