## Thomas' Calculus 13th Edition

$2$
On the interval $[0,\pi]$, the graph of $y=(1-\cos x)\sin x$ is above the graph of $y=0$. The area between the two graphs over [a,b], where one graph is above the other, is given with $\displaystyle \int_{a}^{b}(y_{above}-y_{below})dx$. $A=\displaystyle \int_{0}^{\pi}[(1-\cos x)\sin x -0]dx=\int_{0}^{\pi}(1-\cos x)\sin xdx=$ Apply the substitution: $\left[\begin{array}{ll} u=1-\cos x & du=\sin xdx\\ Borders: & \\ x=0\rightarrow u=0 & x=\pi\rightarrow u=2 \end{array}\right]$ $=\displaystyle \int_{0}^{2}udu$ $=[\displaystyle \frac{u^{2}}{2}]_{0}^{2}$ $=\displaystyle \frac{2^{2}}{2}-0$ $=2$