Answer
The area of the shaded region is $2$.
Work Step by Step
The shaded region is bounded above by the curve $y=(1-\cos x)\sin x$ and below by the line $y=0$ from the point $x=0$ to $x=\pi$.
So the area of the region would be $$A=\int^\pi_0\Big((1-\cos x)\sin x-0\Big)dx=\int^\pi_0(1-\cos x)\sin xdx$$
We set $u=1-\cos x$, which means $$du=-(-\sin x)dx=\sin xdx$$
- For $x=\pi$, we have $u=1-\cos\pi=1-(-1)=2$
- For $x=0$, we have $u=1-\cos0=1-1=0$
Therefore, $$A=\int^2_0udu=\frac{u^2}{2}\Big]^2_0$$ $$A=\frac{1}{2}(2^2-0^2)$$ $$A=\frac{1}{2}(4)=2$$
