Answer
$4$
Work Step by Step
Over the interval $x\in[0,\pi]$,
the graph of $f(x)=2\sin x$ is above the graph of $g(x)=\sin 2x$.
$A=\displaystyle \int_{a}^{b}(y_{above}-y_{below})dx=\int_{0}^{\pi}[2\sin x-\sin 2x]dx$.
$=2\displaystyle \int_{0}^{\pi}\sin xdx -\int_{0}^{\pi}\sin 2xdx\qquad \left[\begin{array}{ll}
u=2x & du=2dx\\
x=0\Rightarrow u=0 & \\
x=\pi\Rightarrow u=2\pi &
\end{array}\right]$
$=2\displaystyle \left[-\cos x \right]_{0}^{\pi}-\int_{0}^{2\pi}\sin u(\frac{du}{2})$
$=-2(-1-1)-\displaystyle \frac{1}{2}\left[-\cos x \right]_{0}^{2\pi}$
$=4-0$
$=4$
