Answer
$4$
Work Step by Step
Consider $f(x)=\int_{0}^{\pi} (2\sin x-\sin 2x) dx$
Use formula such as: $\int x^{n} dx=\dfrac{x^{n+1}}{n+1}+C$
This implies that
$[-2 \cos x+\dfrac{\cos 2x}{2}]_0^{\pi}=-2[\cos \pi-\cos 0]+\dfrac{(\cos 2 \pi-\cos 0)}{2} $
Thus $f(x)=(-2) \cdot (-1)+(-2) \cdot (1)+\dfrac{1}{2}-\dfrac{1}{2}=4$
