## Thomas' Calculus 13th Edition

$4$
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 $f(x)=(-2) \cdot (-1)+(-2) \cdot (1)+\dfrac{1}{2}-\dfrac{1}{2}=4$