Question

If $I={\int }_{-1}^2\left|x\sin \pi x\right|dx,$ then $I$ equals

• A. $1/\pi$
• B. $2/\pi$
• C. $4/\pi$
• D. $5/\pi$

Answer 1

Answer: (D)

$5/\pi$

We can write
$I={\int }_{-1}^1\left|x\sin \pi x\right|dx+{\int }_1^2\left|x\sin \pi x\right|dx$
As
$\left|x\sin \pi x\right|$ is an even function, $\sin \pi x\ge 0$, for $0\le \pi x\le \pi$ and $\sin \pi x\le 0$ for
$\pi x\le 2\pi$ we get
$I=2{\int }_0^{1}x\sin \pi xdx-{\int }_1^{2}x\sin \pi xdx$
Now $\int x\sin \pi xdx=\frac{-x\cos \pi x}{\pi }+\frac{1}{\pi }\int \cos \pi xdx$
$=-\frac{x}{\pi }\cos \pi x+\frac{1}{{\pi }^2}\sin \pi x$
Thus,
$I=2(-\frac{1}{\pi }\cos \pi +0)-(-\frac{2}{\pi }\cos 2\pi +\frac{1}{\pi }\cos \pi )$ $=\frac{5}{\pi }$