Question

The value of $\underset{-1}{\overset{3/2}{\int }}|x\sin \pi x|dx=\frac{k\pi +1}{{\pi }^m}$, then $k+m=$

Answer 1

$I=\underset{-1}{\overset{3/2}{\int }}|x\sin \pi x|dx\Rightarrow I=\underset{-1}{\overset{0}{\int }}(-x)(-\sin \pi x)dx+\underset{0}{\overset{1}{\int }}x\sin \pi xdx+\underset{1}{\overset{3/2}{\int }}x(-\sin \pi x)dx\Rightarrow I=\underset{-1}{\overset{1}{\int }}x\sin \pi xdx-\underset{1}{\overset{3/2}{\int }}x\sin \pi xdx$
$I_1=\underset{-1}{\overset{1}{\int }}x\sin \pi xdx\cdots \left(1\right)\Rightarrow I_1=2\underset{0}{\overset{1}{\int }}x\sin \pi xdx\Rightarrow I_1=2\underset{0}{\overset{1}{\int }}(1-x)\sin \pi xdx\cdots \left(2\right)$
Adding (1) and (2),
$\Rightarrow I_1=\underset{0}{\overset{1}{\int }}\sin \pi xdx\Rightarrow I_1=-{\left[\frac{\cos \pi x}{\pi }\right]}_0^1\Rightarrow I_1=\frac{2}{\pi }$
$I_2=\underset{1}{\overset{3/2}{\int }}x\sin \pi xdx\Rightarrow I_2={\left[\frac{-x\cos \pi x}{\pi }\right]}_1^{3/2}+\underset{1}{\overset{3/2}{\int }}\frac{\cos \pi x}{\pi }dx\Rightarrow I_2=-\frac{1}{\pi }+{\left[\frac{\sin \pi x}{{\pi }^2}\right]}_1^{3/2}\Rightarrow I_2=-\frac{1}{\pi }-\frac{1}{{\pi }^2}$
Therefore,
$I=\frac{2}{\pi }+\frac{1}{\pi }+\frac{1}{{\pi }^2}\Rightarrow I=\frac{k\pi +1}{{\pi }^m}=\frac{3\pi +1}{{\pi }^2}\Rightarrow k+m=5$