Question

Evaluate the following integral ${\int }_{-1}^2|x\sin \pi x|dx$

Answer 1

${\int }_{-1}^2|xsin\pi x|dx={\int }_{-1}^1|x\sin \pi x|dx+{\int }_1^2|x.\sin \pi x|dx$

$=2{\int }_0^{1}x\sin \pi x|dx+{\int }_1^2|x.\sin \pi x|dx$

$=2{\int }_0^{1}x.\sin \pi xdx-{\int }_1^{2}x.\sin \pi xdx=2I_1-I_2$

$I_1={\int }_0^{1}x\sin \pi xdx={\left[\begin{array}{c}-x\frac{\cos \pi x}{\pi }\end{array}\right]}_0^1+{\int }_0^1\frac{\cos \pi x}{\pi }dx$

$={\left[\begin{array}{c}-x\frac{\cos \pi x}{\pi }+\frac{\sin \pi x}{{\pi }^2}\end{array}\right]}_0^1=\frac{1}{\pi }$

and $I_2={\int }_1^{2}x\sin \pi xdx={\left[\begin{array}{c}-x\frac{\cos \pi x}{\pi }+\frac{\sin \pi x}{{\pi }^2}\end{array}\right]}_1^2$

$=\frac{-2}{\pi }+0+\left(\begin{array}{c}-\frac{1}{\pi }\end{array}\right)=-\frac{3}{\pi }$

So, $2I_1-I_2=\frac{2}{\pi }+\frac{3}{\pi }=\frac{5}{\pi }$

$\Rightarrow {\int }_{-1}^2|x\sin \pi x|dx=\frac{5}{\pi }$