Question

The solution for ${\int }_{-1}^{\frac{3}{2}}|x\sin \pi x|dx$, is

• A. $\frac{3}{\pi }-\frac{1}{{\pi }^2}$
• B. $\frac{3}{\pi }+\frac{1}{{\pi }^2}$
• C. $\frac{2}{\pi }+\frac{1}{{\pi }^2}$
• D. $\frac{3}{\pi }+\frac{1}{\pi }$

Answer 1

Answer: (B)

$\frac{3}{\pi }+\frac{1}{{\pi }^2}$

Let $I={\int }_{-1}^{\frac{3}{2}}\left|x\sin \left(\pi x\right)\right|dx={\int }_{-1}^0\left|x\sin \left(\pi x\right)\right|dx+{\int }_0^1\left|x\sin \left(\pi x\right)\right|dx+{\int }_1^{\frac{3}{2}}\left|x\sin \left(\pi x\right)\right|dx$
For,
$-1<x<0;-\pi <\pi x<0\Rightarrow x\sin \left(\pi x\right)>0$
$0<x<1;0<\pi x<\pi \Rightarrow x\sin \left(\pi x\right)>0$
$1<x<\frac{3}{2};\pi <\pi x<\frac{3\pi }{2}\Rightarrow x\sin \left(\pi x\right)<0$
Therefore, on integrating,
$I={\int }_{-1}^{0}x\sin \left(\pi x\right)dx+{\int }_0^{1}x\sin \left(\pi x\right)dx-{\int }_1^{\frac{3}{2}}x\sin \left(\pi x\right)dx$
Let, $\int x\sin \left(\pi x\right)=-\frac{x}{\pi }\cos \left(\pi x\right)+\int \frac{\cos \left(\pi x\right)}{\pi }dx=-\frac{x\cos \left(\pi x\right)}{\pi }+\frac{\sin \left(\pi x\right)}{{\pi }^2}$
$I={[-\frac{x\cos \left(\pi x\right)}{\pi }+\frac{\sin \left(\pi x\right)}{{\pi }^2}]}_{-1}^0+{[-\frac{x\cos \left(\pi x\right)}{\pi }+\frac{\sin \left(\pi x\right)}{{\pi }^2}]}_0^1-{[-\frac{x\cos \left(\pi x\right)}{\pi }+\frac{\sin \left(\pi x\right)}{{\pi }^2}]}_1^{\frac{3}{2}}$

$=(0+0+\frac{1}{\pi }+0)+(\frac{1}{\pi }-0)-(-\frac{1}{{\pi }^2}-\frac{1}{\pi }+0)=\frac{3}{\pi }+\frac{1}{{\pi }^2}$