Question

${\int }_{-1}^{3/2}\left|x\sin \pi x\right|dx$

Answer 1

We have,

${\int }_{-1}^{\frac{3}{2}}\left|x\sin \pi x\right|dx$

Applying rule,

$\int I.IIdx=I.\int IIdx-\int \frac{d}{dx}I\int IIdx$

Now,

${\int }_{-1}^{\frac{3}{2}}\left|x\sin \pi x\right|dx=x{\int }_{-1}^{\frac{3}{2}}\left|\sin \pi x\right|dx-{\int }_{-1}^{\frac{3}{2}}\frac{d}{dx}x{\int }_{-1}^{\frac{3}{2}}\left|\sin \pi x\right|dx$

$=x{{\left[|-\frac{\cos \pi x}{\pi }|\right]}_{-1}}^{\frac{3}{2}}+{{\left[\left|\frac{\sin \pi x}{\pi }\right|\right]}_{-1}}^{\frac{3}{2}}$

$=(\frac{3}{2}+1)\left(\frac{\cos \frac{3\pi }{2}-\cos (-\pi )}{\pi }\right)+\frac{1}{\pi }[\sin \frac{3}{2}\pi -\sin (-\pi )]$

$=\frac{5}{2\pi }[\cos (\pi +\frac{\pi }{2})-\cos \pi ]+\frac{1}{\pi }[\sin (\pi +\frac{\pi }{2})+\sin \pi ]\therefore \sin (-\theta )=-\sin \theta ,\cos (-\theta )=\cos \theta$

$=\frac{5}{2\pi }[-\cos \frac{\pi }{2}-\cos \pi ]+\frac{1}{\pi }[-\sin \frac{\pi }{2}+\sin \pi ]$

$=\frac{5}{2\pi }[-0-(-1)]+\frac{1}{\pi }[-1+0]$

$=\frac{5}{2\pi }\left[1\right]-\frac{1}{\pi }$

$=\frac{5}{2\pi }-\frac{1}{\pi }$

$=\frac{5-2}{2\pi }$

$=\frac{3}{2\pi }$

Hence, this is the answer.