Question

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

Answer 1

We have,

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

Using $ILATE$ and we get,

$\int x.ydx=x\int ydx-\int (\int \frac{dx}{dx}ydx)dx+C$

$I=x{\int }_{-1}^{\frac{1}{2}}\sin \pi xdx-{\int }_{-1}^{\frac{1}{2}}\left({\int }_{-1}^{\frac{1}{2}}\left(\frac{dx}{dx}\sin \pi x\right)dx\right)dx$

$I=x{{\left[\frac{\cos \pi x}{\pi }\right]}_{-1}}^{\frac{1}{2}}-{\int }_{-1}^{\frac{1}{2}}\frac{\left(\cos \pi x\right)}{\pi }dx+C$

$I=x{{\left[\frac{\cos \pi x}{\pi }\right]}_{-1}}^{\frac{1}{2}}+{{\left[\frac{\left(\sin \pi x\right)}{{\pi }^2}\right]}_{-1}}^{\frac{1}{2}}+C$

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

$I=\frac{(\frac{1}{2}-1)}{\pi }[\cos \frac{\pi }{2}-\cos (-\pi )]+\frac{1}{{\pi }^2}[\sin \frac{\pi }{2}-\sin (-\pi )]+C$

$I=-\frac{1}{2\pi }[0-1]+\frac{1}{{\pi }^2}[1+0]\therefore \cos (-\theta )=\cos \theta ,\sin (-\theta )=-\sin \theta$

$I=\frac{1}{2\pi }+\frac{1}{{\pi }^2}$

Hence, this is the answer.