intlimits_{ - 1}^{frac{3}{2}} {|xsin pi x|dx} | vedclass

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Question

Hence,

$|x \sin (\pi x)|=\left\{\begin{array}{cc}x \sin (\pi x) & -1 \leq x \leq 1 \ -x \sin (\pi x) & 1 \leq x \leq \frac{3}{2}\end{array

Vedclass · Accepted Answer

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

Vedclass · Answer

Hence,

$|x \sin (\pi x)|=\left\{\begin{array}{cc}x \sin (\pi x) & -1 \leq x \leq 1 \ -x \sin (\pi x) & 1 \leq x \leq \frac{3}{2}\end{array}ight.$

Therefore, we can write

$	herefore \int_{-1}^{\frac{3}{2}}|x \sin (\pi x)| d x$

$=\int_{-1}^{1} x \sin (\pi x) d x-\int_{1}^{\frac{3}{2}} x \sin (\pi x) d x$

Solving $\int x \sin (\pi x) d x$ separately

$\int x \sin (\pi x) d x$

Using by parts

$\int f(x) g(x) d x=f(x) \int g(x) d x-\int\left(f^{\prime}(x) \int g(x) d xight) d x$

Putting $f(x)=x$ and $g(x)=\sin (\pi x)$

$=x \int \sin (\pi x)-\int\left(\frac{d(x)}{d x} \int \sin (\pi x)ight) d x$

$=x \frac{(-\cos (\pi x))}{\pi}-\int 1\left(\frac{-\cos (\pi x)}{\pi}ight) d x$

$=\frac{-x \cos (\pi x)}{\pi}+\int \frac{\cos (\pi x)}{\pi} d x$

$=\frac{-x \cos (\pi x)}{\pi}+\frac{\sin (\pi x)}{\pi^{2}}$

$\int_{-1}^{1} x \sin \pi x d x$

$=\left[\frac{-x \cos (\pi x)}{\pi}+\frac{\sin (\pi x)}{\pi^{2}}ight]_{-1}^{1}$

$=\left(\frac{-1 \cos \pi}{\pi}+\frac{\sin \pi}{\pi^{2}}ight)-\left(\frac{-(-1) \cos (-\pi)}{\pi}+\frac{\sin (-\pi)}{\pi^{2}}ight)$

$=\left(\frac{-1 	imes(-1)}{\pi}+\frac{0}{\pi^{2}}ight)-\left(\frac{\cos \pi}{\pi}+\frac{-\sin \pi}{\pi^{2}}ight)$

$=\left(\frac{1}{\pi}+0ight)-\left(\frac{-1}{\pi}+0ight)$

$=\frac{1}{\pi}+\frac{1}{\pi}$

$=\frac{2}{\pi}$

$\int_{1}^{\frac{3}{2}} x \sin \pi x d x$

$=\left[\frac{-x \cos (\pi x)}{\pi}+\frac{\sin (\pi x)}{\pi^{2}}ight]_{1}^{\frac{3}{2}}$

$=\left(\frac{\frac{-3}{2} \cos \left(\frac{3 \pi}{2}ight)}{\pi}+\frac{\sin \left(\frac{3 \pi}{2}ight)}{\pi^{2}}ight)-\left(\frac{-1 \cos \pi}{\pi}+\frac{\sin \pi}{\pi^{2}}ight)$

$=\left(0+\frac{(-1)}{\pi^{2}}ight)-\left(\frac{-1 	imes-1}{\pi}+\frac{0}{\pi^{2}}ight)$

$=\frac{-1}{\pi^{2}}-\frac{1}{\pi}$

Now,

$	herefore \int_{-1}^{\frac{3}{2}}|x \sin (\pi x)| d x$

$=\int_{-1}^{1} x \sin (\pi x) d x-\int_{1}^{\frac{3}{2}} x \sin (\pi x) d x$

$=\frac{2}{\pi}-\left(\frac{-1}{\pi^{2}}-\frac{1}{\pi}ight)$

$=\frac{2}{\pi}+\frac{1}{\pi^{2}}+\frac{1}{\pi}$

$=\frac{3}{\pi}+\frac{1}{\pi^{2}}$

$\int\limits_{ - 1}^{\frac{3}{2}} {|x\sin \pi x|dx} $ equals

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