7.2 definite integral — Maths STD 12 Science — Question

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

Therefore, we can write

$\therefore \int_{-1}^{\frac{3}{2}}|x \sin (\pi x)| d x$

$=\int_{-1}^{1} x \sin (\pi x) d x-\int_{1}^{\frac{3}{2}} x \sin (\pi x) d x$

Solving $\int x \sin (\pi x) d x$ separately

$\int x \sin (\pi x) d x$

Using by parts

$\int f(x) g(x) d x=f(x) \int g(x) d x-\int\left(f^{\prime}(x) \int g(x) d x\right) d x$

Putting $f(x)=x$ and $g(x)=\sin (\pi x)$

$=x \int \sin (\pi x)-\int\left(\frac{d(x)}{d x} \int \sin (\pi x)\right) d x$

$=x \frac{(-\cos (\pi x))}{\pi}-\int 1\left(\frac{-\cos (\pi x)}{\pi}\right) d x$

$=\frac{-x \cos (\pi x)}{\pi}+\int \frac{\cos (\pi x)}{\pi} d x$

$=\frac{-x \cos (\pi x)}{\pi}+\frac{\sin (\pi x)}{\pi^{2}}$

$\int_{-1}^{1} x \sin \pi x d x$

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

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

$=\left(\frac{-1 \times(-1)}{\pi}+\frac{0}{\pi^{2}}\right)-\left(\frac{\cos \pi}{\pi}+\frac{-\sin \pi}{\pi^{2}}\right)$

$=\left(\frac{1}{\pi}+0\right)-\left(\frac{-1}{\pi}+0\right)$

$=\frac{1}{\pi}+\frac{1}{\pi}$

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

$\int_{1}^{\frac{3}{2}} x \sin \pi x d x$

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

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

$=\left(0+\frac{(-1)}{\pi^{2}}\right)-\left(\frac{-1 \times-1}{\pi}+\frac{0}{\pi^{2}}\right)$

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

Now,

$\therefore \int_{-1}^{\frac{3}{2}}|x \sin (\pi x)| d x$

$=\int_{-1}^{1} x \sin (\pi x) d x-\int_{1}^{\frac{3}{2}} x \sin (\pi x) d x$

$=\frac{2}{\pi}-\left(\frac{-1}{\pi^{2}}-\frac{1}{\pi}\right)$

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

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