The integral _ -1 ^ 3 2 ( | ^ 2 x ( x) | ) d x is equal to : - Vidyadip

MCQ

The integral $\int_{-1}^{\frac{3}{2}}\left(\left|\pi^{2} x \sin (\pi x)\right|\right) d x$ is equal to :

A
$3+2 \pi$
B
$4+\pi$
✓
$1+3 \pi$
D
$2+3 \pi$

✓

Answer

Correct option: C.

$1+3 \pi$

(C) $1+3 \pi$
Let, $\mathrm{I}=\pi^{2} \int_{-1}^{3 / 2}|\mathrm{x} \sin \pi \mathrm{x}| \mathrm{dx}$
$=\pi^2\left\{\int_{-1}^1 x \sin \pi xdx -\int_1^{3 / 2} x \sin \pi xdx \right\}$
$=\pi^2\left\{2 \int_0^1 x \sin \pi xdx -\int_{-1}^{3 / 2} x \sin \pi xdx \right\}$
Consider
$\int x \sin \pi x d x$
$-\mathrm{x} \cdot \frac{1}{\pi} \cos \pi \mathrm{x}+\int 1 \cdot \frac{1}{\pi} \cos \pi \mathrm{xdx}$
$=-\frac{\mathrm{x}}{\pi} \cos \pi \mathrm{x}+\frac{\sin \pi \mathrm{x}}{\pi^{2}}$
$\mathrm{I}=\pi^{2}\left\{2\left(-\frac{\mathrm{x}}{\pi} \cos \pi \mathrm{x}+\frac{\sin \pi \mathrm{x}}{\pi^{2}}\right)_{0}^{1}-\left(-\frac{\mathrm{x}}{\pi} \cos \pi \mathrm{x}+\frac{\sin \pi \mathrm{x}}{\pi^{2}}\right)_{1}^{3 / 2}\right\}$
$=\pi^{2}\left\{\frac{2}{\pi}-\left(-\frac{1}{\pi^{2}}-\frac{1}{\pi}\right)\right\}$
$=\pi^{2}\left\{\frac{3}{\pi}+\frac{1}{\pi^{2}}\right\}$
$=3 \pi+1$

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