The integral – 1 3 / 2 ( | 2 x sin ( x ) | ) d x is equal to : [2025]

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Solution

Q.
The integral $\int_{- 1}^{3 / 2} (| π^{2} x \sin (π x) |) d x$ is equal to : [2025]

1 $3 + 2 π$

2 $4 + π$

3 $1 + 3 π$

4 $2 + 3 π$

Ans.
(3)

Let $I = \int_{- 1}^{3 / 2}$ $| π^{2} x \sin π x |$ $d x$

$= π^{2} [\int_{- 1}^{1} x \sin π x d x - \int_{1}^{3 / 2} x \sin π x d x]$

$= π^{2} [2 \int_{0}^{1} x \sin π x d x - \int_{1}^{3 / 2} x \sin π x d x]$

$= π^{2} [2 {[\frac{- x \cos π x}{π} + \frac{\sin π x}{π^{2}}]}_{0}^{1} - {[\frac{- x \cos π x}{π} + \frac{\sin π x}{π^{2}}]}_{1}^{3 / 2}]$

$= π^{2} [\frac{2}{π} - (\frac{- 1}{π^{2}} - \frac{1}{π})] = π^{2} [\frac{3}{π} + \frac{1}{π^{2}}]$

$= 3 π + 1$ .

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