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

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$\frac{3}{π} - \frac{1}{π^{2}}$

$\frac{3}{π} + \frac{1}{π^{2}}$

$\frac{3}{π} + \frac{1}{π}$

$\frac{2}{π} + \frac{1}{π^{2}}$

answer is B.

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Detailed Solution

$| x \sin π x | = \{\begin{cases} (- x) (- \sin π x) if - 1 \leq x < 0 \\ x \sin π x if 0 < x \leq 1 \\ x (- \sin π x) if 1 < x \leq 3 / 2 \end{cases}$

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

$\begin{aligned} = \int_{- 1}^{1} x \sin π x d x - \int_{1}^{3 / 2} x \sin π x d x \\ = 2 \int_{0}^{1} x \sin π x d x - \int_{1}^{3 / 2} x \sin π x d x \\ = 2 [{\{x (\frac{- 1}{π}) \cos π x\}}_{0}^{i} - \int_{0}^{1} 1 (\frac{- 1}{π}) \cos π x d x] \\ - {\{x (\frac{- 1}{π}) \cos π x\}}_{1}^{3 / 2} + \int_{1}^{3 / 2} 1 (\frac{- 1}{π}) \cos π x d x \\ = (\frac{2}{π}) + (\frac{2}{π^{2}}) [\sin π x]_{0}^{1} + \{\frac{3}{(2 π)}\} \cos \frac{3}{2} π + (\frac{1}{π}) \\ - (\frac{1}{π^{2}}) [\sin π x]^{3} \\ = (2 / π) + 0 + 0 + (1 / π) + (1 / π^{2}) \end{aligned}$