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$\int_{-1}^{\frac{3}{2}}|x \sin (\pi x)| d x=$
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Verified Answer
The correct answer is:
$\frac{3}{\pi}+\frac{1}{\pi^2}$
Let
$$
\begin{gathered}
I=\int_{-1}^{3 / 2}|x \sin (\pi x)| d x \\
I=\int_{-1}^1 x \sin \pi x d x-\int_1^{3 / 2} x \sin \pi x d x \\
I=2 \int_0^1 x \sin \pi x d x-\int_1^{3 / 2} x \sin \pi x d x \\
I=2\left[\frac{-x \cos \pi x}{\pi}+\frac{\sin \pi x]_0^1}{\pi^2}\right]_0^{-\left[\frac{-x \cos \pi x}{\pi}+\frac{\sin \pi x}{\pi^2}\right]_1^{3 / 2}} \\
I=2\left[\left(\frac{1}{\pi}+0\right)-(0)\right]-\left[0-\frac{1}{\pi^2}-\frac{1}{\pi}\right] \\
I=\frac{3}{\pi}+\frac{1}{\pi^2}
\end{gathered}
$$
$$
\begin{gathered}
I=\int_{-1}^{3 / 2}|x \sin (\pi x)| d x \\
I=\int_{-1}^1 x \sin \pi x d x-\int_1^{3 / 2} x \sin \pi x d x \\
I=2 \int_0^1 x \sin \pi x d x-\int_1^{3 / 2} x \sin \pi x d x \\
I=2\left[\frac{-x \cos \pi x}{\pi}+\frac{\sin \pi x]_0^1}{\pi^2}\right]_0^{-\left[\frac{-x \cos \pi x}{\pi}+\frac{\sin \pi x}{\pi^2}\right]_1^{3 / 2}} \\
I=2\left[\left(\frac{1}{\pi}+0\right)-(0)\right]-\left[0-\frac{1}{\pi^2}-\frac{1}{\pi}\right] \\
I=\frac{3}{\pi}+\frac{1}{\pi^2}
\end{gathered}
$$
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