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\( \int_{-2}^{2}|x \cos \pi x| d x \) is equal to
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Verified Answer
The correct answer is:
\( \frac{8}{\pi} \)
Given that,
\[
\begin{array}{l}
\int_{-2}^{2}|x \cos \Pi x| d x=2 \int_{0}^{2}|x \cos x| d x \\
=2\left[\int_{0}^{1 / 2} x \cos \pi x-\int_{1 / 2}^{3 / 2}(x \cos \Pi x) d x+\int_{3 / 2}^{2}(x \cos \Pi x) d x\right] \\
=2\left[\frac{1}{2 \Pi}-\frac{1}{\Pi^{2}}+\frac{4}{2 \Pi}+\frac{3}{2 \Pi}+\frac{1}{\Pi^{2}}\right] \\
=2\left(\frac{8}{2 \Pi}\right)=\frac{8}{\Pi}
\end{array}
\]
\[
\begin{array}{l}
\int_{-2}^{2}|x \cos \Pi x| d x=2 \int_{0}^{2}|x \cos x| d x \\
=2\left[\int_{0}^{1 / 2} x \cos \pi x-\int_{1 / 2}^{3 / 2}(x \cos \Pi x) d x+\int_{3 / 2}^{2}(x \cos \Pi x) d x\right] \\
=2\left[\frac{1}{2 \Pi}-\frac{1}{\Pi^{2}}+\frac{4}{2 \Pi}+\frac{3}{2 \Pi}+\frac{1}{\Pi^{2}}\right] \\
=2\left(\frac{8}{2 \Pi}\right)=\frac{8}{\Pi}
\end{array}
\]
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