int_{-2}^{2} |x cos pi x| , dx is equal to | vedclass

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Question

(A) Let $I = \int_{-2}^{2} |x \cos \pi x| \, dx$. Since $f(x) = |x \cos \pi x|$ is an even function,we have $I = 2 \int_{0}^{2} |x \cos \pi x| \, dx$.

Vedclass · Accepted Answer

$ \frac{8}{\pi} $

Vedclass · Answer

(A) Let $I = \int_{-2}^{2} |x \cos \pi x| \, dx$. Since $f(x) = |x \cos \pi x|$ is an even function,we have $I = 2 \int_{0}^{2} |x \cos \pi x| \, dx$.The function $x \cos \pi x$ changes sign at $x = \frac{1}{2}$ and $x = \frac{3}{2}$ in the interval $[0, 2]$.Thus,$I = 2 \left[ \int_{0}^{1/2} x \cos \pi x \, dx - \int_{1/2}^{3/2} x \cos \pi x \, dx + \int_{3/2}^{2} x \cos \pi x \, dx \right]$.Using integration by parts,$\int x \cos \pi x \, dx = \frac{x \sin \pi x}{\pi} + \frac{\cos \pi x}{\pi^2}$.Evaluating the integrals:$1$. $\int_{0}^{1/2} x \cos \pi x \, dx = \left[ \frac{x \sin \pi x}{\pi} + \frac{\cos \pi x}{\pi^2} \right]_{0}^{1/2} = (\frac{1}{2\pi} + 0) - (0 + \frac{1}{\pi^2}) = \frac{1}{2\pi} - \frac{1}{\pi^2}$.$2$. $\int_{1/2}^{3/2} x \cos \pi x \, dx = \left[ \frac{x \sin \pi x}{\pi} + \frac{\cos \pi x}{\pi^2} \right]_{1/2}^{3/2} = (-\frac{3}{2\pi} + 0) - (\frac{1}{2\pi} + 0) = -\frac{2}{\pi}$.$3$. $\int_{3/2}^{2} x \cos \pi x \, dx = \left[ \frac{x \sin \pi x}{\pi} + \frac{\cos \pi x}{\pi^2} \right]_{3/2}^{2} = (0 + \frac{1}{\pi^2}) - (-\frac{3}{2\pi} + 0) = \frac{3}{2\pi} + \frac{1}{\pi^2}$.Substituting these values: $I = 2 [(\frac{1}{2\pi} - \frac{1}{\pi^2}) - (-\frac{2}{\pi}) + (\frac{3}{2\pi} + \frac{1}{\pi^2})] = 2 [\frac{1}{2\pi} + \frac{2}{\pi} + \frac{3}{2\pi}] = 2 [\frac{4}{\pi}] = \frac{8}{\pi}$.

$ \int_{-2}^{2} |x \cos \pi x| \, dx $ is equal to

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