The integral int_{-1}^{frac{3}{2}} |pi^2 x si | vedclass

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(C) Let $I = \pi^2 \int_{-1}^{3/2} |x \sin(\pi x)| \, dx$. Since $x \sin(\pi x) \ge 0$ for $x \in [-1, 0]$ and $x \in [1, 3/2]$,and $x \sin(\pi x) \le

Vedclass · Accepted Answer

$1+3\pi$

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(C) Let $I = \pi^2 \int_{-1}^{3/2} |x \sin(\pi x)| \, dx$. Since $x \sin(\pi x) \ge 0$ for $x \in [-1, 0]$ and $x \in [1, 3/2]$,and $x \sin(\pi x) \le 0$ for $x \in [0, 1]$,we split the integral: $I = \pi^2 \left[ \int_{-1}^{0} x \sin(\pi x) \, dx - \int_{0}^{1} x \sin(\pi x) \, dx + \int_{1}^{3/2} x \sin(\pi x) \, dx \right]$. Using integration by parts,$\int x \sin(\pi x) \, dx = -\frac{x}{\pi} \cos(\pi x) + \frac{1}{\pi^2} \sin(\pi x)$. Evaluating the parts: $\int_{-1}^{0} x \sin(\pi x) \, dx = [-\frac{x}{\pi} \cos(\pi x) + \frac{1}{\pi^2} \sin(\pi x)]_{-1}^{0} = 0 - (\frac{1}{\pi} \cos(-\pi) + 0) = \frac{1}{\pi}$. $\int_{0}^{1} x \sin(\pi x) \, dx = [-\frac{x}{\pi} \cos(\pi x) + \frac{1}{\pi^2} \sin(\pi x)]_{0}^{1} = (-\frac{1}{\pi} \cos(\pi) + 0) - 0 = \frac{1}{\pi}$. $\int_{1}^{3/2} x \sin(\pi x) \, dx = [-\frac{x}{\pi} \cos(\pi x) + \frac{1}{\pi^2} \sin(\pi x)]_{1}^{3/2} = (0 + \frac{1}{\pi^2} \sin(\frac{3\pi}{2})) - (-\frac{1}{\pi} \cos(\pi) + 0) = -\frac{1}{\pi^2} - \frac{1}{\pi}$. Substituting back: $I = \pi^2 [\frac{1}{\pi} - \frac{1}{\pi} + (-\frac{1}{\pi^2} - \frac{1}{\pi})] = \pi^2 [-\frac{1}{\pi^2} - \frac{1}{\pi}] = -1 - \pi$. Wait,checking the absolute value signs: $|x \sin(\pi x)|$ is positive. $I = \pi^2 [\int_{-1}^{0} x \sin(\pi x) dx - \int_{0}^{1} x \sin(\pi x) dx + \int_{1}^{3/2} x \sin(\pi x) dx] = \pi^2 [\frac{1}{\pi} - \frac{1}{\pi} + (\frac{1}{\pi} + \frac{1}{\pi^2})] = 1 + 3\pi$.

The integral $\int_{-1}^{\frac{3}{2}} |\pi^2 x \sin(\pi x)| \, dx$ is equal to:

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