int_{-1}^{3/2} |x sin pi x| , dx = | vedclass

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(B) We need to evaluate $I = \int_{-1}^{3/2} |x \sin \pi x| \, dx$. Since $x \sin \pi x \ge 0$ for $x \in [-1, 0]$ and $x \sin \pi x \ge 0$ for $x \in

Vedclass · Accepted Answer

$\frac{3}{\pi} + \frac{1}{\pi^2}$

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(B) We need to evaluate $I = \int_{-1}^{3/2} |x \sin \pi x| \, dx$. Since $x \sin \pi x \ge 0$ for $x \in [-1, 0]$ and $x \sin \pi x \ge 0$ for $x \in [0, 1]$,but $x \sin \pi x \le 0$ for $x \in [1, 3/2]$,we split the integral: $I = \int_{-1}^{1} x \sin \pi x \, dx - \int_{1}^{3/2} x \sin \pi x \, dx$. Using integration by parts $\int u \, dv = uv - \int v \, du$,let $u = x$ and $dv = \sin \pi x \, dx$,so $du = dx$ and $v = -\frac{\cos \pi x}{\pi}$. $\int x \sin \pi x \, dx = -\frac{x \cos \pi x}{\pi} + \frac{1}{\pi} \int \cos \pi x \, dx = -\frac{x \cos \pi x}{\pi} + \frac{\sin \pi x}{\pi^2}$. Evaluating the first part: $\int_{-1}^{1} x \sin \pi x \, dx = [-\frac{x \cos \pi x}{\pi} + \frac{\sin \pi x}{\pi^2}]_{-1}^{1} = (-\frac{1 \cdot (-1)}{\pi} + 0) - (-\frac{-1 \cdot (-1)}{\pi} + 0) = \frac{1}{\pi} - \frac{1}{\pi} = 0$ is incorrect; let's re-evaluate: $[-\frac{1 \cdot (-1)}{\pi} + 0] - [-\frac{-1 \cdot (-1)}{\pi} + 0] = \frac{1}{\pi} - (-\frac{1}{\pi}) = \frac{2}{\pi}$. Evaluating the second part: $\int_{1}^{3/2} x \sin \pi x \, dx = [-\frac{x \cos \pi x}{\pi} + \frac{\sin \pi x}{\pi^2}]_{1}^{3/2} = (0 + \frac{\sin(3\pi/2)}{\pi^2}) - (-\frac{1 \cdot (-1)}{\pi} + 0) = -\frac{1}{\pi^2} - \frac{1}{\pi}$. Thus,$I = \frac{2}{\pi} - (-\frac{1}{\pi^2} - \frac{1}{\pi}) = \frac{2}{\pi} + \frac{1}{\pi} + \frac{1}{\pi^2} = \frac{3}{\pi} + \frac{1}{\pi^2}$.

$\int_{-1}^{3/2} |x \sin \pi x| \, dx =$

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