intlimits_{ - 1}^{frac{3}{2}} {|xsin pi x|dx} | vedclass

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Question

(B) We define the function $f(x) = |x \sin(\pi x)|$. The expression $x \sin(\pi x)$ changes sign at $x = 0$ and $x = 1$. For $x \in [-1, 0]$,$x \leq 0

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) We define the function $f(x) = |x \sin(\pi x)|$. The expression $x \sin(\pi x)$ changes sign at $x = 0$ and $x = 1$. For $x \in [-1, 0]$,$x \leq 0$ and $\sin(\pi x) \leq 0$,so $x \sin(\pi x) \geq 0$. For $x \in [0, 1]$,$x \geq 0$ and $\sin(\pi x) \geq 0$,so $x \sin(\pi x) \geq 0$. For $x \in [1, 3/2]$,$x > 0$ and $\sin(\pi x) Thus,$|x \sin(\pi x)| = x \sin(\pi x)$ for $x \in [-1, 1]$ and $-x \sin(\pi x)$ for $x \in [1, 3/2]$. $\int_{-1}^{3/2} |x \sin(\pi x)| dx = \int_{-1}^{1} x \sin(\pi x) dx - \int_{1}^{3/2} x \sin(\pi x) dx$. Using integration by parts,$\int x \sin(\pi x) dx = -\frac{x \cos(\pi x)}{\pi} + \frac{\sin(\pi x)}{\pi^2} + C$. Evaluating the first part: $\int_{-1}^{1} x \sin(\pi x) dx = \left[ -\frac{x \cos(\pi x)}{\pi} + \frac{\sin(\pi x)}{\pi^2} \right]_{-1}^{1} = \left( -\frac{1 \cdot (-1)}{\pi} + 0 \right) - \left( -\frac{(-1) \cdot (-1)}{\pi} + 0 \right) = \frac{1}{\pi} - (-\frac{1}{\pi}) = \frac{2}{\pi}$. Evaluating the second part: $\int_{1}^{3/2} x \sin(\pi x) dx = \left[ -\frac{x \cos(\pi x)}{\pi} + \frac{\sin(\pi x)}{\pi^2} \right]_{1}^{3/2} = \left( 0 + \frac{\sin(3\pi/2)}{\pi^2} \right) - \left( -\frac{1 \cdot (-1)}{\pi} + 0 \right) = -\frac{1}{\pi^2} - \frac{1}{\pi}$. Total integral = $\frac{2}{\pi} - (-\frac{1}{\pi^2} - \frac{1}{\pi}) = \frac{2}{\pi} + \frac{1}{\pi^2} + \frac{1}{\pi} = \frac{3}{\pi} + \frac{1}{\pi^2}$.

$\int\limits_{ - 1}^{\frac{3}{2}} {|x\sin \pi x|dx} $ equals

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