Evaluate $\int_{-1}^{\frac{3}{2}}|x \sin (\pi x)| d x$

(A) Let $f(x) = |x \sin(\pi x)|$. The function $x \sin(\pi x)$ is non-negative on $[-1, 1]$ and non-positive on $[1, \frac{3}{2}]$.
Thus,$|x \sin(\pi x)| = \begin{cases} x \sin(\pi x) & \text{for } -1 \leq x \leq 1 \\ -x \sin(\pi x) & \text{for } 1 < x \leq \frac{3}{2} \end{cases}$
Therefore,$\int_{-1}^{\frac{3}{2}} |x \sin(\pi x)| dx = \int_{-1}^{1} x \sin(\pi x) dx - \int_{1}^{\frac{3}{2}} x \sin(\pi x) dx$.
Using integration by parts,$\int x \sin(\pi x) dx = -\frac{x \cos(\pi x)}{\pi} + \frac{\sin(\pi x)}{\pi^2}$.
Evaluating the first integral: $\left[ -\frac{x \cos(\pi x)}{\pi} + \frac{\sin(\pi x)}{\pi^2} \right]_{-1}^{1} = (-\frac{1 \cdot (-1)}{\pi} + 0) - (-\frac{-1 \cdot (-1)}{\pi} + 0) = \frac{1}{\pi} - (-\frac{1}{\pi}) = \frac{2}{\pi}$.
Evaluating the second integral: $\left[ -\frac{x \cos(\pi x)}{\pi} + \frac{\sin(\pi x)}{\pi^2} \right]_{1}^{\frac{3}{2}} = (0 + \frac{\sin(3\pi/2)}{\pi^2}) - (-\frac{1 \cdot (-1)}{\pi} + 0) = -\frac{1}{\pi^2} - \frac{1}{\pi}$.
Subtracting the two results: $\frac{2}{\pi} - (-\frac{1}{\pi^2} - \frac{1}{\pi}) = \frac{3}{\pi} + \frac{1}{\pi^2}$.