Evaluate $\displaystyle \int_{-1}^{3/2} |x \sin(\pi x)| \, dx$

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

The correct answer is Option (2) → $\frac{3}{\pi} + \frac{1}{\pi^2}$

Here $f(x) = |x \sin \pi x| = \begin{cases} x \sin \pi x & \text{for } -1 \leq x \leq 1 \\ -x \sin \pi x & \text{for } 1 \leq x \leq \frac{3}{2} \end{cases}$

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

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

Integrating both integrals on righthand side, we get

$\int_{-1}^{3/2} |x \sin \pi x| \, dx = \left[ \frac{-x \cos \pi x}{\pi} + \frac{\sin \pi x}{\pi^2} \right]_{-1}^{1} - \left[ \frac{-x \cos \pi x}{\pi} + \frac{\sin \pi x}{\pi^2} \right]_{1}^{3/2}$

$= \frac{2}{\pi} - \left[ -\frac{1}{\pi^2} - \frac{1}{\pi} \right] = \frac{3}{\pi} + \frac{1}{\pi^2}$