JEE Main 2025 — 8 April, Evening Shift — Question 23

$I=\int_{-1}^{3 / 2}\left|\pi^{2} x \sin (\pi x)\right| d x$

$=\int_{-1}^{1}\left|\pi^{2} x \sin (\pi x)\right| d x+\int_{1}^{3 / 2}\left|\pi^{2} x \sin (\pi x)\right| d x$

$=2 \int_{0}^{1}\left|\pi^{2} x \sin (\pi x)\right| d x-\pi^{2} \int_{1}^{3 / 2}|x \sin (\pi x)| d x$

$=2 \pi^{2} \int_{0}^{1}|x \sin (\pi x)| d x-\pi^{2} \int_{1}^{3 / 2}|x \sin (\pi x)| d x$

$\because \int x \sin (\pi x) d x=x\left(\frac{-\cos \pi x}{\pi}\right)-\int \frac{-\cos \pi x}{\pi} d x$

$=-\frac{x}{\pi} \cos \pi x+\frac{1}{\pi^{2}} \sin \pi x+C$

$\therefore \quad I=2 \pi^{2}\left(\frac{1}{\pi}\right)-\pi^{2}\left(-\frac{1}{\pi^{2}}-\frac{1}{\pi}\right)$

$=2 \pi+1+\pi$

$=3 \pi+1$