JEE Main 2025 Apr 8 Shift 2, Mathematics Q11: Integration by Parts

Since ${\pi }^2>0$, factor it out: $I={\pi }^2{\int }_{-1}^{3/2}|x\sin (\pi x)|dx$.

Observe that $x\sin (\pi x)\ge 0$ on $[-1,0]$ (both factors non-positive) and on $[0,1]$ (both non-negative), while it is non-positive on $[1,3/2]$.

Also $x\sin (\pi x)$ is even, so ${\int }_{-1}^1|x\sin (\pi x)|dx=2{\int }_0^{1}x\sin (\pi x)dx$.

Compute the antiderivative by parts: $\int x\sin (\pi x)dx=-\frac{x\cos (\pi x)}{\pi }+\frac{\sin (\pi x)}{{\pi }^2}$.

Evaluate on $[0,1]$: ${\left[-\frac{x\cos \pi x}{\pi }+\frac{\sin \pi x}{{\pi }^2}\right]}_0^1=\frac{1}{\pi }$.

Evaluate on $[1,3/2]$: ${\left[-\frac{x\cos \pi x}{\pi }+\frac{\sin \pi x}{{\pi }^2}\right]}_1^{3/2}=\left(0-\frac{1}{{\pi }^2}\right)-\left(\frac{1}{\pi }+0\right)=-\frac{1}{{\pi }^2}-\frac{1}{\pi }$.

Combining, and taking absolute values: $I={\pi }^2\left[2\cdot \frac{1}{\pi }+\frac{1}{{\pi }^2}+\frac{1}{\pi }\right]={\pi }^2\left(\frac{3}{\pi }+\frac{1}{{\pi }^2}\right)=3\pi +1$, matching option (3).