JEE Main 2026 · 4 Apr · Shift 1, Mathematics question 18

Symmetry. Work on $x\in[0,\pi]$ and double.

Which curve is lower. $x\sin x$ and $\pi-x$ meet at $x=\frac\pi2$. The region is under $x\sin x$ on $[0,\frac\pi2]$ and under $\pi-x$ on $[\frac\pi2,\pi]$.

Area.
$$2\left[\int_0^{\pi/2}x\sin x\,dx+\frac12\left(\frac\pi2\right)^2\right]=2\left[1+\frac{\pi^2}8\right]=2+\frac{\pi^2}4.$$

(B)