JEE Main 2026 Apr 4 Shift 1, Mathematics Q18: Area as Definite Integral

The region is symmetric in $x$ since $|x\sin x|$ and $\pi -|x|$ are both even, so compute twice the area for $x\ge 0$.

For $x\in [0,\pi ]$, $\sin x\ge 0$ so $|x\sin x|=x\sin x$. The relevant upper boundary switches from $x\sin x$ (lower of the two) to $\pi -x$ at $x=\frac{\pi }{2}$ (where both equal $\frac{\pi }{2}$).

$\text{Area}=2\left[{\int }_0^{\pi /2}x\sin xdx+{\int }_{\pi /2}^{\pi }(\pi -x)dx\right].$

Integration by parts: $\int x\sin xdx=-x\cos x+\sin x$. Evaluated from $0$ to $\frac{\pi }{2}$: $0+1=1$.

${\int }_{\pi /2}^{\pi }(\pi -x)dx={\left[-\frac{(\pi -x{)}^2}{2}\right]}_{\pi /2}^{\pi }=0-\left(-\frac{{\pi }^2/4}{2}\right)=\frac{{\pi }^2}{8}.$

$\text{Area}=2\left(1+\frac{{\pi }^2}{8}\right)=2+\frac{{\pi }^2}{4}.$