JEE Main · 2026 · 04 Apr 2026 · Shift 1 · Area Under Curves

Both $\pi -|x|$ and $|x\sin x|$ are even, so the area is twice that for $x\ge 0$. On $[0,\pi ]$, $|x\sin x|=x\sin x$ lies below $\pi -x$ up to $x=\frac{\pi }{2}$ (the two curves meet there, as $\frac{\pi }{2}\sin \frac{\pi }{2}=\frac{\pi }{2}=\pi -\frac{\pi }{2}$) and above it beyond, so the upper boundary is $x\sin x$ on $\left[0,\frac{\pi }{2}\right]$ and $\pi -x$ on $\left[\frac{\pi }{2},\pi \right]$. Hence the area is $2\left({\int }_0^{\pi /2}x\sin xdx+{\int }_{\pi /2}^{\pi }(\pi -x)dx\right)$. The first integral is $[-x\cos x+\sin x{]}_0^{\pi /2}=1$ and the second is ${\left[-\frac{(\pi -x{)}^2}{2}\right]}_{\pi /2}^{\pi }=\frac{{\pi }^2}{8}$. The area is $2\left(1+\frac{{\pi }^2}{8}\right)=2+\frac{{\pi }^2}{4}$.