JEE Main 2026 — Area Under Curves Question with Solution

The given region is defined by the inequalities $y\le \pi -|x|$, $y\le |x\sin x|$, and $y\ge 0$.

Since replacing $x$ with $-x$ leaves the inequalities unchanged, the region is symmetric with respect to the y-axis. We can find the area of the region in the first quadrant ($x\ge 0$) and multiply it by $2$.

For $x\ge 0$, the inequalities become:
$y\le \pi -x$
$y\le x\sin x$
$y\ge 0$

The condition $y\ge 0$ and $y\le \pi -x$ restricts $x$ to the interval $[0,\pi ]$. In this interval, $\sin x\ge 0$, so $|x\sin x|=x\sin x$.

To find the intersection of the curves $y=\pi -x$ and $y=x\sin x$, we equate them:
$x\sin x=\pi -x\Rightarrow x(1+\sin x)=\pi$

By inspection, $x=\frac{\pi }{2}$ is a solution since $\frac{\pi }{2}(1+1)=\pi$.

For $x\in \left[0,\frac{\pi }{2}\right]$, $x\sin x\le \pi -x$.
For $x\in \left[\frac{\pi }{2},\pi \right]$, $x\sin x\ge \pi -x$.

The area in the first quadrant, $A_1$, is given by:
$A_1={\int }_0^{\pi /2}x\sin xdx+{\int }_{\pi /2}^{\pi }(\pi -x)dx$

Evaluating the first integral using integration by parts:
${\int }_0^{\pi /2}x\sin xdx=[-x\cos x{]}_0^{\pi /2}-{\int }_0^{\pi /2}(-\cos x)dx$
$=0+[\sin x{]}_0^{\pi /2}=1$

Evaluating the second integral:
${\int }_{\pi /2}^{\pi }(\pi -x)dx={\left[-\frac{(\pi -x{)}^2}{2}\right]}_{\pi /2}^{\pi }=0-\left(-\frac{{\pi }^2}{8}\right)=\frac{{\pi }^2}{8}$

Thus, the area in the first quadrant is:
$A_1=1+\frac{{\pi }^2}{8}$

The total area of the region is twice the area in the first quadrant:
$\text{Total Area}=2\times A_1=2\left(1+\frac{{\pi }^2}{8}\right)=2+\frac{{\pi }^2}{4}$

Answer: $2+\frac{{\pi }^2}{4}$