JEE Main 2026 — Mathematics Calculus

The given region is defined by the inequalities $y \leq \pi - |x|$y≤π−∣x∣, $y \leq |x \sin x|$y≤∣xsinx∣, and $y \geq 0$y≥0.

Since replacing $x$x with $-x$−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 ($x \geq 0$x≥0) and multiply it by $2$2.

For $x \geq 0$x≥0, the inequalities become:

$$y \leq \pi - x$$y≤π−x

$$y \leq x \sin x$$y≤xsinx

$$y \geq 0$$y≥0

The condition $y \geq 0$y≥0 and $y \leq \pi - x$y≤π−x restricts $x$x to the interval $[0, \pi]$[0,π]. In this interval, $\sin x \geq 0$sinx≥0, so $|x \sin x| = x \sin x$∣xsinx∣=xsinx.

To find the intersection of the curves $y = \pi - x$y=π−x and $y = x \sin x$y=xsinx, we equate them:

$$x \sin x = \pi - x \Rightarrow x(1 + \sin x) = \pi$$xsinx=π−x⇒x(1+sinx)=π

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

For $x \in \left[0, \dfrac{\pi}{2}\right]$x∈[0,2π​], $x \sin x \leq \pi - x$xsinx≤π−x.

For $x \in \left[\dfrac{\pi}{2}, \pi\right]$x∈[2π​,π], $x \sin x \geq \pi - x$xsinx≥π−x.

The area in the first quadrant, $A_1$A1​, is given by:

$$A_1 = \int_{0}^{\pi/2} x \sin x \, dx + \int_{\pi/2}^{\pi} (\pi - x) \, dx$$A1​=∫0π/2​xsinxdx+∫π/2π​(π−x)dx

Evaluating the first integral using integration by parts:

$$\int_{0}^{\pi/2} x \sin x \, dx = [-x \cos x]_{0}^{\pi/2} - \int_{0}^{\pi/2} (-\cos x) \, dx$$∫0π/2​xsinxdx=[−xcosx]0π/2​−∫0π/2​(−cosx)dx

$$= 0 + [\sin x]_{0}^{\pi/2} = 1$$=0+[sinx]0π/2​=1

Evaluating the second integral:

$$\int_{\pi/2}^{\pi} (\pi - x) \, dx = \left[ -\dfrac{(\pi - x)^2}{2} \right]_{\pi/2}^{\pi} = 0 - \left( -\dfrac{\pi^2}{8} \right) = \dfrac{\pi^2}{8}$$∫π/2π​(π−x)dx=[−2(π−x)2​]π/2π​=0−(−8π2​)=8π2​

Thus, the area in the first quadrant is:

$$A_1 = 1 + \dfrac{\pi^2}{8}$$A1​=1+8π2​

The total area of the region is twice the area in the first quadrant:

$$\text{Total Area} = 2 \times A_1 = 2 \left( 1 + \dfrac{\pi^2}{8} \right) = 2 + \dfrac{\pi^2}{4}$$Total Area=2×A1​=2(1+8π2​)=2+4π2​

Answer: $2 + \dfrac{\pi^2}{4}$2+4π2​