Mathematics · Area under the Curves

JEE Main 2026 — 4 April, Morning Shift — Question 41

The area of the region ${(x,y): y ≤ π – |x|, y ≤ |x sin x|, y ≥ 0}$ is :

Option A:
$\frac{\pi^2}{8}+1$
Option B:
$\frac{\pi^2}{4}+2$
Correct
Option C:
$\frac{\pi^2}{8}-1$
Option D:
$\frac{\pi^2}{4}+2$

Answer: B

Step-by-step solution

The region is symmetric about the y-axis, so area = $2\left(\int_{0}^{\pi/2} x\sin x\,dx + \int_{\pi/2}^{\pi} (\pi - x)\,dx\right)$ . $\int_{0}^{\pi/2} x\sin x\,dx = [-x\cos x + \sin x]_{0}^{\pi/2} = 1$ . $\int_{\pi/2}^{\pi} (\pi - x)\,dx = \left[\pi x - \frac{x^2}{2}\right]_{\pi/2}^{\pi} = \frac{\pi^2}{8}$ . Thus area = $2\left(1 + \frac{\pi^2}{8}\right) = 2 + \frac{\pi^2}{4}$ .

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 4 April, Morning Shift
Subject: Mathematics
Chapter: Area under the Curves
Topic: Area under the Curves