Mathematics · Area under the Curves

JEE Main 2026 — 4 April, Evening Shift — Question 42

The area of the region bounded by the curves $x + 3y^2 = 0$ and $x + 4y^2 = 1$ is equal to:

Option A:
$\frac{1}{3}$
Option B:
$\frac{2}{3}$
Option C:
$\frac{4}{3}$
Correct
Option D:
$\frac{5}{3}$

Answer: C

Step-by-step solution

$x=-3 y^{2}$ and $x=1-4 y^{2}$ $A=\int_{-1}^{1}\left(\left(1-4 y^{2}\right)+3 y^{2}\right) d y$ $A=\int_{-1}^{1}\left(1-y^{2}\right) d y=2 \int_{0}^{1}\left(1-y^{2}\right) d y$ $\mathrm{A}=\frac{4}{3}$

Answer key and solution verified before publishing.

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