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JEE Main 20264 AprilEvening Shift

Exam: JEE Main
Year: 2026
Date: 4 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Area Under Curves

JEE Main 2026 · 4 April · Evening Shift · Mathematics · Area Under Curves

The area of the region bounded by the curves x+3y^2=0 and x+4y^2=1...

The area of the region bounded by the curves

x+3y^2=0

and

x+4y^2=1

is equal to:

Options

A $\dfrac{1}{3}$
B $\dfrac{2}{3}$
C $\dfrac{4}{3}$
D $\dfrac{5}{3}$

Correct answer

C. $\dfrac{4}{3}$

Step-by-step solution

The equations of the given curves are

x = -3y^2

and

x = 1 - 4y^2

.

To find the points of intersection, equate the values of

x

:

-3y^2 = 1 - 4y^2

y^2 = 1 \Rightarrow y = \pm 1

The region is bounded between

y = -1

and

y = 1

. In this interval,

1 - 4y^2 \ge -3y^2

.

The required area

A

is given by:

A = \int_{-1}^{1} \left( (1 - 4y^2) - (-3y^2) \right) dy

A = \int_{-1}^{1} (1 - y^2) dy

Since

1 - y^2

is an even function:

A = 2 \int_{0}^{1} (1 - y^2) dy

A = 2 \left[ y - \dfrac{y^3}{3} \right]_{0}^{1}

A = 2 \left( 1 - \dfrac{1}{3} \right) = 2 \left( \dfrac{2}{3} \right) = \dfrac{4}{3}

Answer:

\dfrac{4}{3}

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