JEE Main 2026 · 4 Apr · Shift 2 · Mathematics Q18

MathematicsSingle correct+4 / -1Question 18 of 75

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

\dfrac13

\dfrac23

\dfrac43

\dfrac53

Answer and worked solution

Official answer: C

Intersections. $-3y^2=1-4y^2$ gives $y=\pm1$ .

Area (integrating in y):
$\int_{-1}^1\left[(1-4y^2)-(-3y^2)\right]dy=\int_{-1}^1(1-y^2)\,dy=\frac43.$

(C)

Topics:Area between curves Parabola

JEE Main 2026 · 4 Apr · Shift 2, Mathematics question 18