Application of IntegralsmediumPYQ · JEE Main · 4 Apr 2026 · Shift 2 (Afternoon)Free

Area Between x+3y²=0 and x+4y²=1 = 4/3 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
The area of the region bounded by the curves x+3y2=0x+3y^2=0 and x+4y2=1x+4y^2=1 is equal to
A13\dfrac13
B23\dfrac23
C43\dfrac43correct
D53\dfrac53
Solution
Step 1: Write x=−3y2x=-3y^2 and x=1−4y2x=1-4y^2. Intersection: −3y2=1−4y2⇒y2=1⇒y=±1-3y^2=1-4y^2\Rightarrow y^2=1\Rightarrow y=\pm1. Step 2: Area (integrating in yy, right curve minus left):
A=∫−11[(1−4y2)−(−3y2)]dy=∫−11(1−y2) dy=2∫01(1−y2) dy.A=\int_{-1}^{1}\big[(1-4y^2)-(-3y^2)\big]dy=\int_{-1}^{1}(1-y^2)\,dy=2\int_0^1(1-y^2)\,dy.
 Step 3: Evaluate:
A=2[y−y33]01=2(1−13)=43.A=2\left[y-\frac{y^3}{3}\right]_0^1=2\left(1-\frac13\right)=\frac43.
 Correct answer: (3)
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