Let A 1 be the bounded area enclosed by the curves y=x 2 +2, x+y=8 and y -axis that lies in the first quadrant. Let A 2 be the bounded area enclosed by the curves y = x 2 +2, y 2 = x , x =2 , and y -axis that lies in the first quadrant. Then A 1 - A 2 is equal to

Mathematics · Area under the Curves

JEE Main 2026 — 24 January, Morning Shift — Question 19

Let $\mathrm{A}_{1}$ be the bounded area enclosed by the curves $y=x^{2}+2, x+y=8$ and $y$ -axis that lies in the first quadrant. Let $\mathrm{A}_{2}$ be the bounded area enclosed by the curves $\mathrm{y}=\mathrm{x}^{2}+2, \mathrm{y}^{2}=\mathrm{x}, \mathrm{x}=2$ , and y -axis that lies in the first quadrant. Then $\mathrm{A}_{1}-\mathrm{A}_{2}$ is equal to

Option A:
$\frac{2}{3}(2 \sqrt{2}+1)$
Correct
Option B:
$\frac{2}{3}(4 \sqrt{2}+1)$
Option C:
$\frac{2}{3}(\sqrt{2}+1)$
Option D:
$\frac{2}{3}(3 \sqrt{2}+1)$

Answer: A

Step-by-step solution

$A_{1}=\int_{0}^{2}\left((8-x)-\left(x^{2}+2\right)\right) d x$

$A_{1}=\int_{0}^{2}\left(6-x-x^{2}\right) d x$

$A_{1}=\left(6 x-\frac{x^{2}}{2}-\frac{x^{3}}{3}\right)_{0}^{2}=12-2-\frac{8}{3}=10-\frac{8}{3}=\frac{22}{3}$

$\mathrm{A}_{2}=\int_{0}^{2}\left(\mathrm{x}^{2}+2\right) \mathrm{dx}-\frac{2}{3}(2 \sqrt{2})$

$A_{2}=\left(\frac{x^{3}}{3}+2 x\right)_{0}^{2}-\frac{4 \sqrt{2}}{3}$

$\mathrm{A}_{2}=\frac{8}{3}+4-\frac{4 \sqrt{2}}{3}=\frac{20}{3}-\frac{4 \sqrt{2}}{3}$

$\mathrm{A}_{1}-\mathrm{A}_{2}=\frac{2}{3}+\frac{4 \sqrt{2}}{3}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 24 January, Morning Shift
Subject: Mathematics
Chapter: Area under the Curves
Topic: Area under the Curves