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 $y=x^{2}+2, y^{2}=x, x=2$, and $y$-axis that lies in the first quadrant. Then $\mathrm{A}_{1}-\mathrm{A}_{2}$ is equal to

(A) $\frac{2}{3}(2 \sqrt{2}+1)$

(B) $\frac{2}{3}(4 \sqrt{2}+1)$

(C) $\frac{2}{3}(\sqrt{2}+1)$

(D) $\frac{2}{3}(3 \sqrt{2}+1)$

JEE Main · 2026 · 24 Jan 2026 · Shift 1 · Area Under Curves

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

Answer: (A)

The parabola

y = x^{2} + 2

meets the line

x + y = 8

where

x^{2} + x - 6 = 0

, i.e.

x = 2

. Hence

A_{1} = \int_{0}^{2} [(8 - x) - (x^{2} + 2)] d x = \int_{0}^{2} (6 - x - x^{2}) d x = 12 - 2 - \frac{8}{3} = \frac{22}{3}

. For

A_{2}

, the region lies between the upper curve

y = x^{2} + 2

and the lower curve

y = x

from

x = 0

x = 2

, so

A_{2} = \int_{0}^{2} [(x^{2} + 2) - x] d x = \frac{8}{3} + 4 - \frac{4 2}{3} = \frac{20 - 4 2}{3}

. Therefore

A_{1} - A_{2} = \frac{22 - ( 20 - 4 2 )}{3} = \frac{2 + 4 2}{3} = \frac{2}{3} (22 + 1)

JEE Main · 2026 · 24 Jan 2026 · Shift 1 · Single correct option