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JEE Main 202624 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
24 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Area Under Curves

JEE Main 2026 · 24 January · Morning Shift · Mathematics · Area Under Curves

Let A_ 1 be the bounded area enclosed by the curves y=x^ 2 +2,...

Let A1\mathrm{A}_{1} be the bounded area enclosed by the curves y=x2+2,x+y=8y=x^{2}+2, x+y=8 and yy-axis that lies in the first quadrant. Let A2\mathrm{A}_{2} be the bounded area enclosed by the curves y=x2+2,y2=x,x=2y=x^{2}+2, y^{2}=x, x=2, and yy-axis that lies in the first quadrant. Then A1−A2\mathrm{A}_{1}-\mathrm{A}_{2} is equal to

Options

  1. A23(32+1)\frac{2}{3}(3 \sqrt{2}+1)
  2. B23(22+1)\frac{2}{3}(2 \sqrt{2}+1)
  3. C23(2+1)\frac{2}{3}(\sqrt{2}+1)
  4. D23(42+1)\frac{2}{3}(4 \sqrt{2}+1)

Correct answer

B. 23(22+1)\frac{2}{3}(2 \sqrt{2}+1)

Step-by-step solution

For A1A_1: intersection of y=x2+2y = x^2 + 2 and x+y=8x + y = 8 at x=2,y=6x = 2, y = 6
A1=∫02(8−x−(x2+2))dx=∫02(6−x−x2)dx=[6x−x22−x33]02=12−2−83=223A_1 = \int_0^2 (8-x-(x^2+2)) dx = \int_0^2 (6-x-x^2) dx = \left[6x - \frac{x^2}{2} - \frac{x^3}{3}\right]_0^2 = 12 - 2 - \frac{8}{3} = \frac{22}{3}
For A2A_2: bounded by y=x2+2y = x^2 + 2, y2=xy^2 = x (i.e., x=y2x = y^2), x=2x = 2, and yy-axis.
In first quadrant, y2=xy^2 = x goes from (0,0) to (2,2\sqrt{2}).
The parabola y=x2+2y = x^2 + 2 passes through (0,2) and (2,6).
A2=∫02(2−y2)dy=[2y−y33]02=22−223=423A_2 = \int_0^{\sqrt{2}} (2 - y^2) dy = \left[2y - \frac{y^3}{3}\right]_0^{\sqrt{2}} = 2\sqrt{2} - \frac{2\sqrt{2}}{3} = \frac{4\sqrt{2}}{3}
A1−A2=223−423=23(22+1)A_1 - A_2 = \frac{22}{3} - \frac{4\sqrt{2}}{3} = \frac{2}{3}(2\sqrt{2}+1)
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