JEE Main 2026 Jan 24 Shift 1, Mathematics Q19: Area as Definite Integral

JEE Main2026Jan 24, Shift 1Mathematics

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

A
$\frac{2}{3} (22 + 1)$
B
$\frac{2}{3} (42 + 1)$
C
$\frac{2}{3} (2 + 1)$
D
$\frac{2}{3} (32 + 1)$

Solution

$A_{1}$ : $y = x^{2} + 2$ meets $y = 8 - x$ when $x^{2} + x - 6 = 0 \Rightarrow x = 2$ (in Q1). The line lies above the parabola on $[0, 2]$ .

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

$= [6 x - \frac{x ^{2}}{2} - \frac{x ^{3}}{3}]_{0}^{2} = 12 - 2 - \frac{8}{3} = \frac{22}{3}$ .

$A_{2}$ : Region between $y = x^{2} + 2$ (top) and $y = x$ (bottom, from $y^{2} = x$ ) on $[0, 2]$ .

$A_{2} = \int_{0}^{2} [(x^{2} + 2) - x] d x = [\frac{x ^{3}}{3} + 2 x - \frac{2}{3} x^{3/2}]_{0}^{2}$

$= \frac{8}{3} + 4 - \frac{4 2}{3} = \frac{20}{3} - \frac{4 2}{3}$ .

Difference:

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

Concept behind this question

Area as Definite IntegralNotes, formulas and examples →

Practice more Mathematics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →

JEE Main 2026 Jan 24 Shift 1, Mathematics Q19: Area as Definite Integral

Concept behind this question

More previous year questions on Area as Definite Integral

Practice more Mathematics