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JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Area Under Curves

The area of the region bounded by

y - x = 2

and

x^{2} = y

is equal to :

(A) $\frac{2}{3}$
(B) $\frac{9}{2}$ ✓ Correct
(C) $\frac{16}{3}$
(D) $\frac{4}{3}$

Answer: (B)

Solution

The curves intersect where

x^{2} = x + 2

, giving

x^{2} - x - 2 = 0

, so

x = - 1

or

x = 2

.

The area is

\int_{- 1}^{2} [(x + 2) - x^{2}] d x = [\frac{x ^{2}}{2} + 2 x - \frac{x ^{3}}{3}]_{- 1}^{2}

= (2 + 4 - \frac{8}{3}) - (\frac{1}{2} - 2 + \frac{1}{3}) = 6 - \frac{8}{3} + \frac{3}{2} - \frac{1}{3} = \frac{9}{2}

.

JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Single correct option

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