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JEE Main · 2020 · 07 Jan 2020 · Shift 2 · Area Under Curves

The area (in sq. units of the region

{(x, y) \in R^{2} ∣ 4 x^{2} \leq y \leq 8 x + 12}

is:

(A) $\frac{124}{3}$
(B) $\frac{125}{3}$
(C) $\frac{128}{3}$ ✓ Correct
(D) $\frac{127}{3}$

Answer: (C)

Solution

The parabola

y = 4 x^{2}

meets the line

y = 8 x + 12

where

4 x^{2} = 8 x + 12

, i.e.

x^{2} - 2 x - 3 = 0

, at

x = - 1

and

x = 3

. The enclosed area is

\int_{- 1}^{3} [(8 x + 12) - 4 x^{2}] d x = [4 x^{2} + 12 x - \frac{4 x ^{3}}{3}]_{- 1}^{3} = 36 - (- 8 + \frac{4}{3}) = \frac{128}{3}

.

JEE Main · 2020 · 07 Jan 2020 · Shift 2 · Single correct option

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