The area of the region $S=\left\{(x, y): 3 x^{2} \leq 4 y \leq 6 x+24\right\}$ is ______.

JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Area Under Curves

The area of the region

S = {(x, y) : 3 x^{2} \leq 4 y \leq 6 x + 24}

is ______.

Answer: 27

The region lies above the parabola

4 y = 3 x^{2}

and below the line

4 y = 6 x + 24

. The two curves meet where

3 x^{2} = 6 x + 24

, i.e.

x^{2} - 2 x - 8 = 0

, so

x = - 2

and

x = 4

, and between these values the line is the upper boundary. Hence

Area = \frac{1}{4} \int_{- 2}^{4} (6 x + 24 - 3 x^{2}) d x = \frac{1}{4} [3 x^{2} + 24 x - x^{3}]_{- 2}^{4} = \frac{1}{4} (80 - (- 28)) = 27.

The area of the region is

27

square units.

JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Integer answer