The area of the region $\mathrm{S}=\left\{(\mathrm{x}, \mathrm{y}): \mathrm{y}^{2} \leq 8 \mathrm{x}, \mathrm{y} \geq \sqrt{2} \mathrm{x}, \mathrm{x} \geq 1\right\}$ is

JEE Main · 2022 · 28 Jun 2022 · Shift 1 · Area Under Curves

The area of the region

S = {(x, y) : y^{2} \leq 8 x, y \geq 2 x, x \geq 1}

Answer: (B)

The region is bounded by the parabola

y^{2} = 8 x

, the line

y = 2 x

, and the vertical line

x = 1

.

Find intersection of

y^{2} = 8 x

and

y = 2 x

2 x^{2} = 8 x \Rightarrow x = 0 or x = 4

For

x \geq 1

, the upper boundary is

y = 2 2 x

(from the parabola) and the lower boundary is

y = 2 x

(the line).

Area:

\int_{1}^{4} (2 2 x - 2 x) d x = 22 [\frac{2 x ^{3/2}}{3}]_{1}^{4} - 2 [\frac{x ^{2}}{2}]_{1}^{4}

= \frac{4 2}{3} (8 - 1) - \frac{2}{2} (16 - 1) = \frac{28 2}{3} - \frac{15 2}{2} = \frac{56 2 - 45 2}{6} = \frac{11 2}{6}

JEE Main · 2022 · 28 Jun 2022 · Shift 1 · Single correct option