The area of the region x , y : x 2 ≤ y ≤ x 2 - 4 , y ≥ 1 is

The area of the region

x, y : x^{2} \leq y \leq |x^{2} - 4|, y \geq 1

is

$\frac{4}{3} (4 \sqrt{2} - 1)$
$\frac{4}{3} (4 \sqrt{2} + 1)$
$\frac{3}{4} (4 \sqrt{2} + 1)$
$\frac{3}{4} (4 \sqrt{2} - 1)$

Solution

The required diagram is:
The shaded region is symmetric about $y -$ axis.
Required area $= 2 [\int_{1}^{2} \sqrt{y} d y + \int_{2}^{4} \sqrt{4 - y} d y]$
$= 2 {[\frac{y^{3 / 2}}{\frac{3}{2}}]}_{1}^{2} - {[\frac{2 (4 - y)^{3 / 2}}{3}]}_{2}^{4}$
$= \frac{4}{3} (4 \sqrt{2} - 1)$
Hence this is the correct option.

Asked in: JEE Main 2023 (13 Apr Shift 2)

Practice more Area Under Curves questions on Aicharya