JEE Main 2024 — 9 April, Shift 1 — Question 2

$\mathrm{y}^{2}=4 \mathrm{x}$

$\mathrm{x}^{2}+\mathrm{y}^{2}=5$

$\therefore$ Area of shaded region as shown in the figure will be

$A_{1}=\int_{0}^{1} \sqrt{4 x} d x+\int_{1}^{\sqrt{5}} \sqrt{5-x^{2}} d x$

$=\frac{4}{3} \cdot\left[x^{\frac{3}{2}}\right]_{0}^{1}+\left[\frac{x}{2} \sqrt{5-x^{2}}+\frac{5}{2} \sin ^{-1} \frac{x}{\sqrt{5}}\right]_{1}^{\sqrt{5}}$

$=\frac{1}{3}+\frac{5 \pi}{4}-\frac{5}{2} \sin ^{-1}\left(\frac{1}{\sqrt{5}}\right)$

$\therefore$ Required Area $=2 \mathrm{~A}_{1}$

$=\frac{2}{3}+\frac{5 \pi}{2}-5 \sin ^{-1}\left(\frac{1}{\sqrt{5}}\right)$ $=\frac{2}{3}+5\left(\frac{\pi}{2}-\sin ^{-1} \frac{1}{\sqrt{5}}\right)$

$=\frac{2}{3}+5 \cos ^{-1} \frac{1}{\sqrt{5}}$

$=\frac{2}{3}+5 \sin ^{-1}\left(\frac{2}{\sqrt{5}}\right)$