$\begin{aligned} & y^2=4 x \\ & x^2+y^2=5\end{aligned}$
$\therefore$ Area of shaded region as shown in the figure will be

$\begin{aligned} & \mathrm{A}_1=\int_0^1 \sqrt{4 \mathrm{x}} \mathrm{dx}+\int_1^{\sqrt{5}} \sqrt{5-\mathrm{x}^2} \mathrm{dx} \\ & =\frac{4}{3} \cdot\left[\mathrm{x}^{\frac{3}{2}}\right]_0^1+\left[\frac{\mathrm{x}}{2} \sqrt{5-\mathrm{x}^2}+\frac{5}{2} \sin ^{-1} \frac{\mathrm{x}}{\sqrt{5}}\right]_1^{\sqrt{5}} \\ & =\frac{1}{3}+\frac{5 \pi}{4}-\frac{5}{2} \sin ^{-1}\left(\frac{1}{\sqrt{5}}\right) \\ & \therefore \text { Required Area }=2 \mathrm{~A}_1 \\ & =\frac{2}{3}+\frac{5 \pi}{2}-5 \sin ^{-1}\left(\frac{1}{\sqrt{5}}\right) \\ & =\frac{2}{3}+5\left(\frac{\pi}{2}-\sin ^{-1} \frac{1}{\sqrt{5}}\right) \\ & =\frac{2}{3}+5 \cos ^{-1} \frac{1}{\sqrt{5}} \\ & =\frac{2}{3}+5 \sin ^{-1}\left(\frac{2}{\sqrt{5}}\right)\end{aligned}$