The parabola y^2=4 x divides the area of the circle x^2+y^2=5 in two parts. The area of the smaller part is equal to:

$\begin{aligned} & y^2=4 x \\ & x^2+y^2=5\end{aligned}$​y2=4xx2+y2=5​ $\therefore$∴ Area of shaded region as shown in the figure will be

$$\begin{aligned} & \mathrm{A}_1=\int_0^1 \sqrt{4 \mathrm{x}} \mathrm{dx}+\int_1^{\sqrt{5}} \sqrt{5-\mathrm{x}^2} \mathrm{dx} \\ & =\frac{4}{3} \cdot\left[\mathrm{x}^{\frac{3}{2}}\right]_0^1+\left[\frac{\mathrm{x}}{2} \sqrt{5-\mathrm{x}^2}+\frac{5}{2} \sin ^{-1} \frac{\mathrm{x}}{\sqrt{5}}\right]_1^{\sqrt{5}} \\ & =\frac{1}{3}+\frac{5 \pi}{4}-\frac{5}{2} \sin ^{-1}\left(\frac{1}{\sqrt{5}}\right) \\ & \therefore \text { Required Area }=2 \mathrm{~A}_1 \\ & =\frac{2}{3}+\frac{5 \pi}{2}-5 \sin ^{-1}\left(\frac{1}{\sqrt{5}}\right) \\ & =\frac{2}{3}+5\left(\frac{\pi}{2}-\sin ^{-1} \frac{1}{\sqrt{5}}\right) \\ & =\frac{2}{3}+5 \cos ^{-1} \frac{1}{\sqrt{5}} \\ & =\frac{2}{3}+5 \sin ^{-1}\left(\frac{2}{\sqrt{5}}\right)\end{aligned}$$​A1​=∫01​4x​dx+∫15​​5−x2​dx=34​⋅[x23​]01​+[2x​5−x2​+25​sin−15​x​]15​​=31​+45π​−25​sin−1(5​1​)∴ Required Area =2 A1​=32​+25π​−5sin−1(5​1​)=32​+5(2π​−sin−15​1​)=32​+5cos−15​1​=32​+5sin−1(5​2​)​