The parabola y2=4x divides the area of the circle x2+y2=5 in two parts. The area of the smaller part is equal to :
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Official solution
The point of intersection of y2=4x and x2+y2=5 are (1,2) and (1,−2).
\[\because\] Area of smaller region bounded by y2=4x and
x2+y2=5
\[=2\{\text {area of } O A C O+\text { area of } C A B C\}\]
\[\begin{aligned}
& =2\left[\int_\limits0^1 2 \sqrt{x} d x+\int_\limits1^{\sqrt{5}} \sqrt{5-x^2} d x\right. \\
& =2\left[\left|\frac{4}{3} x^{\frac{3}{2}}\right|_0^1+\left(\frac{1}{2} x \sqrt{5-x^2}+\frac{5}{2} \sin ^{-1} \frac{x}{\sqrt{5}}\right)\right]_1^{\sqrt{5}} \\
& =2\left[\left(\frac{4}{3}-0\right)+\left(0+\frac{5 \pi}{4}\right)-\left(1+\frac{5}{2} \sin ^{-1} \frac{1}{\sqrt{5}}\right)\right] \\
& =2\left[\frac{1}{3}+\frac{5 \pi}{4}-\frac{5}{2} \sin ^{-1} \frac{1}{\sqrt{5}}\right]=\frac{2}{3}+\frac{5 \pi}{2}-5 \sin ^{-1} \frac{1}{\sqrt{5}} \\
& =\frac{2}{3}+5 \cos ^{-1}\left(\frac{1}{\sqrt{5}}\right)=\frac{2}{3}+5 \sin ^{-1}\left(\frac{2}{\sqrt{5}}\right)
\end{aligned}\]