Mathematics - Area Under Curves Question with Solution | TestHub

$\begin{array}{rl}& y^2=4x\\ & x^2+y^2=5\end{array}$ $\therefore$ Area of shaded region as shown in the figure will be
$\begin{array}{rl}& {\mathrm{A}}_1={\int }_0^1\sqrt{4\mathrm{x}}\mathrm{dx}+{\int }_1^{\sqrt{5}}\sqrt{5-{\mathrm{x}}^2}\mathrm{dx}\\ & =\frac{4}{3}\cdot {\left[{\mathrm{x}}^{\frac{3}{2}}\right]}_0^1+{\left[\frac{\mathrm{x}}{2}\sqrt{5-{\mathrm{x}}^2}+\frac{5}{2}{\sin }^{-1}\frac{\mathrm{x}}{\sqrt{5}}\right]}_1^{\sqrt{5}}\\ & =\frac{1}{3}+\frac{5\pi }{4}-\frac{5}{2}{\sin }^{-1}\left(\frac{1}{\sqrt{5}}\right)\\ & \therefore \text{ Required Area }=2{\mathrm{A}}_1\\ & =\frac{2}{3}+\frac{5\pi }{2}-5{\sin }^{-1}\left(\frac{1}{\sqrt{5}}\right)\\ & =\frac{2}{3}+5\left(\frac{\pi }{2}-{\sin }^{-1}\frac{1}{\sqrt{5}}\right)\\ & =\frac{2}{3}+5{\cos }^{-1}\frac{1}{\sqrt{5}}\\ & =\frac{2}{3}+5{\sin }^{-1}\left(\frac{2}{\sqrt{5}}\right)\end{array}$