Mathematics - Area Under Curves Question with Solution | TestHub

The curves meet where $x^2+8x-9=0$, i.e. $x=1$. Area $=2\left[{\int }_0^{1}2\sqrt{2}\sqrt{x}dx+{\int }_1^3\sqrt{9-x^2}dx\right]=$ $2\left[\frac{4\sqrt{2}}{3}+\frac{1}{2}\left(\frac{9\pi }{2}-2\sqrt{2}-9{\sin }^{-1}\frac{1}{3}\right)\right]=2\left[\frac{\sqrt{2}}{3}+\frac{9\pi }{4}-\frac{9}{2}{\sin }^{-1}\left(\frac{1}{3}\right)\right]$.