Mathematics - Area Under Curves Question with Solution | TestHub

The curve consists of $y={\sin }^{-1}x-\sqrt{x-x^2}$ and $y={\sin }^{-1}x+\sqrt{x-x^2}$ for $x\in [0,1]$. Required area $={\int }_0^{1}2\sqrt{x-x^2}dx={\int }_0^{1}2\sqrt{\frac{1}{4}-{\left(x-\frac{1}{2}\right)}^2}dx=2\times \frac{\pi }{8}=\frac{\pi }{4}$ (area of a semicircle of radius $\frac{1}{2}$ ).