Mathematics - Area Under Curves Question with Solution | TestHub

Required area
$={\int }_0^1\left(x_2-x_1\right)dy$ $={\int }_0^1\left(\sqrt{4-y^2}-\sqrt{3}y\right)dy$ $={\left[\frac{1}{2}y\sqrt{4-y^2}+\frac{1}{2}(4){\sin }^{-1}\frac{y}{2}-\frac{\sqrt{3}{y}^2}{2}\right]}_0^1$ $$\begin{array}{l}=\frac{\sqrt{3}}{2}+2{\sin }^{-1}\left(\frac{1}{2}\right)-\frac{\sqrt{3}}{2}-2{\sin }^{-1}0\\ =\frac{\sqrt{3}}{2}+2\left(\frac{\pi }{6}\right)-\frac{\sqrt{3}}{2}=\frac{\pi }{3}\text{ sq units }\end{array}$$