Mathematics - Area Under Curves Question with Solution | TestHub

The tangent on $x^2+y^2=4$ at $(1,\sqrt{3})$ is $[x+\sqrt{3}y=4]$ and equation of normal at $(1,\sqrt{3})$ is $\mathrm{y}=\mathrm{x}\sqrt{3}$
$\begin{array}{rl}& \text{ Required area }={\int }_0^{1}x\sqrt{3}\mathrm{dx}+{\int }_1^4\frac{4-x}{\sqrt{3}}\mathrm{dx}\\ & =\sqrt{3}{\left[\frac{{\mathrm{x}}^2}{2}\right]}_0^1+\frac{1}{\sqrt{3}}{\left[4\mathrm{x}-\frac{{\mathrm{x}}^2}{2}\right]}_1^4\\ & =\sqrt{3}\times \frac{1}{2}+\frac{1}{\sqrt{3}}\left[4(4-1)-\frac{1}{2}(16-1)\right]\\ & =\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{3}}\left[12-\frac{15}{2}\right]=\frac{\sqrt{3}}{2}+\frac{9}{2\sqrt{3}}\\ & =\frac{\sqrt{3}}{2}+\frac{3\sqrt{3}}{2}=2\sqrt{3}\text{ sq. units }\end{array}$