Mathematics - Area Under Curves Question with Solution | TestHub

Required area is shaded. Point of intersection of $x=a$ and $y^2=4ax$, is $$y^2=4a^2\Rightarrow y=\pm 2a\text{ and }x=a\Rightarrow (a,\pm 2a)$$ $$\begin{array}{rl}& \therefore \mathrm{A}=2{\int }_0^{\mathrm{a}}(2\sqrt{\mathrm{a}}\sqrt{\mathrm{x}})\mathrm{dx}\\ & =4\sqrt{\mathrm{a}}{\int }_0^{\mathrm{a}}{\mathrm{x}}^{\frac{1}{2}}\mathrm{dx}=4\sqrt{\mathrm{a}}{\left[\frac{{\mathrm{x}}^{\frac{3}{2}}}{\left(\frac{3}{2}\right)}\right]}_0^{\mathrm{a}}=(4\sqrt{\mathrm{a}})\left(\frac{2}{3}\right)(\mathrm{a}\sqrt{\mathrm{a}})=\frac{8}{3}{\mathrm{a}}^2\text{ sq. units }\end{array}$$