Mathematics - Area Under Curves Question with Solution | TestHub

$y=3x,2y=27-3x\&y=3x-x\sqrt{x}$
$\begin{array}{rl}& \mathrm{A}={\int }_0^{3}3\mathrm{x}-(3\mathrm{x}-\mathrm{x}\sqrt{\mathrm{x}})\mathrm{dx}+{\int }_3^9\left(\frac{27-3\mathrm{x}}{2}-(3\mathrm{x}-\mathrm{x}\sqrt{\mathrm{x}})\right)\mathrm{dx}\\ & \mathrm{A}={\int }_0^3{\mathrm{x}}^{3/2}\mathrm{dx}+{\int }_3^9\frac{27}{2}-\frac{9\mathrm{x}}{2}+{\mathrm{x}}^{3/2}\mathrm{dx}\\ & \mathrm{A}={\left[\frac{2{\mathrm{x}}^{5/2}}{5}\right]}_0^3+\frac{27}{2}[\mathrm{x}{]}_3^9-\frac{9}{2}{\left[\frac{{\mathrm{x}}^2}{2}\right]}_3^9+{\left[\frac{2{\mathrm{x}}^{5/2}}{5}\right]}_3^9\\ & \mathrm{A}=\frac{2}{5}\left(3^{5/2}\right)+\frac{27}{2}(6)-\frac{9}{4}(72)+\frac{2}{5}\left(9^{5/2}-3^{5/2}\right)\\ & \mathrm{A}=\frac{2}{5}\left(3^{5/2}\right)+81-162+\frac{2}{5}\times 3^5-\frac{2}{5}\times 3^{5/2}\\ & \mathrm{A}=\frac{486}{5}-81=\frac{81}{5}\\ & 10\mathrm{A}=162\\ & \text{Ans. }=(2)\end{array}$