Mathematics - Area Under Curves Question with Solution | TestHub

$\begin{array}{rl}& |x-y|\le y\le 4\sqrt{x}\\ & \text{ Now }y=|x-y|\\ & y^2=(x-y{)}^2\\ & \Rightarrow y=\frac{x}{2}\text{ and }x=0\end{array}$
$\begin{array}{rl}& \text{ Now area }={\int }_0^{64}\left(4\sqrt{\mathrm{x}}-\frac{\mathrm{x}}{2}\right)\mathrm{dx}\\ & ={\left[\frac{4{\mathrm{x}}^{3/2}}{3/2}-\frac{{\mathrm{x}}^2}{4}\right]}_0^{64}=\frac{8}{3}\cdot 8^3-\frac{64^2}{4}=64^2\left(\frac{1}{12}\right)\\ & =\frac{1024}{3}\end{array}$