Mathematics - Area Under Curves Question with Solution | TestHub

We have,

$y^2=8x$

$y=x$

Point of intersection of curve and a line is $A\left(8,8\right)$.

Required area

$\alpha ={\int }_2^8\left(\sqrt{8x}-x\right)dx$

$\Rightarrow \alpha ={\left[\frac{4\sqrt{2}{x}^{{\displaystyle \frac{3}{2}}}}{3}-\frac{x^2}{2}\right]}_2^8$

$\Rightarrow \alpha =\left[\left(\frac{4\sqrt{2}{\left(2\sqrt{2}\right)}^3}{3}-\frac{64}{2}\right)-\left(\frac{4\sqrt{2}{\left(\sqrt{2}\right)}^3}{3}-\frac{4}{2}\right)\right]$

$\Rightarrow \alpha =\frac{112}{3}-30$

$\Rightarrow \alpha =\frac{22}{3}$

So, $3\alpha =22$