Mathematics - Area Under Curves Question with Solution | TestHub

Given curves, $$y^2=8x...(i)$$ and $$y=x...(ii)$$
On solving Eqs. (i) and (ii), we get $\begin{array}{rl}{x}^2-8x& =0\\ x(x-8)& =0\\ \Rightarrow x& =0,8\end{array}$ and $y=0,8$ $\begin{array}{rl}\therefore \text{ Required area, }(OPA)& ={\int }_0^8(\sqrt{8x}-x)dx\\ & ={\left[2\sqrt{2}\cdot \frac{2}{3}\cdot x^{3/2}-\frac{x^2}{2}\right]}_0^8\end{array}$ $=\frac{4\sqrt{2}}{3}\cdot (8{)}^{3/2}-\frac{(8{)}^2}{2}$ $=\frac{4\sqrt{2}}{3}\cdot (2{)}^3\cdot 2\sqrt{2}-\frac{64}{2}$ $=\frac{16}{3}\times 8-\frac{64}{2}=64\left(\frac{2}{3}-\frac{1}{2}\right)$ $=64\times \frac{(4-3)}{6}=64\times \frac{1}{6}=\frac{32}{3}$