Question

What is the area bounded by the curve $y^2=4x$ and the double ordinate $x=1$?

Answer 1

$\begin{array}{c}A=2{\int }_0^{1}ydx\\ =2{\int }_0^{1}2\sqrt{x}dx\\ =2\times 2{\int }_0^1\sqrt{x}dx\\ =4{\left[\frac{2}{3}{x}^{\frac{3}{2}}\right]}_0^1\\ =4\times \frac{2}{3}=\frac{8}{3}sq.units\end{array}$