Area lying between the curve $y^2=4x\text{ and }y=2x$

The area lying between the curve

$y^2=4x\text{ and }y=2x$ is represented by the shaded

area $OBAO$ as 

The points of intersection of the curves are O(0, 0) and A(1, 2).

We draw AC perpendicular to x-axis such that coordinate of C is (1, 0). 

Area of $OBAO=\text{ Area of }\mathrm{△}OCA$

                            $\text{ -Area of }OCABO$

$\begin{array}{l}A={\int }_0^1 2xdx-{\int }_0^1 2\sqrt{x}dx\\ =2{\left[\frac{x^2}{2}\right]}_0^1-2{\left[\frac{x^{3/2}}{\frac{3}{2}}\right]}_0^1\\ =\left|\left[1-\frac{4}{3}\right]\right|\\ =\left|-\frac{1}{3}\right|\\ =\frac{1}{3}\text{ sq. unit }\end{array}$