Solution:
The area lying between the curve, $y^{2}=4 x$ and $y=2 x$, is represented by the shaded area OBAO as

Points of intersection of these curves are $O(0,0)$ and $A(1,2)$.
Draw AC perpendicular to $x$-axis such that the coordinates of $C$ are $(1,0) .$
$\therefore \text { Area } \mathrm{OBAO}=\text { Area }(\triangle O C A)-\text { Area }(\mathrm{OCABO})$
$=\int_{0}^{1} 2 x d x-\int_{0}^{1} 2 \sqrt{x} d x$
$\begin{array}{l}
=2\left[\frac{x^{2}}{2}\right]_{0}^{1}-2\left[\frac{x^{\frac{3}{2}}}{3}\right]_{0}^{1} \\
=\left|1-\frac{4}{3}\right| \\
=\left|-\frac{1}{3}\right|
\end{array}$
$=\frac{1}{3}$ units