The area bounded by the curve $y^2=4x$ with the lines $x=2$ and $x=5$ is 

The given curves are $y^2=4x$ and $x=2,x=5$  is

The rough sketch of the graph of the required region is as below

The required area is the area of the shaded region

The curve $y^2=4x$ is symmetric about $X-$ axis in the shaded region. 

The required region is double the region above the $X-$ axis. 

We know that the area of the  region bounded by the curve $y=f\left(x\right)$, $x=a,x=b$ and $X-$axis is $A={\int }_a^{b}f\left(x\right)dx$

Hence, the area of the shaded region is $A=2{\int }_2^{5}2\sqrt{x}dx$

$\begin{array}{rcl}A& =& 2{\int }_2^{5}2\sqrt{x}dx\\ & =& 4{\int }_2^5\sqrt{x}dx\\ & =& 4{\left[\frac{x^{{\displaystyle \frac{3}{2}}}}{{\displaystyle \frac{3}{2}}}\right]}_2^5\\ & =& \frac{8}{3}\left(5^{\frac{3}{2}}-2^{\frac{3}{2}}\right)\\ & =& \frac{8}{3}\left(5\sqrt{5}-2\sqrt{2}\right)\end{array}$

Therefore, the area of the required region is $\begin{array}{l}\boxed{\frac{8}{3}\left(5\sqrt{5}-2\sqrt{2}\right)\mathrm{square} \mathrm{units}}\end{array}$