The point of intersection of $y^{2}=4 x$ and $y=2 x-4$ is
$(2 x-4)^{2}=4 x$
$\Rightarrow x^{2}-5 x+4=0$
$\Rightarrow (x-1)(x-4)=0$
$\Rightarrow x=1,4$
$\Rightarrow y=-2,4$
$\therefore $ Required area
$=\displaystyle\int_{-2}^{4}\left(\frac{y+4}{2}\right) d y-\displaystyle\int_{-2}^{4} \frac{y^{2}}{4} d y$
$=\frac{1}{2}\left[\frac{y^{2}}{2}+4 y\right]_{-2}^{4}-\frac{1}{4}\left[\frac{y^{3}}{3}\right]_{-2}^{4}$
$=\frac{1}{2}[8+16-(2-8)]-\frac{1}{12}[64+8]$
$=\frac{1}{2}[30]-\frac{1}{12}(72)$
$=15-6=9$ sq unit
