The area (in sq. units) of the region bounded by

$x=4 y-2 \& x^{2}=4 y$

$\Rightarrow x^{2}=x+2 \Rightarrow x^{2}-x-2=0$

$x=2,-1$

So, $\int_{-1}^{2}\left(\frac{x+2}{4}-\frac{x^{2}}{4}\right) d x=\frac{9}{8}$