The area of the region bounded by the curve $x^2=4y$ and the straight-line $x=4y-2$ is

$\begin{array}{l}{x}^2=x+2\\ x^2-x-2=0\\ (x-2)(x+1)=0\\ x=-1,2\end{array}$

For $x=-1,y=\frac{1}{4}\text{ and for }x=2,y=1$

Points of intersection are $\left(-1,\frac{1}{4}\right)\text{ and }(2,1)$

Graphs of parabola $x^2=4y\text{ and }x=4y-2$ are shown in the following figure : 

$\begin{array}{l}A={\int }_{-1}^2 \left[\frac{x+2}{4}-\frac{x^2}{4}\right]dx\\ =\frac{1}{4}{\left[\frac{x^2}{2}+2x-\frac{x^3}{3}\right]}_{-1}^2\\ =\frac{1}{4}\left[8-\frac{1}{2}-3\right]\\ =\frac{1}{4}\left[8-\frac{1}{2}-3\right]\\ =\frac{9}{8}\text{ sq. units }\end{array}$