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

The given curves are $x^2=4y\text{ and }x=4y-2$

$\Rightarrow x^2=x+2\Rightarrow x^2-x-2=0,\Rightarrow (x-2)(x+1)=0\Rightarrow x=-1,2$

The area enclosed between the. curves 

$\begin{array}{l}={\int }_{-1}^2 \left[\frac{x+2}{4}-\frac{x^2}{4}\right]dx=\frac{1}{4}{\left[\frac{x^2}{2}+2x-\frac{x^3}{3}\right]}_{-1}^2\\ =\frac{1}{4}\left[\right(2+4-8/3)-(1/2-2+1/3\left)\right]=\frac{1}{4}[5-1/2]=\frac{9}{8}\text{ sq.uni }\end{array}$