Solution:

The area bounded by the curve, $x^{2}=4 y$, and line, $x=4 y-2$, is represented by the shaded area OBAO.

Let’s say $A$ and $B$ be the points of intersection of the line and parabola.
$\mathrm{A}$ are $\left(-1, \frac{1}{4}\right)$
Coordinates of point $B$ are $(2,1)$.
Draw AL and BM perpendicular to $x$-axis.
It is observed that,
Area OBAO $=$ Area OBCO+ Area OACO $\ldots$ (1)
Therefore, Area of OBCO = Area of OMBC – Area of OMBO
$\begin{array}{l}
=\int_{0}^{2} \frac{x+2}{4} d x-\int_{0}^{2} \frac{x^{2}}{4} d x \\
=\frac{1}{4}\left[\frac{x^{2}}{2}+2 x\right]_{0}^{2}-\frac{1}{4}\left[\frac{x^{3}}{3}\right]_{0}^{2} \\
=\frac{1}{4}[2+4]-\frac{1}{4}\left[\frac{8}{3}\right] \\
=\frac{3}{2}-\frac{2}{3} \\
=\frac{5}{6}
\end{array}$
In the similar way, Area of OACO = Area of OLAC – Area of OLAO
$\begin{array}{l}
=\int_{-1}^{0} \frac{x+2}{4} d x-\int_{-1}^{0} \frac{x^{2}}{4} d x \\
=\frac{1}{4}\left[\frac{x^{2}}{2}+2 x\right]_{-1}^{0}-\frac{1}{4}\left[\frac{x^{3}}{3}\right]_{-1}^{0} \\
=-\frac{1}{4}\left[\frac{(-1)^{2}}{2}+2(-1)\right]-\left[-\frac{1}{4}\left(\frac{(-1)^{3}}{3}\right)\right] \\
=-\frac{1}{4}\left[\frac{1}{2}-2\right]-\frac{1}{12} \\
=\frac{1}{2}-\frac{1}{8}-\frac{1}{12} \\
=\frac{7}{24}
\end{array}$
As a result, required area $=\left(\frac{5}{6}+\frac{7}{24}\right)=\frac{9}{8}$ units