Solution:
Area enclosed between the parabola, $4 y=3 x^{2}$, and the line, $2 y=3 x+12$, is represented by the shaded area OBAO as

Points of intersection of the given curves are $A(-2,3)$ and $(4,12)$. We draw $A C$ and BD perpendicular to $x$-axis.
$\begin{array}{l}
\therefore \text { Area } O B A O=\text { Area } C D B A-(\text { Area } O D B O+\text { Area } O A C O) \\
=\int_{-2}^{1} \frac{1}{2}(3 x+12) d x-\int_{-2}^{1} \frac{3 x^{2}}{4} d x \\
=\frac{1}{2}\left[\frac{3 x^{2}}{2}+12 x\right]_{-2}^{4}-\frac{3}{4}\left[\frac{x^{3}}{3}\right]_{-2}^{4} \\
=\frac{1}{2}[24+48-6+24]-\frac{1}{4}[64+8] \\
=\frac{1}{2}[90]-\frac{1}{4}[72] \\
=45-18 \\
=27 \text { sq. units }
\end{array}$