Solution:
The eqs. of the sides of the triangle are $y=2 x+1, y=3 x+1$, and $x=4$. On solving these eqs., we get the vertices of triangle as $\mathrm{A}(0,1), \mathrm{B}(4,13)$, and $\mathrm{C}$ $(4,9)$

It is observed that
It can be observed that,
$\begin{array}{l}
\text { Area }(\triangle \mathrm{ACB})=\text { Area }(\mathrm{OLBAO})-\text { Area }(\mathrm{OLCAO}) \\
=\int_{0}^{+}(3 x+1) d x-\int_{0}^{4}(2 x+1) d x \\
=\left[\frac{3 x^{2}}{2}+x\right]_{0}^{4}-\left[\frac{2 x^{2}}{2}+x\right]_{0}^{4} \\
=(24+4)-(16+4) \\
=28-20 \\
=8 \text { units }
\end{array}$