Solution:

$\mathrm{BL}$ and $\mathrm{CM}$ are drawn perpendicular to $x$-axis.
It is observed in the following figure that,
Area $(\triangle \mathrm{ACB})=$ Area (ALBA) $+$ Area (BLMCB) – Area (AMCA) … (1)

Eq. of the line segment AB is
$\begin{array}{l}
y-0=\frac{3-0}{1+1}(x+1) \\
y=\frac{3}{2}(x+1)
\end{array}$
$\therefore$ Area $(\mathrm{ALBA})=\int_{1}^{1} \frac{3}{2}(x+1) d x=\frac{3}{2}\left[\frac{x^{2}}{2}+x\right]_{-1}^{1}=\frac{3}{2}\left[\frac{1}{2}+1-\frac{1}{2}+1\right]=3$ units
Eq. of the line segment BC is
$\begin{array}{l}
y-3=\frac{2-3}{3-1}(x-1) \\
y=\frac{1}{2}(-x+7)
\end{array}$
$\therefore \text { Area }(\mathrm{BLMCB})$$=\int_{1}^{3} \frac{1}{2}(-x+7) d x=\frac{1}{2}\left[-\frac{x^{2}}{2}+7 x\right]_{1}^{3}=\frac{1}{2}\left[-\frac{9}{2}+21+\frac{1}{2}-7\right]=5 \text { units }$
Eq. of the line segment AC is
$\begin{array}{l}
y-0=\frac{2-0}{3+1}(x+1) \\
y=\frac{1}{2}(x+1)
\end{array}$
$\therefore$ Area $(\mathrm{AMCA})=\frac{1}{2} \int_{-1}^{3}(x+1) d x=\frac{1}{2}\left[\frac{x^{2}}{2}+x\right]_{-1}^{3}=\frac{1}{2}\left[\frac{9}{2}+3-\frac{1}{2}+1\right]=4$ units
As a result, from eq. (1), we get
Area of $(\triangle A B C)=(3+5-4)=4$ units