Question

Using integration find the area of the region bounded by the triangle whose vertices are $(-1,0),(1,3)$ and $(3,2)$.

Answer 1

$BL$ and $CM$ are drawn perpendicular to x-axis.
It can be observed in the following figure that,
$Area\left(\Delta ACB\right)=Area\left(ALBA\right)+Area\left(BLMCB\right)-Area\left(AMCA\right)$ ......... (1)
Equating of line segment $AB$ is
$y-0=\frac{3-0}{1+1}(x+1)$
$y=\frac{3}{2}(x+1)$

$\therefore Area\left(ALBA\right)={\int }_{-1}^1\frac{3}{2}(x+1)dx=\frac{3}{2}{[\frac{x^2}{2}+x]}_{-1}^1=\frac{3}{2}[\frac{1}{2}+1-\frac{1}{2}+1]=3$sq. units
Equating of line segment $BC$ is
$y-3=\frac{2-3}{3-1}(x-1)$
$y=\frac{1}{2}(-x+7)$

$\therefore Area\left(BLMCB\right)={\int }_1^3\frac{1}{2}(-x+7)dx=\frac{1}{2}{[-\frac{x^2}{2}+7x]}_1^3=\frac{1}{2}[-\frac{9}{2}+21+\frac{1}{2}-7]=5$sq. units
Equation of line segment $AC$ is
$y-0=\frac{2-0}{3+1}(x+1)$
$y=\frac{1}{2}(x+1)$

$\therefore Area\left(AMCA\right)=\frac{1}{2}{\int }_{-1}^3(x+1)dx=\frac{1}{2}{[\frac{x^2}{2}+x]}_{-1}^3=\frac{1}{2}[\frac{9}{2}+3-\frac{1}{2}+1]=4$sq.units

Therefore, from equation (1), we obtain
$Area\left(\Delta ABC\right)=(3+5-4)=4$ sq. units