Using the method of integration find the area of the triangle

Solution:

The vertices of $\triangle \mathrm{ABC}$ are $\mathrm{A}(2,0), \mathrm{B}(4,5)$, and $\mathrm{C}(6,3)$.

Equation of line segment AB is

$y-0=\frac{5-0}{4-2}(x-2)$

$2 y=5 x-10$

 

$y=\frac{5}{2}(x-2)$    ...(1)

Equation of line segment BC is

$y-5=\frac{3-5}{6-4}(x-4)$

$2 y-10=-2 x+8$

$2 y=-2 x+18$

$y=-x+9$               ...(2)

Equation of line segment CA is

$y-3=\frac{0-3}{2-6}(x-6)$

$-4 y+12=-3 x+18$

$4 y=3 x-6$

$y=\frac{3}{4}(x-2)$            ...(3)

Area (ΔABC) = Area (ABLA) + Area (BLMCB) – Area (ACMA)

$=\int_{2}^{4} \frac{5}{2}(x-2) d x+\int_{4}^{6}(-x+9) d x-\int_{2}^{6} \frac{3}{4}(x-2) d x$

$=\frac{5}{2}\left[\frac{x^{2}}{2}-2 x\right]_{2}^{4}+\left[\frac{-x^{2}}{2}+9 x\right]_{4}^{6}-\frac{3}{4}\left[\frac{x^{2}}{2}-2 x\right]_{2}^{6}$

$=\frac{5}{2}[8-8-2+4]+[-18+54+8-36]-\frac{3}{4}[18-12-2+4]$

$=5+8-\frac{3}{4}(8)$

$=13-6$

$=7$ units