Question

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

Answer 1

Given the vertices of the $△ABC$ are $A(1,0),B(2,2)$ and $C(3,1)$.

By plotting these points on the graph, we find the required area is the shaded portion $ABC$.

To find the equation of the lines,from the given points.use the information given

Segment $AB$ is $\frac{y-2}{0-2}=\frac{x-2}{1-2}$

$⟹-y+2=-2x+4⟹2x-2=y$

Similarly segment $BC$ is $\frac{y-0}{1-0}=\frac{x-1}{3-1}$

$⟹2y=x-1⟹y=\frac{x-1}{2}$

Segment $CA$ is $\frac{y-2}{1-2}=\frac{x-2}{3-2}$

$⟹y-2=-x+2$

$⟹y=-x+4$

Hence the required area is the area of the triangle enclosed by these three lines.

$\therefore A=$(area enclosed by line AB and x-axis )+(area enclosed by line BC and x-axis )+

(area enclosed by line AC and x-axis )

$={\int }_1^2(2x-2)dx+{\int }_1^3\frac{(x-1)}{2}dx+{\int }_2^3(-x+4)dx$

$={|\frac{2x^2}{2}-2x|}_1^2+{|\frac{x^2}{4}-\frac{x}{2}|}_1^3+{|\frac{-x^2}{2}+4x|}_2^3$

$=(0-1+2)+(\frac{9}{4}-\frac{3}{2}+\frac{1}{4})+(\frac{-9}{2}+12-(-2+8\left)\right)$

$=1+1+\frac{3}{2}=\frac{7}{2}$

Hence, Area bounded by these three points is $\frac{7}{2}$