Question

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

Answer 1

Area of $△ABC=$ Area of $△ABD+$ Area of trapezium $BDEC-$ Area of $△ACE$

$AreaABD$

Area ABD $={\int }_1^{2}ydx$

Equation of line between $A(1,0)B(2,2)$ is

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

$\frac{y}{x-1}=2$

$y=2(x-1)$

Area ABD $={\int }_1^{2}2(x-1)dx$

$=2{[\frac{x^2}{2}-x]}_1^2$

$=2[2-2-\frac{1}{2}+1]$

$=1$

$AreaBDEC$

Area $BDEC$ $={\int }_2^{3}ydx$

Equation of line between B(2,2) & C(3,1) is

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

$y-2=-(x-2)$

$y=4-x$

Area BDEC $={\int }_2^3(4-x)dx$

$={[4x-\frac{x^2}{2}]}_2^3$

$=4(3-2)-\frac{1}{2}(3^2-2^2)$

$=4-\frac{1}{2}\times 5$

$=\frac{3}{2}$

$AreaACE$

Area ACE $={\int }_1^{3}ydx$

Equation of line between A(1,0) & C(3,1) is

$\frac{y-0}{x-1}=\frac{1-0}{3-1}$

$\frac{y}{x-1}=\frac{1}{2}$

$y=\frac{1}{2}(x-1)$

Area ACE $={\int }_1^3\frac{1}{2}(x-1)dx$

$=\frac{1}{2}{[\frac{x^2}{2}-x]}_1^3$

$=\frac{1}{2}[\frac{9}{2}-3-\frac{1}{2}+1]$

$=1$

Hence,

Area of $△ABC=$ Area of $△ABD+$ Area of trapezium $BDEC-$ Area of $△ACE$

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

$=\frac{3}{2}$