Question

Using the method of integraton find the area of the region bounded by lines: $2x+y=4,3x-2y=6$ and $x-3y+5=0$

Answer 1

The given equations of lines are
$2x+y=4.........\left(1\right)$
$3x-2y=6.........\left(2\right)$
And, $x-3y+5=0..........\left(3\right)$
The area of the region bounded by the lines is the area of $\Delta ABC$. $AL$ and $CM$ are the perpendicular on $x$-axis.

$Area\left(\Delta ABC\right)=Area\left(ALMCA\right)-Area\left(ALB\right)-Area\left(CMB\right)$

$={\int }_1^4\left(\frac{x+5}{3}\right)dx-{\int }_1^2(4-2x)dx-{\int }_2^4\left(\frac{3x-6}{2}\right)dx$

$=\frac{1}{3}{[\frac{x^2}{2}+5x]}_1^4-[4x-x^2{]}_1^2-\frac{1}{2}{[\frac{3x^2}{2}-6x]}_2^4$

$=\frac{1}{3}[8+20-\frac{1}{2}-5]-[8-4-4+1]-\frac{1}{2}[24-24-6+12]$

$=(\frac{1}{3}\times \frac{45}{2})-\left(1\right)-\frac{1}{2}\left(6\right)$

$=\frac{15}{2}-1-3$

$=\frac{15}{2}-4=\frac{15-8}{2}=\frac{7}{2}$ sq. units