Question

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

Answer 1

The given lines are 2x + y = 4 ...(i)
3y - 2y = 6 ...(ii)
x - 3y + 5 = 0 ...(iii)
From Eqs. (i) and (ii), we get
$4-2x=\frac{3x-6}{2}$ (Eliminating y)
$\Rightarrow 8-4x=3x-6\Rightarrow 7x=14\Rightarrow x=2$
Using Eq. (i), we get $y=4-2x=4-2\times 2=0$
$\therefore$ Eq. (i) and (ii) intersect at the point (2, 0).
From Eqs. (ii) and (iii), we get
$\frac{3x-6}{2}=\frac{x÷5}{3}$
$\Rightarrow 9x-18=2x+10\Rightarrow 7x=28\Rightarrow x=4$
Using Eq. (ii), we get $y=\frac{3\times 4-6}{2}=3$
$\therefore$ Eq. (i) and (iii) intersect at the point (4, 3).
From lines Eqs. (i) and (iii), we get
$4-2x=\frac{x+5}{3}\Rightarrow 7x=7\Rightarrow x=1$
Using Eq.(i), we get $y=4=2x=4-2\times 1=2$
Eqs. (i) and (iii) intersect at the point (1, 2)
$\therefore$ Required area = (Area under line segment BA) - (Area under line segment BC) - (Area under line segment AC)
=${\int }_1^4\left(\frac{x+5}{3}\right)dx-{\int }_1^2(4-2x)dx-{\int }_2^4\left(\frac{3x-6}{2}\right)dx=\frac{1}{3}[\frac{(x+5{)}^2}{2}{]}_1^4-[4x-x^2{]}_1^2-\frac{1}{2}[\frac{3x^2}{2}-6x{]}_2^4=\frac{1}{6}[9^2-6^2]-\{(8-4)-(4-1)\}-\frac{1}{2}[(\frac{3\times 4^2}{2}-6\times 4)-(3\times \frac{2^2}{2}-6\times 2)]=\frac{45}{6}-1-\frac{1}{2}(0-6+12)=\frac{15}{2}-1-3=\frac{7}{2}squnit$