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 correct answer is:$\frac{7}{2}units$
The given equations of lines are
$2x+y=4…(1)$
$3x–2y=6…(2)$
And,$x–3y+5=0…(3)$

The area of the region bounded by the lines is the area of $∆ABC$.AL and CM are the
perpendiculars on $x-axis.$
Area(∆ABC)=Area(ALMCA)–Area(ALB)–Area(CMB)
$=∫^4_1(\frac{x+5}{3})dx-∫^2_1(4-2x)dx-∫^4_2(\frac{3x-6}{2})dx$
$=\frac{1}{3}\bigg[\frac{x^2}{2}+5x\bigg]^4_1-\bigg[4x-x^2\bigg]^2_1-\frac{1}{2}\bigg[\frac{3x^2}{2}-6x\bigg]^4_2$
$=\frac{1}{3}[8+20-\frac{1}{2}-5]-[8-4-4+1]-\frac{1}{2}[24-24-6+12]$
$=(\frac{1}{3}\times\frac{45}{2})-(1)-\frac{1}{2}(6)$
$=\frac{15}{2}-1-3$
$=\frac{15}{2}-4=\frac{15-8}{2}=\frac{7}{2}units$