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)
\(=∫^4_1(\frac{x+5}{3})dx-∫^2_1(4-2x)dx-∫^4_2(\frac{3x-6}{2})dx\)
\(=\frac{1}{3}\bigg[\frac{x^2}{2}+5x\bigg]^4_1-\bigg[4x-x^2\bigg]^2_1-\frac{1}{2}\bigg[\frac{3x^2}{2}-6x\bigg]^4_2\)
\(=\frac{1}{3}[8+20-\frac{1}{2}-5]-[8-4-4+1]-\frac{1}{2}[24-24-6+12]\)
\(=(\frac{1}{3}\times\frac{45}{2})-(1)-\frac{1}{2}(6)\)
\(=\frac{15}{2}-1-3\)
\(=\frac{15}{2}-4=\frac{15-8}{2}=\frac{7}{2}units\)