Question

Using integration, find the area of the triangular region whose sides have the equations y = 2x + 1, y = 3x + 1 and x = 4.

Answer 1

Given equation of sides of the triangle are y=2x + 1, y = 3x + 1 and x = 4. On solving these equations, we obtain the vertices of triangle as A(0,1), B (4, 13) and C(4, 9).

$\therefore$ Required area (shown in shaded region)

= Area (OLBAO)-Area (OLCAO)

$={\int }_0^4(3x+1)dx-{\int }_0^4(2x+1)dx=[\frac{3x^2}{2}+x{]}_0^4-[x^2+x{]}_0^4=\frac{3\times 4^2}{2}+4-0-[4^2+4-0]=24+4-20=8squnit$