Question

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

Answer 1

The equations of sides of the triangle are $y=2x+1,y=3x+1,x=4$.

On solving these equations, we obtain the vertices of triangle as $A(0,1),B(4,13),C(4,9)$.

$Area\left(△ACB\right)=Area\left(OLBAO\right)-Area\left(OLCAO\right)$

$=\underset{0}{\overset{4}{\int }}(3x+1)dx-\underset{0}{\overset{4}{\int }}(2x+1)dx$

$=(24+4)-(16+4)=28-20=8sq.units$