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

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

To solving these equations, we obtain the vertices of triangle as A(0,1), B(4,13), and C

(4,9).

It can be observe that,

Area(ΔACB)=Area(OLBAO)-Area(OLCAO) =

\int_{0}^{4} (3x+1) \,dx$$$$-\int_{0}^{4} (2x+1) \,dx

=[$\frac{3x^2}{2}$+x]40-[$\frac{2x^2}{2}$+x]40

=(24+4)-(16+4)

=28-20

=8 units.