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 sides of the triangles are $y=2x+1,y=3x+1$, and $x=4$.
On solving these question, we obtain the vertices of triangle as $A(0,1),B(4,13)$, and $C(4,9)$.
It can be observed that,
$Area\left(\Delta ACB\right)=Area\left(OLBAO\right)-Area\left(OLCAO\right)$
$={\int }_0^4(3x+1)dx-{\int }_0^4(2x+1)dx$
$={\left[\frac{3x^2}{2}+x\right]}_0^4-{[\frac{2x^2}{2}+x]}_0^4$
$=(24+4)-(16+4)$
$=28-20=8$sq. units.