Question

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

Answer 1

The equations given are

$y=2x+1$

$y=3x+1$

$x=4$

Solving these, we get the coordinates of the vertices of the triangle formed by these.

$(0,1),(4,13),(4,9)$

Area enclosed by these lines forming triangle is:

$A={\int }_0^4(3x+1-(2x+1\left)\right)dx$

$A={\int }_0^4\left(x\right)dx$

$A={\left[\frac{x^2}{2}\right]}_0^4=8sq.units$