Question

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

Answer 1

Given line equations
$y=2x+1,y=3x+1\text{and}x=4$
Required Area
$=\text{Area OABDO -OACDO}$
$={\int }_0^4{y}_{1}dx-{\int }_0^4{y}_{2}dx={\int }_0^4(y_1-y_2)dx$

Where $y_1=3x+1$ & $y_2=2x+1$

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

$={\int }_0^4\left(x\right)dx={\left[\frac{x^2}{2}\right]}_0^4$

$=\frac{16-0}{2}=8$

Therefore, required area is $8$ square units.