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

$y=2x+1$ --------- ( 1 )

$y=3x+1$ --------- ( 2 )

$x=4$ ---------- ( 3 )

To find the vertex $A$,let us solve equation ( 1 ) & ( 2 ),

$2x+1=3x+1$

We get, $x=0$ and $y=1$

Hence, vertex of $A=(0,1)$

To find the vertex $B$ let us solve the equation ( 2 ) & ( 3 ),

$y=3x+1$

$x=4$

$y=12+1=13$.

Hence, vertex $B$ is $(4,13)$

To find the vertex $C$,let us solve equation $\left(3\right)$ & $( 1 )

$x=4$

$y=2x+1$

$y=8+1=9$

Hence, vertex $C$ is $(4,9).$

Now the required area of the triangle is the shaded portion as shown in the fig.

Hence,

$Ar\left(△ACD\right)=Ar\left(OLBAO\right)-Ar\left(OLCAO\right)$

$Ar\left(△ACD\right)={\int }_0^4(3x+1)dxdx-{\int }_0^4(2x+1)dx.$

On integrating we get,

$Ar\left(△ACD\right)={[\frac{3x^2}{2}+x]}_0^4-{[\frac{2x^2}{2}+x]}_0^4$

On applying limits we get,

$Ar\left(△ACD\right)=(24+4)-(16+4)$

$Ar\left(△ACD\right)=28-20$

$\therefore$ $A=8uni{t}^2$

Hence, the required area is $8uni{t}^2$