The required area is the triangle bonded between the three lines\[y+2x+1----(1),y=3x+1-----(2),and \;x=4------(3).\]

Let us solve these equations to get the vertices of the triangle formed.

To find the vertex A,let us solve equ(1)&(2),\[y=2x+1,y=3x+1\]

2x+1=3x+1 $\Rightarrow x=0,y=1.$

Hence vertex A is (0,1)

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

y=3x+1

x=4

y=12+1=13.

vertex B is (4,13)

To find the vertex C,let us solve equ(3)&(1)

x=4

y=2x+1

y=8+1=9

vertex C is (4,9).

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

Hence $A=\int_0^4(3x+1)dx-\int_0^4(2x+1)dx.$

On integrating we get,

$A=\begin{bmatrix}\frac{3x^2}{2}+x\end{bmatrix}_0^4-\begin{bmatrix}\frac{2x^2}{2}+x\end{bmatrix}_0^4$

On applying limits we get,

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

$\;\;\;=28-20$

$\;\;\;=8$ sq.units.

Hence the required area is 8 sq. units.