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

Toolbox:
• If we are given two or more curves represented by y=f(x) and y=g(x),where $f(x)\geq g(x)$ in [a,b],the point of intersection of two curves are given by x=a and x=b by taking the common values of y from the equation of the two curves.
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.
edited Sep 16, 2014