Question

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

Answer 1

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+1y=2x+1,y=3x+1

2x+1=3x+1 ⇒x=0,y=1.⇒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

Hence A=∫40(3x+1)dx−∫40(2x+1)dx.A=∫04(3x+1)dx−∫04(2x+1)dx.

On integrating we get,

A=[3x22+x]40−[2x22+x]40A=[3x22+x]04−[2x22+x]04

On applying limits we get,

A=(24+4)−(16+4)A=(24+4)−(16+4)

=28−20=28−20

=8=8 sq.units.

Hence the required area is 8 sq. units.