Definite Integral – Finding area between a polynomial and 2 lines – Exercise 7015

Exercise

Find the area of the region bounded by the graphs of the equations:

y=x22x3,y=x+3,x=0,x0y=x^2-2x-3, y=-x+3, x=0, x\geq 0

Final Answer


S=1312S=13\frac{1}{2}

Solution

First, we find out how the area looks like:

x22x3=x+3x^2-2x-3=-x+3

x2x6=0x^2-x-6=0

(x+2)(x3)=0(x+2)(x-3)=0

x=2,x=3x=-2, x=3

But the requested area is only for x positive, therefore only the point x = 3 is relevant.

The area looks like this:

S=03x+3(x22x3)dx=S=\int_{0}^3 -x+3-(x^2-2x-3) dx=

S=03x2+x+6dx=S=\int_{0}^3 -x^2+x+6 dx=

=[x33+x22+6x]03==[-\frac{x^3}{3}+\frac{x^2}{2}+6x]_0^3=

=333+322+63(033+022+60)==-\frac{3^3}{3}+\frac{3^2}{2}+6\cdot 3-(-\frac{0^3}{3}+\frac{0^2}{2}+6\cdot 0)=

=9+92+180==-9+\frac{9}{2}+18-0=

=9+92==9+\frac{9}{2}=

=1312=13\frac{1}{2}

Have a question? Found a mistake? – Write a comment below!
Was it helpful? You can buy me a cup of coffee here, which will make me very happy and will help me upload more solutions! 

Share with Friends

Leave a Reply