Exercise
Find the area of the region bounded by the graphs of the equations:
y=x^2-2x-3, y=-x+3, x=0, x\geq 0
Final Answer
Solution
First, we find out how the area looks like:
x^2-2x-3=-x+3
x^2-x-6=0
(x+2)(x-3)=0
x=-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=\int_{0}^3 -x+3-(x^2-2x-3) dx=
S=\int_{0}^3 -x^2+x+6 dx=
=[-\frac{x^3}{3}+\frac{x^2}{2}+6x]_0^3=
=-\frac{3^3}{3}+\frac{3^2}{2}+6\cdot 3-(-\frac{0^3}{3}+\frac{0^2}{2}+6\cdot 0)=
=-9+\frac{9}{2}+18-0=
=9+\frac{9}{2}=
=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!