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

Final Answer


S=13\frac{1}{2}

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}

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