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

Show 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:

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

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

$=[-\frac{x^3}{3}+\frac{x^2}{2}+6x]_0^3=$

$=-\frac{3^3}{3}+\frac{3^2}{2}+6\cdot 3-(-\frac{0^3}{3}+\frac{0^2}{2}+6\cdot 0)=$

$=-9+\frac{9}{2}+18-0=$

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

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

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