Definite Integral – Finding area between 2 polynomials – Exercise 7009

Exercise

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

$y=6-x^2, y=x^2-2x-6$

Final Answer

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Solution

First, we find out how the area looks like:

$6-x^2=x^2-2x-6$

$2x^2-2x-12=0$

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

$x=-2, x=3$

The area looks like this:

$S=\int_{-2}^3 6-x^2-(x^2-2x-6) dx=$

$=\int_{-2}^3 -2x^2+2x+12 dx=$

$=[-2\cdot\frac{x^3}{3}+2\cdot\frac{x^2}{2}+12x]_{-2}^3=$

$=[-2\cdot\frac{x^3}{3}+x^2+12x]_{-2}^3=$

$=-2\cdot\frac{3^3}{3}+3^2+12\cdot 3-(-2\cdot\frac{{(-2)}^3}{3}+{(-2)}^2+12\cdot (-2))=$

$=-18+9+36-\frac{16}{3}-4+24=$

$=27+14\frac{2}{3}=$

$=41\frac{2}{3}$

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