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# 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

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

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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