# Find the area of $y = 12-x^2, y=x^2-6$

I found the x-intercepts via my graphing calculator to be $$(-3,3)$$ and $$(3,3)$$ and from it I formed the Area equation,

$$A = \int_{-3}^{3} (12-x^2 - x^2 - 6)dx = \int_{-3}^{3}(-2x^2+6)dx \\ = -\frac{2}{3}x^3 + 6x \bigg]_{-3}^{3}$$ $$A = \bigg(3\frac{2}{3}\cdot 3^3 + 6(3) \bigg) - \bigg(-\frac{2}{3} \cdot -3^3 + 6(-3)\bigg) \\ = 0 - 0$$

I calculated the integral and solved for the definite integral, but I got 0....I did the calculations correctly, but the answer doesn't seem to make sense for what the graphs look like.

Did I set the problem up wrong? or something?

Edit: I think I need to split the integral bounds up to $$\int_{-3}^{0}$$ and $$\int_{0}^{3}$$ to split the positive/negative area portions

• You missed a pair of parentheses: $(12-x^2 -( x^2 - 6))$ – Bernard Massé Mar 20 at 23:42

$$A = \int_{-3}^{3} (12-x^2 - (x^2 - 6))dx = \int_{-3}^{3}(-2x^2+18)dx = -\frac{2}{3}x^3 + 18x \bigg]_{-3}^{3}$$ $$A = \bigg(-\frac{2}{3}\cdot 3^3 + 18(3) \bigg) - \bigg(-\frac{2}{3} \cdot -3^3 + 18(-3)\bigg)=36-(-36)=72.$$
$$12-x^2-\color{red}(x^2-6\color{red})=\color{red}{18}-2x^2\implies A=\left.18x-2\frac{x^3}3\right]_{-3}^3=72.$$