<s>[INST] 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
 [/INST] $$A = \int_{-3}^{3} (12-x^2 - (x^2 - 6))dx = \int_{-3}^{3}(-2x^2+18)dx
= -\frac{2}{3}x^3 + 18x \bigg]_{-3}^{3}$$
$$A = \bigg(-\frac{2}{3}\cdot 3^3 + 18(3) \bigg) - \bigg(-\frac{2}{3} \cdot -3^3 + 18(-3)\bigg)=36-(-36)=72.$$
You forgot the second set of parentheses inside the first integral.  You don't need to split up the limits of integration.
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