Integration: Area between curves Let $f(x)=x^2−c^2$ and $g(x)=c^2−x^2.$ Find $c>0$ such that the area of the region enclosed by the parabolas $f(x)$ and $g(x)$ is 9. 
The question above is what I am having trouble with. In order to solve this problem I use the formula given as:
$\int_a^b f(x) - g(x) dx$
Here is what I have done so far:
$9 = \int x^2−c^2 - (c^2−x^2) dx$
$9 = \int 2x^2−2c^2 dx$
$9 = 2\int x^2−c^2 dx$
$\frac 9 2 = \int x^2−c^2 dx$
To find a and b:
$x^2 - c^2 = (x-c)(x+c)$ ; therefore  $a = -c$ and $b = c$
Back to the original question, sub in a and b:
$\frac 9 2 = \int_{-c}^c x^2−c^2 dx$
This is as far as I have gotten. I am not sure what to do next. Thanks in advance to anyone who can help.

Draw a picture. I would use symmetry and say that the area is $4$ times the integral from $0$ to $c$ of $c^2-x^2$. So we want
$$\int_0^c (c^2-x^2)\,dx=\frac{9}{4}.$$
The integral is equal to $\frac{2}{3}c^3$. Now solve for $c$.
Remark: Your approach will work, except that we want $\int_{-c}^c (c^2-x^2)\,dx=\frac{9}{2}$. Note that $\int_{-c}^c (x^2-c^2)\,dx$ is negative. 
