Variable Endpoint of definite integral. My book asks to find the number $c$ such that the region bounded by the curves $y = x^2$, and $y = c$ has area 36. I understand the a definite integral will solve the question but am having a hard time applying the right interval. Right now I have the proper integral with constants moved out: $$2c - \int_0^{x} x^2 dx = 36$$
The resulting $2c - \frac{x^3}{3} = 36$ is confusing. I assume that using the correct interval will change the expression to be in terms of c. Any direction would be greatly appreciated.
 A: You need to find where $y = x^2 = c$:
$y = x^2 = c \implies x = \pm \sqrt c$. Use those points as your interval endpoints. 
And you need to integrate the constant. Recall that $\int c\,dx = cx + \text{constant}$.
$$\int_{-\sqrt c}^{\sqrt c} (c - x^2) \,dx = 36.$$
Integrate, then evaluate at the bounds of integration, and set equal to $36$. Then you can solve for $c$ to complete the task.
$$\int_{-\sqrt c}^{\sqrt c} (c - x^2) \,dx = 36 $$
$$\iff cx - \frac{x^3}{3}\Big|_{-\sqrt c}^{\sqrt c} = 36.$$
Since $y = x^2$ and $y = c$ are both symmetric about the y-axis, and the interval of integration is centered at $x = 0$, it suffices to compute:
$$2\int_{0}^{\sqrt c} (c - x^2) \,dx = 36 $$
$$\iff 2\Big(cx - \frac{x^3}{3}\Big)\Big|_0^{\sqrt c} = 2c\cdot \sqrt c - \frac 23(\sqrt c)^3 = 36$$ Simplify and solve for $c$.
A: You have some problems in your setup. Let's address those first.
Suppose $x = \pm a$ are the points where $y = x^2$ and $y = c$ intersect (assume $a > 0$ for concreteness). Then the integral in question is
$$
\int_{-a}^a c - x^2 \,dx = 36
$$
so
$$
2ac - \frac{2}{3}a^3 = 36.
$$
Now, you need to find $a$ in terms of $c$: remember that $x=a$ is where $y =x^2$ and $y = c$ intersect to do this.
