Calculus area between graph and x axis 
How to solve this question ? 
The integral would be 0.3x^3+xc^2 + c
0.3(64)^3 +(64)c^2 +c =0 
c= 78643.2/( -64+c) 
am I right ? 
 A: Draw a picture. You will get a familiar parabola, with axis the $y$-axis. Note that the region we want the area of is below the $x$-axis, and extends from $x=-c$ to $x=c$. By symmetry, the area is 
$$2\int_0^c (0-(x^2-c^2))\,dx.$$
So we are integrating $c^2-x^2$, and then doubling.  Calculation gives $\frac{4}{3}c^3$.
We want this to be $64$. It is now not hard to solve for $c$.
A: The problem should be restated because the area "between" the function $y=x^2-c^2$ and $y=0$ is always equal to $\infty$.
I would restated the problem as follows

Find the exact positive value of $c$ if the area of the region bounded by $y=x^2-c^2$ and the x-axis is 64.


Next, make some sketches.
Clearly the region bounded by the curve and the x-axis is the region in blue. The area between the 2 curves is the |pink area|+|blue area|.
The area bounded would be
$$\bigg|\int_{-c}^c(x^2-c^2)dx\bigg|$$
$$= \bigg|\frac{x^3}{3}-c^2x\bigg|_{-c}^c\bigg| = \bigg|\bigg((\frac{c^3}{3}-c^3) - (-\frac{c^3}{3}-(-c^3))\bigg)\bigg|$$
$$= \bigg|-\frac{2}{3}c^3- \frac{2}{3}c^3\bigg|$$
$$= \bigg|-\frac{4}{3}c^3\bigg| = \frac{4}{3}c^3$$
But
$$\bigg|\int_{-c}^c(x^2-c^2)dx\bigg| = 64$$
so
$$\frac{4}{3}c^3 = 64$$
