Find the total area between region and x-axis? I'm having some difficulty with this problem out of my Calculus Book: 


  Find the total area between region and the $x$-axis: $y=x^3-x^2-6x$,  $-2 \le x \le 3.$


I know I start with setting the function equal to $0$:

$$0=x^3-x^2-6x$$
$$0=x(x-3)(x-2)$$
Hence $x=3$ or $x=2$.

I then need to take the integral of the Top Function minus the bottom function of each respective area:

$$\text{Area} = \int\limits_{a}^{b} (\text{Top}_f - \text{Btm}_f)\, dx$$

The problem I'm having is getting the initial function split in half? I need the function for the portion above the $x$-axis to the $x$-axis, PLUS the portion below the $x$-axis to the $x$-axis added together to get the entire area...?

Can someone walk me through what I am missing here?
You use thr Rienmann integral. Then $A=\int_{-2 }^3(x^3-x^2-6x) dx=[x^4/4-x^3/3-3x^2]_{x-2}^{x=3}$. This is the area with sign. If you want the area without sign: $A= \int_{-2} ^0(x^3-x^2-6x) dx- \int_{0} ^3(x^3-x^2-6x) dx $. The polynomial is negative in $[0,3]$.
You factored incorrectly. Note that $$f(x) = x^3 - x^2 - 6x = x(x-3)(x+2)$$ so the roots lie at $-2,0,3$. The plot looks like



If you want the real area between the curve and the $x$-axis, the region to the right of $0$ must be negated, since it is under the $x$-axis and you get $$\int_{-2}^0 f(x) dx - \int_0^3 f(x) dx$$