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?