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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?

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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 $[0,3]$.

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  • $\begingroup$ Thanks Vin! I guess I was over complicating things trying to separate out the shaded areas but then not having the necessary other function to do such things. I went with the area with sign. $\endgroup$ – Analytic Lunatic Dec 14 '15 at 20:40
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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

enter image description here

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 $$\int_{-2}^0 f(x) dx - \int_0^3 f(x) dx$$

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