0
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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 $[0,3]$.

$\endgroup$
  • $\begingroup$ 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
-1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.