# Thread: Improper integral with infinite discontinuity?

1. ## Improper integral with infinite discontinuity?

Hi,

Me and my friend tried to solve:

integral from 1 to 9 of -2/(x-9)^(1/3) dx

So we set it as a lim t --> 9- and got -12 but the homework system says it's wrong. I checked wolfram and it gives the answer as 12(-1)^(2/3) so it's a bit confusing. Not to mention that the graph's area is positive from 1 to 9. Any clarification is appreciated

2. Originally Posted by DannyMath
Hi,

Me and my friend tried to solve:

integral from 1 to 9 of -2/(x-9)^(1/3) dx

So we set it as a lim t --> 9- and got -12 but the homework system says it's wrong. I checked wolfram and it gives the answer as 12(-1)^(2/3) so it's a bit confusing. Not to mention that the graph's area is positive from 1 to 9. Any clarification is appreciated
Well, you've run into a weird part of Mathematics. The integral comes out to be $\displaystyle 3(-8)^{2/3}$. You might think that, since -2 is the cube root of -8 and then you square it that $\displaystyle (-8)^{2/3} = 4$. But you'd be wrong. We must take that negative out and leave it on its own:
$\displaystyle (-8)^{2/3} = 8^{2/3}(-1)^{2/3} = 4(-1)^{2/3}$.

This is a similar rule to why you have to factor -1s out of the following: $\displaystyle \sqrt{-2}\sqrt{-8}$. This expression is not $\displaystyle \sqrt{-2}\sqrt{-8} = \sqrt{-2 \cdot -8} = \sqrt{16} = 4$. This is

$\displaystyle \sqrt{-2}\sqrt{-8} = \sqrt{-1}\sqrt{2} \cdot \sqrt{-1}\sqrt{8} = i^2 \cdot \sqrt{16} = -\sqrt{16} = -4$.

-Dan

In case it makes you feel better, playing around in the complex plane for a bit gives
$\displaystyle \displaystyle (-1)^{2/3} = -\frac{1}{2} + i~\frac{ \sqrt{3}}{2}$

3. I've not learned about imaginary numbers but I plan on it in the summer. From your post I get that you also say that it's -12? It just seemed odd to me that it would be negative since the function is above the x-axis on the given limits of integration.

4. here is a link to some discussion over Wolfram alpha's response to roots of negative values ...

Wolfram|Alpha Community &bull; View topic - Odd Roots of Negative Numbers

since you're dealing with real values, you'd be safe in saying the value of the improper integral is +4.

5. It would not be 4 x 3 = 12?

6. Originally Posted by DannyMath
It would not be 4 x 3 = 12?
yes, 12 ... forgot about the 3 since the discussion centered around $\displaystyle (-8)^{\frac{2}{3}}$