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Math Help - Improper integral with infinite discontinuity?

  1. #1
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    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
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  2. #2
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    Quote Originally Posted by DannyMath View Post
    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 3(-8)^{2/3}. You might think that, since -2 is the cube root of -8 and then you square it that (-8)^{2/3} = 4. But you'd be wrong. We must take that negative out and leave it on its own:
    (-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: \sqrt{-2}\sqrt{-8}. This expression is not \sqrt{-2}\sqrt{-8} = \sqrt{-2 \cdot -8} = \sqrt{16} = 4. This is

    \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 (-1)^{2/3} = -\frac{1}{2} + i~\frac{ \sqrt{3}}{2}
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    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.
    Last edited by DannyMath; March 17th 2011 at 08:04 PM.
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  4. #4
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    here is a link to some discussion over Wolfram alpha's response to roots of negative values ...

    Wolfram|Alpha Community • 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.
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    It would not be 4 x 3 = 12?
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  6. #6
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    Quote Originally Posted by DannyMath View Post
    It would not be 4 x 3 = 12?
    yes, 12 ... forgot about the 3 since the discussion centered around (-8)^{\frac{2}{3}}
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