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Improper integral 1/x^(1/3) from -1 to 8

  1. Mar 25, 2014 #1
    Hi there,
    I am stuck on this problem: the integral of 1/x^(1/3) from -1 to 8.

    I have broken it up into the integral from -1 to 0 and 0 to 8. I am confused as to how the negative values under a cubed root affect things and whether or not I need to break it up further.
    I am not sure whether the limit as n goes to -1 of 3/2(x)^(2/3) is real or imaginary. As if -1 is squared first it becomes real but if not then it is imaginary. We haven't done anything on imaginary integrals in class
     
  2. jcsd
  3. Mar 25, 2014 #2

    Dick

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    Cube roots of negative numbers are negative, not imaginary. E.g. (-2)^(1/3)=-(2)^(1/3).
     
  4. Mar 25, 2014 #3
    thanks so is 3/2(-1)^2/3 a cubed root squared and therefore =3/2?
     
  5. Mar 25, 2014 #4

    Dick

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    Yes, it is.
     
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