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Improper integrals

  1. Oct 3, 2011 #1
    1. The problem statement, all variables and given/known data
    integral 1/x^(2/3)dx from -1 to 1

    2. Relevant equations

    3. The attempt at a solution
    so i split it up into two integrals, one with limits going from -1 to b and the other with limits going from c to 1, and taking the limits as b and c go to 0
    i know my antiderivative is 3x^(1/3) and i plugged in my limits of integration and then took the limit as both b and c went to 0 and i got an answer of 3-3(-1)^(1/3) i know this is wrong simply because the cube root of -1 will be give me a complex number but i tried it on wolfram alpha and it gave me the same answer. i looked at the back of my book and it says the answer is 6, i dont understand how the book got an answer of 6
  2. jcsd
  3. Oct 3, 2011 #2


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    Wolfram Alpha is wrong and so is plugging into the antiderivative (which it's likely WA did) for an improper integral. 1/x^(2/3) is actually an even function if you take it to be (1/x^(1/3))^2, isn't it? That should mean you can take the improper integral over (0,1] and double it. What's that integral? BTW (-1)^(1/3) is not necessarily a complex number.
    Last edited: Oct 3, 2011
  4. Oct 3, 2011 #3
    ohh thanks! ill watch out for even and odd functions

    one more quick question though
    i have another improper integral, its 2dx/(x^2-1) from -infinity to -2
    so i broke it up into two integrals, the first one has limits from a to -1 with the limit as a approaches -infinity and the second is from -1 to -2
    am i doing this right?
  5. Oct 3, 2011 #4


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    Why do you need the integral from -1 to -2? -1 isn't in (-infinity,-2). There's no need to break it there. Just integrate from a to -2.
  6. Oct 3, 2011 #5
    ohh alright
    thanks again
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