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Comparison theorem

  1. Dec 9, 2007 #1
    1. The problem statement, all variables and given/known data
    Use the Comparison Theorem to determine whether the integral below is convergent or divergent:

    e^-x / sqrt x dx integrated from 0-->1

    3. The attempt at a solution
    I think it is convergent because 1/e^x is convergent. I set the original integral less than or equal to the integral of 1/(e^x) dx

    When I solved for it, I got -1/e + 1, therefore it is convergent. However, my professor marked my paper as saying it's not true. He set the original integral less than or equal to 1/sqrt x, and solving for that, got 2. Why can't my comparison hold true?
  2. jcsd
  3. Dec 9, 2007 #2


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    Because it's not true in (0,1] that e^-x / sqrt x <e^-x
  4. Dec 9, 2007 #3
    Your comparison doesn't hold true because on (0,1), we have that

    [tex] \frac{e^{-x}}{\sqrt{x}} > e^{-x} [/tex]

    This follows from the fact that on (0,1), [itex] \sqrt{x}< 1 \Rightarrow \frac{1}{\sqrt{x}} > 1} [/itex]
  5. Dec 9, 2007 #4
    Haha, barely beaten to it. I knew I shouldn't have wasted my time previewing the post :tongue:
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