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Math Help - Definite integral question

  1. #1
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    Definite integral question

    Integral from 0 to 1 (1-x)9

    I did u substitution
    u=1-x
    du=-1dx
    du/-1 =dx

    get new intervals
    1-1=0
    1-0=1

    -1 integral u^9 du
    -1 u10/10 interval 1 to 0

    -1 (010/10 - 110/10)

    =1/10

    Am I correct?
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  2. #2
    Member sbhatnagar's Avatar
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    Re: Definite integral question

    \int_0^1 (1-x)^9 dx = \frac{\Gamma(1)\Gamma(10)}{\Gamma(11)}=\frac{9!}{1  0!}=\frac{1}{10}

    Yes, your answer is correct!
    Last edited by sbhatnagar; March 3rd 2013 at 11:54 PM.
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  3. #3
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    Re: Definite integral question

    Quote Originally Posted by Steelers72 View Post
    Integral from 0 to 1 (1-x)9

    I did u substitution
    u=1-x
    du=-1dx
    du/-1 =dx

    get new intervals
    1-1=0
    1-0=1

    -1 integral 1/u^9 du <-- unfortunately wrong
    -1 u10/10 interval 1 to 0

    -1 (010/10 - 110/10)

    =1/10

    Am I correct?
    You certainely noticed that \frac1{(1-x)^9} is not defined for x = 1, that means the graph of f(x)=\frac1{(1-x)^9} has a vertical asymptote x = 1. Therefore the definite integral doesn't exist.

    EDIT: Sorry, I misunderstood your question completely!
    Last edited by earboth; March 3rd 2013 at 11:54 PM.
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