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Math Help - division of sections ,bounds definition question

  1. #1
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    division of sections ,bounds definition question

    f(x)=1 for x=\frac{1}{n}
    f(x)=0 for otherwise
    A)
    prove that for 0<a<1 f is integrabile an \int_{a}^{1}f(x)dx=0
    B)prove that for every \epsilon>0 there is a division P of the [0,1] section so
    S(P)<\epsilon
    C)prove that f is integrabile on [0,1] and \int_{0}^{1}f(x)dx=0

    how i tried:
    part A i have solved
    its very easy i just couddnt see that 1/n is behind a and till 1 we have n-1 fractures
    so its 0 exsept finite number of points so the integral is zero


    i am stuck on part B:


    i was told to look at [\frac{\epsilon}{2},1]section so from part A \int_{\frac{\epsilon}{2}}^{1}f(x)dx=0and because f is integrabile on the section then \int_{\frac{\epsilon}{2}}^{1}f(x)dx=inf\{S(P)|P=di  viding[\frac{\epsilon}{2},1]}
    “there exists deviding Q of [\frac{\epsilon}{2},1] so S(Q)<\frac{\epsilon}{2} (other wise \frac{\epsilon}{2} would be lower bound of the upper sums and the integral will not be zero.”

    why the integral would be zero???

    i cant see how to get it
    i know that if its integrabile then the integral equals the lower bound of the upper sums
    but why it will be zero?
    Last edited by transgalactic; September 17th 2011 at 01:58 AM.
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  2. #2
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    Re: division of sections ,bounds definition question

    Choose your intervals so that the endpoints are 1/2, 1/3, 1/4, etc.
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  3. #3
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    Re: division of sections ,bounds definition question

    why??
    how it will show that the integral will not be zero
    ?
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  4. #4
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    Re: division of sections ,bounds definition question

    It won't! The integral is 0!!!
    (C) specifically says that!
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  5. #5
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    Re: division of sections ,bounds definition question

    Quote Originally Posted by transgalactic View Post
    f(x)=1 for x=\frac{1}{n}
    f(x)=0 for otherwise
    B)prove that for every \epsilon>0 there is a division P of the [0,1] section so
    S(P)<\epsilon
    C)prove that f is integrabile on [0,1] and \int_{0}^{1}f(x)dx=0
    i am stuck on part B:
    Let \delta=\frac{\epsilon}{2} then the a positive integer, J, such that if n\ge J then \frac{1}{n}<\delta.
    Now note that the interval [0,\delta] has an upper sum which is less that \delta^2.
    Now cover the J-1 points 1,\frac{1}{2},\cdots,\frac{1}{J-1} that add up to a number less than \frac{\epsilon}{2}.
    Now we have a partition and the upper sum is less than \epsilon
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  6. #6
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    Re: division of sections ,bounds definition question

    A similar thing in slightly different words.
    Quote Originally Posted by transgalactic View Post
    i was told to look at [\frac{\epsilon}{2},1]section so from part A \int_{\frac{\epsilon}{2}}^{1}f(x)dx=0and because f is integrabile on the section then \int_{\frac{\epsilon}{2}}^{1}f(x)dx=inf\{S(P)|P=di  viding[\frac{\epsilon}{2},1]}
    “there exists deviding Q of [\frac{\epsilon}{2},1] so S(Q)<\frac{\epsilon}{2} (other wise \frac{\epsilon}{2} would be lower bound of the upper sums and the integral will not be zero.”
    Since \inf\{S(P)\mid P\ \mbox{is a division of}\ [\epsilon/2,1]\}=0, there exists a division Q such that S(Q) <\epsilon/2. This follows from the definition of infimum. For the segment [0,\epsilon/2], choose any division. Since the value of f(x) does not exceed 1, the upper sum of the segment [0,\epsilon/2] does not exceed \epsilon/2. For a division for the whole segment [0, 1] take the union of any division on [0,\epsilon/2] and Q. Then the corresponding sum will be < \epsilon.
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