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Math Help - function integrable?

  1. #1
    Super Member Anonymous1's Avatar
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    function integrable?

    Let n be a fixed integer >0; consider the function


    f_{n}(x) = \{0, if (x = \frac{1}{n}, \frac{2}{n},...,\frac{n-1}{n}),  1, otherwise.\}
    Prove directly from the definition of integrability that f_{n}(x) is integrable on [0,1].

    All I've got so far is that: M_{i} = (\frac{i}{n}); m_{i} = (\frac{i-1}{n}); \Delta x_{i} = \frac{1}{n}.

    Any help? Thanks
    Last edited by Anonymous1; November 8th 2009 at 01:42 AM.
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by Anonymous1 View Post
    Let n be a fixed integer >0; consider the function


    f_{n}(x) = \{0, if (x = \frac{1}{n}, \frac{2}{n},...,\frac{n-1}{n}),  1, otherwise.\}

    Prove directly from the definition of integrability that f_{n}(x) is integrable on [0,1].

    All I've got so far is that: M_{i} = (\frac{i}{n}); m_{i} = (\frac{i-1}{n}); \Delta x_{i} = \frac{1}{n}.

    Any help? Thanks
    f_n(x)=\left\{\begin{array}{lr}0:&x=\frac{k}{n},~k  \in\mathbb{N},~1\leq k<n\\1:&otherwise\end{array}\right\}

    One thing that should be obvious is that U(P,f)=1, since for all intervals in any partition you choose, M_i will always equal 1. Proving that L(P,f)=1 is a matter of choosing a clever partition. I don't think the "standard" partition (i.e. the one you're using) will work. You want a partition such that the total length of the n-1 intervals containing the \frac{k}{n} can be made as small as desired.
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  3. #3
    Super Member Anonymous1's Avatar
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    I don't think this function is Riemann integrable. Anyone disagree?


    [edit] Wait yes it is, since it only has a finite number of discontinuities.
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    Quote Originally Posted by Anonymous1 View Post
    I don't think this function is Riemann integrable. Anyone disagree?
    I definitely disagree!
    Because I know that \left( {\forall n} \right)\left[ {f_n } \right] has only n-1 discontinuities.
    Therefore, it is Riemann intergrable.
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  5. #5
    Super Member Anonymous1's Avatar
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    Hey just came back to look at this problem.

    So, when I use the above partition I get \frac{1}{2} as the limit for both the upper and lower sum. This seems to violate my intuitive answer of 1.

    So my question is: can I just say, for some N > \frac{n-1}{n} the sequence will get and stay at 1, hence the limit of the upper sum is 1 which implies the value of the integral is 1.

    Any faulty logic going on here?

    Thanks
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