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Math Help - Integrate this function

  1. #1
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    Integrate this function

    Let f: \mathbb {R} \rightarrow \mathbb {R} be defined by  f(x) = 1 if x= \frac {1}{n} \ \ \ n \in \mathbb {N} , and f(x) = 0 otherwise.

    Prove that  \int ^1 _0 f exist and find its value.

    Question so far:

    Define a partition P on [0,1] with P = \{ a_0=0 , a_1,a_2,...,a_{N-1},a_N=1 \}

    Define M_j = sup \{ f(x) : x \in [x_j,x_{j+1}]
    m_j = inf \{ f(x) : x \in [x_j,x_{j+1}]

    So I need to show that for any given  \epsilon > 0 , we will have U(f,P)-L(f,P)< \epsilon , but how should I process? Thanks.
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  2. #2
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    Quote Originally Posted by tttcomrader View Post
    Let f: \mathbb {R} \rightarrow \mathbb {R} be defined by  f(x) = 1 if x= \frac {1}{n} \ \ \ n \in \mathbb {N} , and f(x) = 0 otherwise.

    Prove that  \int ^1 _0 f exist and find its value.
    obviously L(f,P)=0, for any partition P of [0,1]. so: \int_{-} f = 0. suppose n \geq 2. let r= \frac{1}{2n^2} and define the partition P_n by:

    x_0=0, \ \ x_1=\frac{1}{n}, \ \ x_2=\frac{1}{n} \ + \ r, \ \ x_{2i+1}=\frac{1}{n-i} \ - \ r, \ \ x_{2i+2}= \frac{1}{n-i} \ + \ r, \ \ 1 \leq i \leq n-2, and x_{2n-1}=1-r, \ \ x_{2n}=1. see that

    U(f,P_n)=\frac{2n-1}{n^2} and thus: 0 \leq \int^{-} f - \int_{-} f = \int^{-} f \leq U(f,P_n) = \frac{2n-1}{n^2}. therefore f is integrable and \int_0^1 f = 0. \ \ \ \Box
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  3. #3
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    Thanks for your help, but there is something that I ain't clear about here.

    Now, I know that
    <br />
U(f,P_n) = \sum ^{2n-1}_{j=0}[sup \{ f(x) : x \in [x_j,x_{j+1}](x_{j+1}-x_j), but how does this equals to  \frac{2n-1}{n^2}<br />
?

    Thanks again!
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