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Thread: Integrate this function

  1. #1
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    Integrate this function

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

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

    Question so far:

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

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

    So I need to show that for any given $\displaystyle \epsilon > 0 $, we will have $\displaystyle 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 $\displaystyle f: \mathbb {R} \rightarrow \mathbb {R} $ be defined by $\displaystyle f(x) = 1 $ if $\displaystyle x= \frac {1}{n} \ \ \ n \in \mathbb {N} $, and $\displaystyle f(x) = 0$ otherwise.

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

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

    $\displaystyle U(f,P_n)=\frac{2n-1}{n^2}$ and thus: $\displaystyle 0 \leq \int^{-} f - \int_{-} f = \int^{-} f \leq U(f,P_n) = \frac{2n-1}{n^2}.$ therefore $\displaystyle f$ is integrable and $\displaystyle \int_0^1 f = 0. \ \ \ \Box$
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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
    $\displaystyle
    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 $\displaystyle \frac{2n-1}{n^2}
    $?

    Thanks again!
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