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Math Help - Darboux Sums

  1. #1
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    Darboux Sums

    So we were doing Darboux sums in Analysis, and this was a given sample question and answer, but I don't really understand where part of this solution comes from:

    "Suppose f is increasing on [0,1]. Show that f is integrable on [0,1]

    Let P = \left\{ x_{0},x_{1},...,x_{n} \right\} be a partition of [0,1] into n subintervals of the same length.
    Since f(x) is increasing, we have
    \inf_{[x_{k+1},x_{k}]} f(x) = f(x_{k}) for 0 \leqslant k \leqslant n-1
    and this implies that

    S^{-}(f,P) = \sum_{k = 0}^{n-1} \inf_{[x_{k+1},x_{k}]} f(x).(x_{k+1}-x_{k})

    = \frac{1}{n}(f(x_{0})+f(x_{1})+...+f(x_{n-1})

    But I don't understand how those last two statements are equal; for example, how is

    \sum_{k = 0}^{2} f(x).(x_{k+1}-x_{k})

    = (f(x_{0})(x_{1}-x_{0})+f(x_{1})(x_{2}-x_{1})+f(x_{2})(x_{3}-x_{2}))

    equal to

    \frac{1}{3} (f(x_{0})+f(x_{1})+f(x_{2})?
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  2. #2
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    Quote Originally Posted by Conn View Post
    So we were doing Darboux sums in Analysis, and this was a given sample question and answer, but I don't really understand where part of this solution comes from:

    "Suppose f is increasing on [0,1]. Show that f is integrable on [0,1]

    Let P = \left\{ x_{0},x_{1},...,x_{n} \right\} be a partition of [0,1] into n subintervals of the same length.
    Since f(x) is increasing, we have
    \inf_{[x_{k+1},x_{k}]} f(x) = f(x_{k}) for 0 \leqslant k \leqslant n-1
    and this implies that

    S^{-}(f,P) = \sum_{k = 0}^{n-1} \inf_{[x_{k+1},x_{k}]} f(x).(x_{k+1}-x_{k})

    = \frac{1}{n}(f(x_{0})+f(x_{1})+...+f(x_{n-1})

    But I don't understand how those last two statements are equal; for example, how is

    \sum_{k = 0}^{2} f(x).(x_{k+1}-x_{k})

    = (f(x_{0})(x_{1}-x_{0})+f(x_{1})(x_{2}-x_{1})+f(x_{2})(x_{3}-x_{2}))

    equal to

    \frac{1}{3} (f(x_{0})+f(x_{1})+f(x_{2})?
    It follows from the fact that

    (x_{k+1}-x_k)=\frac{1}{n}, \quad \forall \, k

    Then just factor it out of each term!
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