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Math Help - Riemann sum: Limit

  1. #1
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    Riemann sum: Limit

    I do not understand how

    \lim_{n \rightarrow \infty} \sum_{i=1}^n \, c \frac{b-a}{n}  = c(b-a) \, .

    Can somebody show the steps?
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Riemann sum: Limit

    Note that \sum_{i=1}^{n} \frac{1}{n} = 1, thus we obtain

    \lim_{n \to \infty}\sum_{i=1}^{n} c \frac{b-a}{n} = c(b-a) \lim_{n \to \infty} \sum_{i=1}^{n} \frac{1}{n} = c(b-a) \lim_{n \to \infty} 1 = c(b-a)
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    Re: Riemann sum: Limit

    Quote Originally Posted by Siron View Post
    Note that \sum_{i=1}^{n} \frac{1}{n} = 1
    Where did that come from, never seen that before. And how exactly does it equal one? The first two terms make up 1 + 0.5 = 1.5.
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  4. #4
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    Re: Riemann sum: Limit

    Quote Originally Posted by Siron View Post
    Note that \sum_{i=1}^{n} \frac{1}{n} = 1[/tex]
    That is a false statement.

    Quote Originally Posted by MathCrusader View Post
    I do not understand how
    \lim_{n \rightarrow \infty} \sum_{i=1}^n \, c \frac{b-a}{n}  = c(b-a) \, .
    That is an approximating sum for the integral \int_a^b {cdx} .

    You can see that \forall n,~\Delta x=\frac{b-a}{n} and f(a+k\Delta x)\Delta x=c~\frac{b-a}{n}
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