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Math Help - Integral of a function as limit of a sequence

  1. #1
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    Integral of a function as limit of a sequence

    Hey people.
    Given a continuous function f(x) in the interval [a,b] , I need to prove that
    \lim_{n\to\infty} \frac{b-a}{n}\sum_{k=1}^{n}f(a+\frac{k(b-a)}{n}) = \int_a^b f(x)dx

    I think I should do it somehow through Riemann's Integral defnition and/or use Riemann's sums , but I tried to play with it for a while and I haven't gotten anywhere....

    Any ideas how should I start proving this?


    Thanks in advance!
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  2. #2
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    Quote Originally Posted by Gok2 View Post
    Hey people.
    Given a continuous function f(x) in the interval [a,b] , I need to prove that
    \lim_{n\to\infty} \frac{b-a}{n}\sum_{k=1}^{n}f(a+\frac{k(b-a)}{n}) = \int_a^b f(x)dx

    I think I should do it somehow through Riemann's Integral defnition and/or use Riemann's sums , but I tried to play with it for a while and I haven't gotten anywhere....
    I am not sure what is meant by “to prove” in this case.
    You have given a Riemann approximating sum for the integral of f\text{ on }[a,b].
    \lim_{n\to\infty} \frac{b-a}{n}\sum_{k=1}^{n}f(a+\frac{k(b-a)}{n}) in that expression \Delta x = \frac{{b - a}}<br />
{n}\,\& \,t_k  = a + t_k \Delta x\,,\,k = 1,2, \cdots ,n.

    Because f\text{ is continuous on }[a,b] it is Riemann integrable and all valid approximating sums converge to the integral.

    So I don’t know what there is to prove.
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  3. #3
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    Ohhh, I should have looked about this limit as a Riemann sum.
    I see.

    Thanks a lot .
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  4. #4
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    Hey again,
    In my task , after this question, I need to calculate a few limits using the formula
    \lim_{n\to \infty} \frac{1}{n} \sum_{k=1}^n f(\frac{k}{n})  =\int_0^1 f(x)dx

    I managed to calculate all the limits in the task except of this limit

    \lim_{n\to \infty} \frac{\ln(2n+1)+\ln(2n+2)+...+\ln 3n}{n}-\ln n

    I tried many times but I couldn't write this expression as a sum of f(k/n)...
    The problem is what's inside the ln function , i.e 2n+1 , 2n+2 , ... , 2n+k , can't be written as an expression of k/n , so I am kinda stuck....

    Any ideas people ?
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