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Math Help - Riemann Sum Help

  1. #1
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    Riemann Sum Help

    I am having trouble with Riemann sum for a geometric series. The question prompted me to find the area of the region beneath the curve of e^x from x = 0 to x = 1.

    Here is what I have so far:

    change in x = 1/n
    xi = i/n
    f(x) = e^x
    f(
    xi ) = e^(i/n)
    Area = lim as n goes to infinity of the sum from i = 1 to n of e^(i/n) times 1/n

    then, since this is a geometric sequence, it can be simplified to:
    Area = lim as n goes to infinity of 1/n * e^(1/n) * (e-1)/(e^(1/n)-1)
    i have no clue how to solve this limit. I tried l'Hopital's rule, but got nowhere. Any help is very much appreciated.
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  2. #2
    Grand Panjandrum
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    someplace
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    Quote Originally Posted by cyber_ninja83 View Post
    I am having trouble with Riemann sum for a geometric series. The question prompted me to find the area of the region beneath the curve of e^x from x = 0 to x = 1.

    Here is what I have so far:

    change in x = 1/n
    xi = i/n
    f(x) = e^x
    f(
    xi ) = e^(i/n)
    Area = lim as n goes to infinity of the sum from i = 1 to n of e^(i/n) times 1/n

    then, since this is a geometric sequence, it can be simplified to:
    Area = lim as n goes to infinity of 1/n * e^(1/n) * (e-1)/(e^(1/n)-1)
    i have no clue how to solve this limit. I tried l'Hopital's rule, but got nowhere. Any help is very much appreciated.
    You need to find:

    \lim_{n \to \infty} \frac{e^{1/n}(e-1)}{n(e^{1/n}-1)}

    .......... = (e-1)\lim_{n \to \infty} \frac{(1/n)}{1-e^{-1/n}}

    Now one application of L'Hopital's rule will show that the limit above is 1, and so:

    \lim_{n \to \infty} \frac{e^{1/n}(e-1)}{n(e^{1/n}-1)}=e-1

    RonL
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  3. #3
    Newbie
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    Sep 2008
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    Alright, I was forgetting to apply the chain rule to e^(1/n), so I messed up the derivative. Thanks for your help!
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