1. ## 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.

2. Originally Posted by cyber_ninja83
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

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