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This is my second time posting. Thanks again for answering my first question.
This is my question:

Find the area above y = 0, from x = -1 to x = 1 of y = 2^x. Use subintervals of equal length. I started with this summation:

$$\sum_{i=1}^n\frac{2}{n}\left(2^{2i/n}\right)$$

I am not sure where to go next. I was trying to factor using exponents laws to get the sum into a form so that a summation formula could be applied to get the sum into closed form. I do have the solution, but I don't understand how they got to the step right after this:

Here is the solution:

enter image description here

If anyone could help me, that would be fantastic.
Thank you guys for your time.

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  • $\begingroup$ In the given solution, do you have trouble with the first line? Or going from the first line to the second? Or in going from the second to the third? $\endgroup$ – André Nicolas Jan 15 '15 at 4:45
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You've made an error in representing a Riemann sum for $2^x$. The $2^{2i/n}$ in your summands should be $2^{-1 + 2i/n}$. Remember, you're partitioning the interval $[-1,1]$, not $[0,2]$. So your sum should be $$\sum_{j = 1}^n \frac{2}{n}\cdot2^{-1+2j/n}$$ which is the same as $$\frac{2}{n}\sum_{j = 1}^n 2^{-1}(2^{2/n})^j = \frac{1}{n}\sum_{j = 1}^n (2^{2/n})^j$$ The last sum can be written $$\frac{2^{2/n}}{n}\sum_{j = 1}^n (2^{2/n})^{j-1}$$ which, by the way, is the same as $$\frac{2^{2/n}}{n}[1 + 2^{2/n} + (2^{2/n})^2 + \cdots + (2^{2/n})^{n-1}]$$ as expressed in the solution.

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  • $\begingroup$ After that, the sum is rearranged in such a way as to use the geometric series; then computing the limit of this geometric series divided by $n$ requires a L'Hopital's rule problem. (It might be good to include this in the answer.) $\endgroup$ – Ian Jan 15 '15 at 4:44

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