Riemann sum of $e^x$ I understand the how to sum the area under $e^x$ from, say, $[0,1]$ — but how do you sum from $[-1,1]$?
 A: For the left Riemann sums, evaluate $e^x$ at $x=-1+\frac{2k}{n}$, for $k=0$ to $n-1$. 
The same method that you used for $[0,1]$ then works, for we can take the $e^{-1}$ "out."
Added: If we use the left Riemann sum mentioned above, we want
$$\lim_{n\to\infty}\frac{2}{n}\sum_{k=0}^{n-1} e^{-1+2k/n}=e^{-1}\lim_{n\to\infty}\frac{2}{n}\sum_{k=0}^{n-1}e^{2k/n}.$$
The inner sum is a geometric series with common ratio $e^{2/n}$. Summing in the usual way, we find that we want 
$$e^{-1}\lim_{n\to\infty}\frac{2}{n}\frac{e^2-1}{e^{2/n}-1}.$$
We need to evaluate 
$$\lim_{n\to\infty}\frac{2}{n}\frac{1}{e^{2/n}-1}.$$
One way is to make the substitution $\frac{2}{n}=t$. So we want $\lim_{t\to 0^+}\frac{t}{e^t-1}$. 
There are various ways to evaluate this limit, such as L'Hospital's Rule. The limit is $1$. Thus the required Riemann integral is $e^{-1}(e^2-1)$. 
A: You need to compute the Riemann sum over $[-1,0]$ as you already have the other answer. But $e^{-x}=\frac{1}{e^x}=\frac{1}{e}^x$, so you should be able to apply the same procedure to this one as you did before.
