TITLE: The integral: $\int_0^1(e^x)dx$ via Darboux's sums
QUESTION [6 upvotes]: Need to calculate the integral: $\int_0^1(e^x)dx$ via Darboux's sums.

My attempt:
First, we calculate the lower summation: $$L(p,f)=\sum_{i=1}^n m_i\Delta x_i=\sum_{i=1}^n(e^{\frac{i-1}{n}}\cdot\frac{1}{n})=\frac{1}{n}\cdot e^{\frac{n(n-1)}{2n}}=\frac{1}{n}\cdot e^{\frac{n-1}{2}}$$
Second, we calculate the upper summation: $$U(p,f)=\sum_{i=1}^n M_i\Delta x_i=\sum_{i=1}^n(e^{\frac{i}{n}}\cdot\frac{1}{n})=\frac{1}{n}\cdot e^{\frac{n(n+1)}{2n}}=\frac{1}{n}\cdot e^{\frac{n+1}{2}}$$
From the result, it easy to see that both of the limits are $\infty$.

I know that there is a mistake over here, I would appreciate that, if you can enlighten me.

REPLY [3 votes]: Elaborating on projectilemotion's comment,
$$\begin{align*}L(p_n,f)=\frac{1}{n}\sum_{i=1}^n e^{\frac{i-1}{n}}&=\frac{1}{n}\sum_{i=1}^{n}\left(e^{1/n}\right)^{i-1}=\frac{1}{n}\left(\frac{e-1}{e^{1/n}-1}\right)=(e-1)\left(\frac{1/n}{e^{1/n}-1}\right)\end{align*}$$
Then $$\lim_{x\to 0}\frac{x}{e^x-1}=1$$ where we replace $1/n$ with $x$ so that $$\lim_{n\to\infty}L(p_n,f)=e-1$$
Similarly, we have
$$U(p_n,f) = \sum_{i=1}^{n}e^{i/n}=\frac{e^{1/n}}{n}\sum_{i=1}^{n}\left(e^{1/n}\right)^{i-1}=\frac{e^{1/n}}{n}\left(\frac{e-1}{e^{1/n}-1}\right)=(e-1)\left(\frac{e^{1/n}}{e^{1/n}-1}\right)\cdot\frac{1}{n}$$
Then notice that $$\left(\frac{e^{1/n}}{e^{1/n}-1}\right)=\frac{1}{1-e^{-1/n}}$$ and $$\lim_{x\to 0}\frac{x}{1-e^{-x}}=1$$ where we again, replace $1/n$ with $x$.
Then we have $$\lim_{n\to\infty}U(p_n,f)=e-1$$
Note that if you are writing a proof, you may want to justify the limits a bit more rigorously than I did; this is only a sketch of an answer.