Use a suitable sequence of step functions, with the partition $P=\{0, \frac{1}{n}, \cdots, \frac{n-1}{n}, 1 \}$ to show that $\int_0^1 e^x = e-1$
So I know that for any regulated function $f:[a,b] \rightarrow \mathbb{R}$ (which $e^x$ is since it is continuous), $$\int_a^b f(x) dx = \lim_{n\rightarrow \infty} \int_a^b \phi_n (x) dx$$ where $\phi_n$ is a sequence of step functions that converge uniformly to $f$.
So on the interval $[\frac{k-1}{n}, \frac{k}{n}]$, I have chosen $\phi_n = e^{\frac{k}{n}}$, and then $$\int_0^1 e^x dx = \lim_{n\rightarrow \infty} \sum_{k=1}^n \frac{1}{n} e^{\frac{k}{n}} = \lim_{n\rightarrow \infty} \frac{1-e}{n-ne^n}$$ but unfortunately this does not convegre to $e-1$.
In the solution, they use $\phi_n = e^{\frac{k-1}{n}}$, but I don't see why their choice of $\phi_n$ is correct and my choice is wrong.
You have a typo at $$\int_0^1 e^x dx = \lim_{n\rightarrow \infty} \sum_{k=1}^n \frac{1}{n} e^{\frac{k}{n}} = \color{red}{ \lim_{n\rightarrow \infty} \frac{1-e}{n-ne^n}}$$ It should be $$\int_0^1 e^x dx = \lim_{n\rightarrow \infty} \sum_{k=1}^n \frac{1}{n} e^{\frac{k}{n}} = \color{blue}{ \lim_{n\rightarrow \infty} \frac{(1 - e) e^{\frac{1}{n}}}{n\left(1 - e^{\frac{1}{n}}\right)}}$$ where $$\lim_{n\rightarrow \infty}e^{\frac{1}{n}}=1$$ $$\lim_{n\rightarrow \infty} n\left(1 - e^{\frac{1}{n}}\right)=\lim_{n\rightarrow \infty} \frac{1 - e^{\frac{1}{n}}}{\frac{1}{n}}=-1$$ from $$\lim_{x\rightarrow 0} \frac{e^{x}-1}{x}=1$$