Let $a$ and $b$ be any two real numbers such that $a < b$, and let $f$ be the real-valued function defined on $[a, b]$ by the formula $$ f(x) = \mathrm{e}^x \ \mbox{ for all } \ x \in [a, b]. $$
Then how to evaluate $$ \int_a^b f(x) \ \mathrm{d} x, $$ using the definition of the integral given here?
My Attempt:
Since $f$ is strictly increasing on $[a, b]$ [Can we show this fact using the machinery developed in the first seven chapters of Baby Rudin?], so it is Riemann-integrable on $[a, b]$; that is, the integral $\int_a^b f(x) \ \mathrm{d} x$ exists as a real number.
Thus the upper integral $\overline{\int}_a^b f $ and the lower integral $\underline{\int}_a^b f $ are equal, where $$ \overline{\int}_a^b f \colon= \inf \left\{ \ U(P, f) \ \colon P \mbox{ is a partition of the interval } [a, b] \ \right\}, $$ and $$ \underline{\int}_a^b f \colon= \sup \left\{ \ L(P, f) \ \colon P \mbox{ is a partition of the interval } [a, b] \ \right\}, $$
Thus, for every partition $P$ of $[a, b]$, we have $$ L(P, f) \leq \int_a^b f \leq U(P, f). $$
Again as $f$ is Riemann-integrable on $[a, b]$, so, corresponding to every real number $\varepsilon > 0$, we can find a partition $P_\varepsilon$ of $[a, b]$ such that $$ U \left( P_\varepsilon, f \right) - L \left( P_\varepsilon, f \right) < \varepsilon. $$
Now let $P \colon= \left\{ \ x_0, x_1, \ldots, x_{n-1}, x_n \ \right\}$ be any partition of the interval $[a, b]$, where $$ a = x_0 < x_1 < \cdots < x_{n-1} < x_n = b. $$ As $f$ is a monotonically increasing function on $[a, b]$, so for each $i = 1, \ldots, n$, we see that $$ m_i \colon= \inf \left\{ \ f(x) \ \colon \ x_{i-1} \leq x \leq x_i \ \right\} = f \left( x_{i-1} \right) = \mathrm{e}^{x_{i-1}}, $$ and $$ M_i \colon= \sup \left\{ \ f(x) \ \colon \ x_{i-1} \leq x \leq x_i \ \right\} = f \left( x_{i} \right) = \mathrm{e}^{x_{i}}; $$ therefore we have $$ L(P, f) \colon= \sum_{i=1}^n m_i \left( x_i - x_{i-1} \right) = \sum_{i=1}^n \mathrm{e}^{x_{i-1}} \left( x_i - x_{i-1} \right), $$ and $$ U(P, f) \colon= \sum_{i=1}^n M_i \left( x_i - x_{i-1} \right) = \sum_{i=1}^n \mathrm{e}^{x_{i}} \left( x_i - x_{i-1} \right). $$
Now for each positive integer $n$, let $P_n$ be the partition of $[a, b]$ given by $$ P_n \colon= \left\{ \ a, a + \frac{b-a}{n}, a + \frac{2(b-a)}{n}, \ldots, a + \frac{ (n-1) ( b-a ) }{n}, b \ \right\}; $$ that is, $P_n$ partitions the interval $[a, b]$ into $n$ equal subintervals. Thus $$ P_n = \left\{ x_0, x_1, \ldots, x_n \ \right\}, $$ where $$ x_i \colon= a + \frac{ i(b-a)}{n} $$ for each $i = 1, \ldots, n$. Then we have $$ \begin{align} L \left( P_n, f \right) &= \sum_{i=1}^n \mathrm{e}^{a + \frac{ (i-1)(b-a)}{n} } \frac{b-a}{n} \\ &= \frac{b-a}{n} \mathrm{e}^a \sum_{i=1}^n \mathrm{e}^{ \frac{ (i-1)(b-a)}{n} } \\ &= \frac{b-a}{n} \mathrm{e}^a \sum_{i=1}^n \left( \mathrm{e}^{ \frac{ b-a}{n} } \right)^{i-1} \\ &= \frac{b-a}{n} \mathrm{e}^a \sum_{i=0}^{n-1} \left( \mathrm{e}^{ \frac{ b-a}{n} } \right)^{i} \\ &= \frac{b-a}{n} \mathrm{e}^a \frac{ \left( \mathrm{e}^{ \frac{ b-a}{n} } \right)^{n} - 1 }{ \mathrm{e}^{ \frac{ b-a}{n} } - 1 } \\ &= \frac{b-a}{n} \mathrm{e}^a \frac{ \mathrm{e}^{ \frac{ n (b-a) }{n} } - 1 }{ \mathrm{e}^{ \frac{ b-a}{n} } - 1 } \\ &= \frac{b-a}{n} \mathrm{e}^a \frac{ \mathrm{e}^{b-a } - 1 }{ \mathrm{e}^{ \frac{ b-a}{n} } - 1 } \\ &= \frac{b-a}{n} \frac{ \mathrm{e}^{b } - \mathrm{e}^a }{ \mathrm{e}^{ \frac{ b-a}{n} } - 1 } \\ &= \frac{ \frac{b-a}{n} }{ \mathrm{e}^{ \frac{ b-a}{n} } - 1 } \left( \mathrm{e}^{b } - \mathrm{e}^a \right), \end{align} $$ and $$ \begin{align} U \left( P_n, f \right) &= \sum_{i=1}^n \mathrm{e}^{a + \frac{ i(b-a)}{n} } \frac{b-a}{n} \\ &= \frac{b-a}{n} \mathrm{e}^a \sum_{i=1}^n \mathrm{e}^{ \frac{ i(b-a)}{n} } \\ &= \frac{b-a}{n} \mathrm{e}^a \sum_{i=1}^n \left( \mathrm{e}^{ \frac{ b-a}{n} } \right)^{i} \\ &= \frac{b-a}{n} \mathrm{e}^a \mathrm{e}^{ \frac{ b-a}{n} } \frac{ \left( \mathrm{e}^{ \frac{ b-a}{n} } \right)^{n} - 1 }{ \mathrm{e}^{ \frac{ b-a}{n} } - 1 } \\ &= \frac{b-a}{n} \mathrm{e}^a \mathrm{e}^{ \frac{ b-a}{n} } \frac{ \mathrm{e}^{ \frac{ n (b-a) }{n} } - 1 }{ \mathrm{e}^{ \frac{ b-a}{n} } - 1 } \\ &= \frac{b-a}{n} \mathrm{e}^a \mathrm{e}^{ \frac{ b-a}{n} } \frac{ \mathrm{e}^{b-a } - 1 }{ \mathrm{e}^{ \frac{ b-a}{n} } - 1 } \\ &= \frac{b-a}{n} \mathrm{e}^{ \frac{ b-a}{n} } \frac{ \mathrm{e}^{b } - \mathrm{e}^a }{ \mathrm{e}^{ \frac{ b-a}{n} } - 1 } \\ &= \frac{b-a}{n} \frac{ \mathrm{e}^{ \frac{ b-a}{n} } }{ \mathrm{e}^{ \frac{ b-a}{n} } - 1 } \left( \mathrm{e}^{b } - \mathrm{e}^a \right) \\ &= \frac{b-a}{n} \left\{ 1 + \frac{1 }{ \mathrm{e}^{ \frac{ b-a}{n} } - 1 } \right\} \left( \mathrm{e}^{b } - \mathrm{e}^a \right) \\ &= \left\{ \frac{b-a}{n} + \frac{ \frac{b-a}{n} }{ \mathrm{e}^{ \frac{ b-a}{n} } - 1 } \right\} \left( \mathrm{e}^{b } - \mathrm{e}^a \right) . \end{align} $$
Is what I have done so far correct? If so, then what next? How to proceed from here, preferably using only the machinery developed by Walter Rudin as far as Chapter 7 in the book Principles of Mathematical Analysis, third edition?
P.S.:
We note that, for every partition $P$ of $[a, b]$, we have $$ L(P, f) \leq \underline{\int}_a^b f \leq \overline{\int}_a^b f \leq U(P, f). \tag{A} $$
Now from (A), we can conclude that, for each $n \in \mathbb{N}$, by using the partitions $P_n$, we have $$ L \left( P_n, f \right) \leq \underline{\int}_a^b f \leq \overline{\int}_a^b f \leq U \left( P_n, f \right); $$ that is, $$ \frac{ \frac{b-a}{n} }{ \mathrm{e}^{ \frac{ b-a}{n} } - 1 } \left( \mathrm{e}^{b } - \mathrm{e}^a \right) \leq \underline{\int}_a^b f \leq \overline{\int}_a^b f \leq \left\{ \frac{b-a}{n} + \frac{ \frac{b-a}{n} }{ \mathrm{e}^{ \frac{ b-a}{n} } - 1 } \right\} \left( \mathrm{e}^{b } - \mathrm{e}^a \right) \tag{B} $$ for each $n \in \mathbb{N}$.
Now we see that $$ \begin{align} \lim_{r \to 0} \frac{ r }{ \mathrm{e}^r - 1 } &= \lim_{r \to 0} \frac{1}{\mathrm{e}^r - 0} \qquad \mbox{ [ using the L'Hosptial's rule ] } \\ &= \frac{1}{1} \\ &= 1. \tag{1} \end{align} $$
Now as $n \to \infty$, $(b-a)/n \to 0$, and so from (1) we can conclude that $$ \lim_{n \to \infty} \frac{ \frac{b-a}{n} }{ \mathrm{e}^{ \frac{b-a}{n} } - 1 } = 1. \tag{2} $$
Finally letting $n \to \infty$ in (B) and using (2), we find that $$ \mathrm{e}^b - \mathrm{e}^a \leq \underline{\int}_a^b f \leq \overline{\int}_a^b f \leq \mathrm{e}^{b } - \mathrm{e}^a. $$ Therefore $$ \underline{\int}_a^b f = \mathrm{e}^b - \mathrm{e}^a = \overline{\int}_a^b f. $$ That is, $$ \int_a^b f = \mathrm{e}^b - \mathrm{e}^a. $$
I hope I've completed the proof satisfacorily enough in the P.S., haven't I?
