Can an integral always be written as an infinite sum? I'm currently reading about computing the surface area of revolution and encountered the following:
The author first approximates the area by partitioning the interval into a finite number of subintervals to obtain
$$
A \approx \sum_{i = 1}^{n} 2\pi f(x_i^{*})\sqrt{1 + \left[ f'(x_i^{*}) \right]^2 } \ \Delta x
\quad (\Delta x = x_{i}-x_{i-1},\ x_i^{*} \in [x_{i-1}, x_i])
$$
and then proceeds by making the subintervals "smaller and smaller" to obtain
$$
\lim _{n \rightarrow \infty} \sum_{i=1}^{n} 2 \pi f\left(x_{i}^{*}\right) \sqrt{1+\left[f^{\prime}\left(x_{i}^{*}\right)\right]^{2}} \Delta x=\int_{a}^{b} 2 \pi f(x) \sqrt{1+\left[f^{\prime}(x)\right]^{2}} d x
.$$
Intuitively, I understand the last step, but isn't the Riemann integral defined in terms of finite partitions, so the limit isn't really valid here? (Wouldn't one have to use the mean value theorem in order to make this rigorous?)
 A: Here the theorem s.t. it becomes rigorous.
Suppose that $f$ is integrable on $[a, b]$. Then for every $\varepsilon>0$ there is some $\delta>0$ such that, if $P=\left\{t_{0}, \ldots, t_{n}\right\}$ is any partition of $[a, b]$ with all lengths $t_{i}-t_{i-1}<\delta$, then
$$
\left|\sum_{i=1}^{n} f\left(x_{i}\right)\left(t_{i}-t_{i-1}\right)-\int_{a}^{b} f(x) d x\right|<\varepsilon
$$
for any Riemann sum formed by choosing $x_{i} \in\left[t_{i-1}, t_{i}\right]$. Notice that a Riemann sum is just any sum
$$
\sum_{i=1}^{n} f\left(x_{i}\right)\left(t_{i}-t_{i-1}\right)
$$ s.t.
\begin{align*}
\underbrace{\sum_{i=1}^{n} \inf\left\{f\left(x_{i}\right)\mid x_i \in [x_{i-1}, x_i]\right\}\left(t_{i}-t_{i-1}\right)}_{\large:= L(f, P)}  &\leq \sum_{i=1}^{n} f\left(x_{i}\right)\left(t_{i}-t_{i-1}\right) \\[10pt]
&\leq\underbrace{\sum_{i=1}^{n} \sup\left\{f\left(x_{i}\right)\mid x_i \in [x_{i-1}, x_i]\right\}\left(t_{i}-t_{i-1}\right)}_{\large := U(f, P)} .
\end{align*}
Here, $L(f, P)$ and $U(f, P)$ denote the lower and upper sums and a Riemann sum is just any sum of the above form that lies between the lower and upper sum.
Since the Riemann sum and the integral both lie between $L(f, P)$ and $U(f, P)$, this amounts to showing that for any given $\varepsilon$ we can make $U(f, P)-L(f, P)<\varepsilon$ by choosing a $\delta$ such that $U(f, P)-L(f, P)<\varepsilon$ for any partition with all lengths $t_{i}-t_{i-1}<\delta$
The definition of $f$ being integrable on $[a, b]$ includes the condition that $|f| \leq$ $M$ for some $M .$ First choose some particular partition $P^{*}=\left\{u_{0}, \ldots \ldots u_{K}\right\}$ for which
$$
U\left(f, P^{*}\right)-L\left(f, P^{*}\right)<\varepsilon / 2
$$
and then choose a $\delta$ such that
$$
\delta<\frac{\varepsilon}{4 M K}
$$
For any partition $P$ with all $t_{i}-t_{i-1}<\delta$, we can break the sum
$$
U(f, P)-L(f, P)=\sum_{i=1}^{n}\left(M_{i}-m_{i}\right)\left(t_{i}-t_{i-1}\right)
$$
into two sums. The first involves those $i$ for which the interval $\left[t_{i-1}, t_{i}\right]$ is completely contained within one of the intervals $\left[u_{j-1}, u_{j}\right] .$ This sum is clearly $\leq$ $U\left(f, P^{*}\right)-L\left(f, P^{*}\right)<\varepsilon / 2$. For all other $i$ we will have $t_{i-1}<u_{j}<t_{i}$ for some $j=1, \ldots, K-1$, so there are at most $K-1$ of them. Consequently, the sum for these terms is $<(K-1) \cdot 2 M \cdot \delta<\varepsilon / 2.$



Spivak, Michael - Calculus 4th edition
