I recently found myself rather delighted after I had sort of proved the arc length formula for a curve $x \mapsto f(x)$, $x \in [a;b]$. However, I realized that I didn't really know how to argue rigorously why the limit yields the definite integral (I just need a good argument). What I have on my paper looks something like this:
$$\lim_{max \Delta x_i \to 0} \sum_{i=0}^{n-1} \sqrt{1+(\frac{f(x_i+\Delta x_i)-f(x_i) }{\Delta x_i})^2}\Delta x_i = \int_a^b\sqrt{1+(\frac{df(x)}{d x})^2}dx$$
It is obvious that this is true. At first I thought it was just an incidence of a Riemann sum/integral conversion - until I looked it up. A Riemann sum looks like this: $$\lim_{max \Delta x_i \to 0} \sum_{i=0}^{n-1} g(x_i) \Delta x_i = \int_a^b g(x) dx$$
while the result above can be put in this form I guess:
$$\lim_{max \Delta x_i \to 0} \sum_{i=0}^{n-1} g(x_i, \Delta x_i) \Delta x_i = \int_a^b \lim_{max \Delta x_i \to 0} (g(x, \Delta x_i)) \ dx$$
My question is (1) whether there is a 'concise and rigorous' way to argue why the result in the first expression is true, which perhaps relies on other results, and not only 'geometrical intuition'. (2) Whether there is a theorem that states when the last statement holds/not holds.
All relevant answers or links will be appreciated:)
Note that
$$\frac{f(x_i+\Delta x_i) - f(x_i)}{\Delta x_1} = f'(c(i,n))$$
by the mean value theorem. If $f'$ is Riemann integrable on $[a,b],$ then so is $\sqrt {1+(f')^2}.$ It follows that
$$\tag 1 \lim \sum_{i=1}^{n}\sqrt {1+(f'(c(i,n))^2} \Delta x_i = \int_a^b \sqrt {1+f'(x)^2}\, dx$$
from the standard theory of the Riemann integral. Since the left side of $(1)$ equals the left side of your first displayed equation, we've got it.