We defined the arc length of a function as
$$L_I(f):=\int\limits_a^b\sqrt{1+(f'(x))^2}dx$$
for $I=[a,b]$ and $f\in C^1(I)$. We arrived at this formula by approximating the graph of $f$ by a series of straight lines (see here or here). I think the length of the curve is always greater or equal to the combined length of all the straight lines as a straight line is the shortest connection between two points.
My question: Why is the length of the curve equal to $L_I(f)$. I get that it can't be smaller but why is it exactly this limit? Why can't it be a number bigger than $L_I(f)$?
I think this is more of a comment than a concrete answer, but I can't comment yet.
I think your question is similar to asking how the integral $\int_a^b dx f(x)$ gives the exact area under the curve in the interval (a,b). One might say, thinking of the integral as a Riemann sum, $\sum_{i = 0}^{N-1} f(x_i) \Delta x_i$, the integral should give only an approximate answer to the area. However, this is not the case, of course. We get closer and closer to the answer as we reduce the length of the intervals $\Delta x_i$'s, and in fact get the exact answer as $\Delta x_i$'s $\rightarrow 0$. $\lim\limits_{\Delta x_i \rightarrow 0} \sum_{i = 0}^{N-1} f(x_i) \Delta x_i = \int_a^b dx f(x)$, which is something I assume you are already familiar with. Now my point is that the same argument can carry over for the definition of the arc length. Even though we think about the definition in terms of small straight lines, the integral is the limit where the length of these lines goes to zero, hence giving us an exact answer.