Maybe my question is a duplicate, but I guess I don't know the right terminology to find it elsewhere. I would be happy to delete it if someone can point out a duplicate.
From elementary calculus, the length of a curve defined by a 'nice' function $f(x)$ between $x=a$ and $x=b$ is given by: $$s = \int\limits_{a}^{b} \left( 1+[f^\prime (x)]^2 \right)^{\frac{1}{2}}dx$$
One derivation of this formula is to split the inverval $[a,b]$ into small segments of width $dx$, such that the length of $f(x)$ in each segment is given by $ds^2=dx^2+dy^2$, and then taking the limit as $dx \rightarrow 0$ of the sum of these lengths.
My question is: if we are eventually going to take the limit as $dx\rightarrow 0$, why won't it do just as well to approximate the length of $f(x)$ in a segment as being simply $dx$? (this would lead to the wrong expression).
I'm sure there must be a very rigorous reason for this in analysis, but I wonder if there might also be a more intuitive explanation?
(For some context, this question has been bugging me since high school, so by "intuitive" I guess I mean something that a high school student would be able to digest. Although now I might also be able to follow more formal explanations, if this could help in my understanding.)
The following is an excerpt of an earlier answer of mine:
The "area under the curve" $\gamma$ corresponding to a certain $\Delta x>0$ is roughly $f(\xi)\cdot \Delta x$, independently of the exact slope of the curve at $\xi$. Making $\Delta x$ smaller will decrease the relative area error committed here, since this error is only caused by the bending of the curve. But the length $\Delta s$ of the short arc corresponding to $\Delta x$ is roughly $={\Delta x\over\cos\phi}$, and making $\Delta x$ smaller does not make the factor ${1\over\cos\phi}$ go away. It follows that the final formula for the total length will have to incorporate the value ${1\over\cos\phi}=\sqrt{1+f'(\xi)^2}$.