Arc length definition I've always felt that the notion of 'area under the curve' is more precise than 'length of a curve', because of the Riemann integral and the reasoning used there:
First, we assume the quantity we call area $A$ between the graph of a function and x-axis eixsts. Next, we show that it must necessarily be the limit of upper and lower Riemann sums, if their limits (as we take finer partitions) are equal.
It would be less obvious if we only considered lower Riemann sums, because the fact that it converges doesn't necessarily mean it converges to the quantity we call area $A$ under the graph.
But by using both lower and upper sums, we know that the sum of areas of rectangles of lower sums will always be less or equal to $A$ (because we basically defined it to be like that), similarly upper sums will be greater or equal to $A$, so $A$ is 'trapped' between them.
In the case of arc length it's less certain that $L=\int_{a}^{b} \sqrt { 1 + [f'(x)]^2 }$ is the length of the graph of $f(x)$ on $[a,b]$, assuming $f'$ is continuous on $[a,b]$. Even if it converges, how can we prove that it converges to what we could call the length of a curve? It's like defining area using lower sums only - fine, lower sums might converge to a certain value, but it doesn't mean the area exists (upper sums can have a different limit).
You will probably say we cannot prove it, because it's a definition. However, mathematicians have chosen to define it in this, not some other way for some reason (the could have, right?), thus it is considered more 'correct'. So the question - why should I treat it as the 'right' definition?
 A: Actually that integral is not really the definition. The definition
of the length of the graph of $f$ for $a\le x\le b$ is the sup of the sums $$\sum_{j=1}^n\left((x_j-x_{j-1})^2+(f(x_j)-f(x_{j-1})^2\right)^{1/2}$$over all choices $a+x_0<\dots<x_n=b$. That sum is just the length of a polygonal path joining points of the graph of $f$, so it makes sense as the definition of the length.
Now if we assume that $f'$ is continuous it's possible to prove that the length is equal to that integral.
EDIT: Regarding the question of what makes this definition "correct": Of course that's meaningless in a mathematical sense, a definition is a definition. It's not meaningless to ask why it makes sense as a definition of arc length. 
One way to look at it is this: What is the "length" of a curve anyway? It's the net distance you travel if you walk along the curve. But your pedometer doesn't have a magic arc-lenght measurer - all it can do is measure the distance between successive steps and add. That's what that sum does. Except of course oops, measuring the distance between this step and the next underestimates the arc length because it ignores wiggles on a scale smaller than the step size. Hence the "sup".
A: Rather than write a new answer, I'll just link to an answer I wrote a while back that touches on this issue:
False proof: $\pi=4$, but why?
It's surface area rather than perimeter, but it's basically the same thing. The relevant point is the last paragraph: If everything's convex, we can get an upper bound and use the squeeze theorem, like with Riemann sums. (To get a lower bound, we can use polygonal approximations, or just use convexity again.) We can then find the perimeters of almost all curves, by cutting them into convex pieces and finding the lengths of those. Eventually you can prove that this is equivalent to the usual definition.
(Exception: Cantor's staircase isn't convex anywhere, so this won't work. I'm not quite sure what to do with this. It has length $2$ on the unit interval.)
