# Arclength integral

Suppose that $f: [a,b] \rightarrow \mathrm{R}^n$ is continuous with a derivative $f'$ whose norm is Riemann-integrable. To demonstrate the arclength integral formula, I'm trying to prove that, for every $n \in \mathrm N$, there exists a tagged partition $(P_n,\xi)$ of $[a,b]$ such that

$$|P_n| < \frac 1n\text{ and }|f(t_i) - f(t_{i-1}) - f'(\xi_i)(t_i - t_{i-1})| < \frac{t_i - t_{i-1}}n$$

for every interval $[t_{i-1},t_i]$ of the partition, where $\xi_i$ is the corresponding tag of the interval.

If $f'$ is continuous, the statement is a consequence of the uniform differentiability of $f$. However, I don't think this will be useful in proving the general case. Any suggestions on how this can be proved?

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I'm probably missing some subtlety here, but doesn't the mean value theorem give for any interval $[r,s] \subset [a,b]$ a point $r \lt \xi \lt s$ such that $f(s) - f(r) = f'(\xi)(r-s)$ so that any fine enough partition will work? – t.b. Nov 19 '11 at 7:05
The mean value theorem works when $n = 1$. However, only the inequality $|f(s) - f(r)| \leq M(s-r)$, where $M = \sup\limits_{t \in [r,s]} |f'(t)|$ is true when $n > 1$. Consider for example $f: [0,2\pi] \rightarrow \mathrm R^2$, $f(t) = (\cos{t}, \sin{t})$. – Daniel Nov 19 '11 at 13:34