I am self studying a vector calculus book. I am having trouble proving the following fact:
Theoreom:
Let $g(t):[a,b]\to\mathbb{R}^n$ be a piecewise-${C}^1$ parametrized curve. Then $||g'(t)||=\lim_{h\to 0}\frac{1}{h}\int_t^{t+h}||g'(u)||du$.
This is equivalent to showing $$\lim_{h\to 0}\left|\int_t^{t+h}||g'(u)||du-h||g'(t)||\right|=0$$
My idea was to argue that the integral can be approximated by $$\int_t^{t+h}||g'(u)||du \approx h||g'(t)||+O(h^2)$$ and then substitute this expression into the limit. However, I was only able to derive this approximation for the integral of a $\mathcal{C}^\infty$ function by applying the Taylor series expansion. We can't necessarily do this because $g(t)$ is only $\mathcal{C}^1$.
Any hints would be appreciated.
Notice that the integral on the RHS is just the mean value of $\|g'(t)\|$ over the interval $[t,t+h]$ (integral divided by length of interval). Since the curve is $C^1$ on $[a,b]$, we know that $\|g'(\cdot)\|$ is continuous in $t$, so as $h\to0$, the interval size shrinks to 0 and the mean value, and hence the value of the integral, converges to $\|g'(\cdot)\|$.
Slightly more rigorously, by applying the Mean Value Theorem to the integral, we get that $$\frac{1}{h}\int_t^{t+h}\|g'(u)\|du = \|g'(t+h')\|,$$ where $h'\in[0,h]$. As $h\to0$, $h'\to0$ also, so $\frac{1}{h}\int_t^{t+h}\|g'(u)\|du\to\|g'(t)\|$, as desired.