Show the equivalence of arc length definitions Definition 1:

Let $r: [a,b] \to \Bbb R^d$ be a continuous differentiable function. Then the arc length is given by $$L(r) = \int_a^b || r'(t) || \, dt$$

Definition 2:

Let $r: [a,b] \to \Bbb R^d$ be a continuous function. Then the arc length is given by $$ V(r) = \sup_P \sum_{k=1}^n || r(x_k)-r(x_{k-1}) ||$$ where the supremum is taken over all partitions $P = \{a=x_0 \lt x_1 \lt \ldots \lt x_n = b \}$ of $[a,b]$.

How can I show that for a continuous differentiable $r(t)$ the two definitions are equivalent, i.e. $L(r)=V(r)$?
What I've done so far:
I found this question, which shows that I can convert the supremum to a limit
$$V(r) = \sup_P \sum_{k=1}^n || r(x_k)-r(x_{k-1}) || = \lim_{n \to \infty} \sum_{k=1}^n || r(x_k)-r(x_{k-1}) ||$$
by choosing an appropriate sequence of partitions $P_n$ of which I take the $x_k$'s. This gives
$$ \lim_{n \to \infty} \sum_{k=1}^n || r(x_k)-r(x_{k-1}) || = \lim_{n \to \infty} \sum_{k=1}^n || \frac{r(x_k)-r(x_{k-1})}{x_k-x_{k-1}} || (x_k-x_{k-1})$$
Now I somehow need to show that
$$\lim_{n \to \infty} \sum_{k=1}^n || \frac{r(x_k)-r(x_{k-1})}{x_k-x_{k-1}} || (x_k-x_{k-1}) = \int_a^b ||r'(t)|| \, dt$$
How can I justify this step of converting the sum to an intergral and taking the limit of the inside simultaneously?
 A: Claim. For a parametrized curve $\gamma \in \mathcal C^1([a,b], \Bbb R^d)$, we have 
$$\bbox[5px,border:2px solid #C0A000]{
\lim_{n\to\infty} \sum_{i=1}^n \| \gamma(x_{i,n})-\gamma(x_{i-1,n})\| = \int_a^b \|\gamma'(x)\| \,\mathrm dx,
}$$
where $x_{i,n} := a \cdot (1-\frac in) + b\cdot\frac in$.
Proof. Note that the left hand side equals 
\begin{equation}
\lim_{n\to\infty} 
\underbrace{
\sum_{i=1}^n \left\| \frac{\gamma(x_{i,n})-\gamma(x_{i-1,n})}{\frac{b-a}n}\right\|\cdot {\frac{b-a}n}
}_{=: \kappa_n}
\end{equation} 
and observe that $x_{i,n}-x_{i-1,n}=\frac{b-a}n$.
By definition of the derivative, which is continuous on a compact interval and thus also uniformly contiuous, there exists a $\delta$ for every $\varepsilon > 0$ such that whenever $\frac{b-a}n < \delta$, 
$$\left\|\bigg\|\frac{\gamma(x_{i,n})-\gamma(x_{i-1,n})}{\frac{b-a}n}\bigg\|-\bigg\|\gamma'(x_{i,n})\bigg\|\right\| < \varepsilon.$$
In particular, for any $\varepsilon > 0$, we have 
\begin{equation}\tag{*} \label{*}
\left\|\rule{0cm}{1cm}
\underbrace{
  \sum_{i=1}^n 
  \left\|
  \frac{\gamma(x_{i,n})-\gamma(x_{i-1,n})}{\frac{b-a}n}
  \right\|
  \cdot\frac{b-a}n
}_{=\kappa_n}
-\underbrace{
  \sum_{i=1}^n
  \Big\|\gamma'(x_{i,n})\Big\|\cdot\frac{b-a}n 
}_{=:\rho_n}
\right\|
< \varepsilon \cdot (b-a)
\end{equation}
whenever $n>\frac{b-a}\delta$ (note that $\delta$ depends on $\varepsilon$).
Since $\rho_n$ are just Riemann approximation sums, we have $\lim_{n\to\infty} \rho_n = \int_a^b \|\gamma'(x)\| \,\mathrm dx$.
By \eqref{*}, we can conclude that $\lim_{n\to\infty} \kappa_n = \lim_{n\to\infty} \rho_n$, which proves our claim. $\square$
