Line integral definition and mean value theorem If you take a look at the derivation section of line integral article, you can see the following statement:

$$I = \lim_{\Delta s_i \rightarrow 0} \sum_{i=1}^n f(\mathbf{r}(t_i)) \, \Delta s_i. \tag{1}$$
We note that, by the mean value theorem, the distance between
subsequent points on the curve, is
$$\Delta s_i = |\mathbf{r}(t_i+\Delta t)-\mathbf{r} (t_i)| \approx |\mathbf{r}' (t_i)| \,\Delta t.$$
Substituting this in the above Riemann sum yields
$$I = \lim_{\Delta t  \rightarrow 0} \sum_{i=1}^n f(\mathbf{r}(t_i))|\mathbf{r}'(t_i)| \, \Delta t \tag{2}$$

I'm wondering how to more rigorously derive eq. 2 from eq. 1, because I have a few doubts about the method used above:
Note that it is a vector valued function. The mean value theorem article that it refers to doesn't really give any explanation, it even states there is no direct analog of mean value theorem for vector valued function (actually there's one involving an inequality).

There is no exact analog of the mean value theorem for vector-valued functions.

I will be grateful for a clear explanation as I'm not a math expert.
 A: Let me first clearly state all assumptions made in the article:


*

*$U\subset\mathbb R^m$ is an open set, $C\subset U$ is a subset of $U$

*$\mathbf r: [a,b]\to C$ is a bijective, parametrized curve that is also continuously differentiable. Note that this implies that $C$ is compact.

*$f:U\to\mathbb R$ is a function such that $f\circ\mathbf r$ is absolutely Riemann  integrable over $[a,b]$. It follows that $(f\circ\mathbf r)\cdot\lvert \mathbf r'\rvert$ is Riemann integrable over $[a,b]$.

*For $n\in\mathbb N$, we define $\Delta t=\frac{b-a}n$. For $i=1,2,\dots, n$, we define $t_i=a+(i-1)\Delta t$ and $\Delta s_i = \lvert\mathbf r(t_i+\Delta t)-\mathbf r(t_i)\rvert$.


Now we notice: $$\Delta t\to 0\iff n\to\infty$$ and $$\Delta s_i \to 0\iff \mathbf r(t_i+\Delta t)\to 
\mathbf r(t_i)\overset{\mathbf r\text{ bijective}}\iff\Delta t\to 0\iff n\to\infty.$$
So we need to prove the following 
Theorem. We have $$\bbox[15px,border:1px groove navy]{\lim_{n\to\infty} \sum_{i=1}^n f(\mathbf{r}(t_i)) \Delta s_i=\lim_{n\to\infty} \sum_{i=1}^n f(\mathbf{r}(t_i))|\mathbf{r}'(t_i)| \Delta t.}$$
Proof. Notice that, by definition of the derivative, we have $$\mathbf r'(t_i)=\lim_{\Delta t\to0} \frac{\mathbf r(t_i+\Delta t)-\mathbf r(t_i)}{\Delta t}.$$
It follows that $$|\mathbf r'(t_i)|=\left|\lim_{\Delta t\to0} \frac{\mathbf r(t_i+\Delta t)-\mathbf r(t_i)}{\Delta t}\right|=\lim_{\Delta t\to0}\left|\frac{\mathbf r(t_i+\Delta t)-\mathbf r(t_i)}{\Delta t}\right|=\lim_{\Delta t\to0}\frac{\overbrace{\left|\mathbf r(t_i+\Delta t)-\mathbf r(t_i)\right|}^{\Delta s_i}}{\Delta t},$$
where I have used continuity of the absolute value in the second equality and the fact that $\Delta t>0$ in the third equality.
Indeed, this means that for any $\varepsilon>0$, there is a $\delta>0$ such that for all $\Delta t<\delta$, we have 
$$\big\lvert\Delta s_i-\lvert \mathbf r'(t_i)\rvert\Delta t\big\rvert\le\varepsilon\Delta t.$$ 
Since $\Delta t\to0\iff n\to\infty$, this means that for any $\varepsilon>0$ there is an $N\in\mathbb N$ such that for all $n\geq N$ we have \begin{split} \left|\sum_{i=1}^n f(\mathbf{r}(t_i)) \Delta s_i-\sum_{i=1}^n f(\mathbf{r}(t_i))|\mathbf r'(t_i)|\Delta t\right|
&=\left\lvert\sum_{i=1}^n f(\mathbf{r}(t_i)) (\Delta s_i-\lvert\mathbf r'(t_i)\rvert\Delta t)\right\rvert
\\& \le \sum_{i=1}^n \Big\lvert f(\mathbf{r}(t_i))\Big\rvert \Big\lvert\Delta s_i-\lvert\mathbf r'(t_i)\rvert\Delta t\Big\rvert
\\&\le\varepsilon\underbrace{\sum_{i=1}^n \lvert f(\mathbf r(t_i))\rvert \Delta t}_{\text{Riemann sums of } \lvert f\circ\mathbf r\rvert}
\\&\xrightarrow{n\to\infty} \varepsilon \int_a^b \lvert f\circ\mathbf r\rvert.\end{split}
It follows that, for every $\varepsilon>0$, $$\lim_{n\to\infty} \left|\sum_{i=1}^n f(\mathbf{r}(t_i)) \Delta s_i-\sum_{i=1}^n f(\mathbf{r}(t_i))|\mathbf r'(t_i)|\Delta t\right|\le\varepsilon\int_a^b \lvert f\circ\mathbf r\rvert,$$ which immediately implies that $$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to\infty} \left|\sum_{i=1}^n f(\mathbf{r}(t_i)) \Delta s_i-\sum_{i=1}^n f(\mathbf{r}(t_i))|\mathbf r'(t_i)|\Delta t\right|=0}$$ which proves the Theorem. $\square$
