How is the Mean Value Theorem relevant to the formula for Arc length? I am following the arc length tutorial found here. To derive the arc length formula, the author uses the Mean Value Theorem to express $\Delta y$
 in terms of $x$. 

"By the Mean Value Theorem we know that on the interval $[x_{i-1}, x_1]$ there is a point $x^*_i$ so that,"

$$
\Delta y_i = f'(x_i^*)\Delta x \\
L = \lim_{n \to \infty} \sum_{i=1}^n \sqrt{1 + [f'(x^*_i)]^2}\Delta x
$$
The formula is then written as a definite integral, like so
$$
L = \int_a^b \sqrt{1 + [f'(x)]^2}d x
$$
Why does $x_i^*$ get simplified to a single $x$? Is the integral implying that because of the infinitely-small $\Delta x$, that the special point that makes the MVT true converges to $x$? If so, then why is the MVT even needed for the definition of arc length?
 A: It is worth recalling Riemann integral as a
$$\int_{a}^{b}f(x)dx=\lim_{\max\Delta x_k \rightarrow 0}\sum_{k=1}^nf(x_k^*)\Delta x_k \tag{1}$$
where $\{x_1,x_2,...,x_{n+1}\}$ is a partition of $[a,b]$ and $x_k^* \in [x_k,x_{k+1}]$.
The Euclidean distance between 2 points $(x_1,f(x_1)),(x_2,f(x_2)), x_1<x_2$ is 
$$\sqrt{\left(x_2-x_1\right)^2+\left(f(x_2)-f(x_1)\right)^2}=\left(x_2-x_1\right)\sqrt{1+\left(\frac{f(x_2)-f(x_1)}{x_2-x_1}\right)^2}$$
or using MVT, $\exists x_1^{*}\in (x_1,x_2)$:
$$\sqrt{\left(x_2-x_1\right)^2+\left(f(x_2)-f(x_1)^2\right)}=\left(x_2-x_1\right)\sqrt{1+\left(f'(x_1^{*})\right)^2}=\sqrt{1+\left(f'(x_1^{*})\right)^2}\Delta x_1$$
The arc length is initially approximated with the sum of distances of $(x_1,f(x_1)),(x_2,f(x_2)),...,(x_{n+1},f(x_{n+1}))$, applying MVT for each adjacent pair of points:
$$\sum_{k=1}^n \sqrt{\left(x_{k+1}-x_{k}\right)^2+\left(f(x_{k+1})-f(x_{k})\right)^2}=\sum_{k=1}^n \sqrt{1+\left(f'(x_k^{*})\right)^2}\Delta x_k=...$$
thinking of $g(x)=\sqrt{1+\left(f'(x)\right)^2}$ , we have 
$$...=\sum_{k=1}^n g(x_k^{*})\Delta x_k$$
taking the limit $\max\Delta x_k \rightarrow 0$ we have the Riemann integral $(1)$ for $g(x)$
$$L=\int_{a}^{b}g(x)dx=\int_{a}^{b}\sqrt{1+\left(f'(x)\right)^2}dx$$
