The mean value theorem is used to derive the arc length formula and have questions on the value of $x_i^*$, which is the argument to a function $f$ where equals the mean value.
Deriving Arc Length: $$S = \lim_{n \rightarrow \infty} \sum_{i=1}^{n} ||\vec P_n - \vec P_{n-1}||_2$$
where $P_i$ are points along the arc.
$$ ||\vec P_n - \vec P_{n-1}||_2 = \sqrt{(x_i - x_{i-1})^2 + (y_i - y_{i-1})^2} $$
Mean Value Theorem applied: $$ \frac {f(x_i) - f(x_{i-1})}{x_i - x_{i-1}} = f'(x_i^*) $$ where $x_i^* \in (x_{i-1}, x_i)$. $f'(x_i^*)$ is the mean value of $\int_{x_{i-1}}^{x_i} f'(x)$.
Substitute back in: $$ f(x_i) - f(x_{i-1}) = f'(x_i^*)(x_i - x_{i-1}) $$ $$ \Delta y_i = f'(x_i^*) \Delta x $$ $$ ||\vec P_n - \vec P_{n-1}||_2 = \sqrt{\Delta x^2+f'(x_i^*)^2\Delta x^2 } = \sqrt{1 + f'(x_i^*)^2}\Delta x $$
By definition of the definite integral: $$ S = \int_a^b{\sqrt{1 + f'(x)^2} dx} = \lim_{n \rightarrow \infty} \sum_{i=1}^{n} \sqrt{1 + f'(x_i^*)^2}\Delta x $$
Question why is the mean value theorem required? Is it to find an mean for the derivative, but why?
Question how is the definition of the definite integral applied taking $x_i^*$ in account? Do we just use the definition that $x_i^* \in (x_{i-1}, x_i)$, even though it may be a specific value, and then the equality holds?
Appreciate your guidance