Continuously Differentiable Curve has Finite Arc Length

Theorem
Let $y = f\left({x}\right)$ be a real function such that $D_xf$ is continuous on the closed interval $\left[a..b\right]$.

The definite integral:


 * $s= \displaystyle \int_a^b \sqrt{1 + \left({\frac {\mathrm dy}{\mathrm dx}}\right)^2}\ \mathrm d x$

exists, and is called the arc length of $f$ between $a$ and $b$.

Proof
It intuitively makes sense to define the length of a line segment to be the distance between the two end points, as given by the Distance Formula:


 * $\sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$

Similarly, it is reasonable to assume that the actual length of the curve would be approximately equal to the sum of the lengths of each of the line segments, as shown:



To calculate the sum of the length of these line segments, divide $\left[{a .. b}\right]$ into any number of closed subintervals of the form $\left[{x_{i-1} .. x_i}\right]$ where:


 * $a = x_0 < x_1 \cdots < x_{k-1} < x_k = b$

Define:


 * $\Delta x_i = x_i - x_{i-1}$


 * $\Delta y_i = y_i - y_{i-1}$

As the length of the $i$th line segment is $\sqrt{\left(\Delta x_i\right)^2 + \left(\Delta y_i\right)^2}$, the sum of all these line segments is given by:


 * $\displaystyle \sum_{i=1}^{k}\ \sqrt{\left(\Delta x_i\right)^2 + \left(\Delta y_i\right)^2}$

Thus the approximate arc length is given by the sum:


 * $s \approx \displaystyle \sum_{i=1}^{k}\ \sqrt{1 + \left(\frac{\Delta y_i}{\Delta x_i}\right)^2}\Delta x_i$

Recall that by hypothesis, $D_xf$ exists and is continuous on the entirety of $\left[a..b\right]$.

From Differentiable Function is Continuous, $f$ is continuous on $\left(a..b\right)$.

Thus the Mean Value Theorem can be applied.

In every open interval $\left(x_i..x_{i-1}\right]$ there exists some $c_i$ such that:


 * $D_xf\left({c_i}\right) = \dfrac {\Delta y_i}{\Delta x_i}$

Plugging this into the above sum we have:


 * $s \approx \displaystyle \sum_{i=1}^{k}\ \sqrt{1 + \left(\frac{\mathrm dy}{\mathrm dx}\right)^2}\Delta x_i$

We have already established that $D_xf$ is continuous.

We also know, from Even Powers are Positive, that the radicand is always positive.

From Continuity of Root Function and Limit of Composite Function, $\sqrt{1 + \left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2}\Delta x_i$ is continuous as well.

Because Continuous Function is Riemann Integrable, there exists a definite integral that confirms the intuitive notion that there is a value that represents the exact length of the curve as the limit of the above sum.

This integral is:


 * $s = \displaystyle \int_a^b \sqrt{1 + \left({\frac {\mathrm dy}{\mathrm dx}}\right)^2}\ \mathrm d x$