# Arc Length

In this post I want to explain how a definite integral can be used to find the arc length of a curve because I think that it is fascinating that integrals have functions other than just finding the area under curves. Let us start with a definition:

A RECTIFIABLE curve is one that has a finite arc length. A sufficient condition for the graph of a function $f$ to be rectifiable between $(a, f(a))$ and $(b, f(b))$ is that  $f'$ be continuous on $[a, b]$. Such a function is continuously differentiable on $[a, b]$, and its graph on the interval $[a, b]$ is a smooth curve.

Now let’s consider a function $y = f(x)$ that is continuously differentiable on the interval $[a, b]$. We can approximate the graph of $f$ by line segments whose endpoints are determined by the partition:

$a = x_{0} < x_{1} < x_{2} <....< x_{n} = b$

By letting $\Delta x_{i} = x_{i} - x_{i-1}$ and $\Delta y_{i} = y_{i} - y_{i-1}$, we can approximate the length of the graph by the following:

$\displaystyle S \approx \sum_{i=1}^{n} \sqrt{(x_{i} - x_{i-1})^2 +(y_{i} - y_{i-1})^2}$

$= \displaystyle \sum_{i=1}^{n} \sqrt{(\Delta x_{i})^2 + (\Delta y_{i})^2}$

$= \displaystyle \sum_{i=1}^{n} \sqrt{(\Delta x_{i})^2 +(\frac{\Delta y_{i}}{\Delta x_{i}})^2(\Delta x_{i})^2}$

$= \displaystyle \sum_{i=1}^{n} \sqrt{1 + (\frac{\Delta y_{i}}{\Delta x_{i}})^2}(\Delta x_{i})$

This approximation becomes more and more precise as $||\Delta|| \to 0 (n \to \infty)$. Therefore the length of the graph is $S = \displaystyle \lim_{||\Delta|| \to 0} \displaystyle \sum_{i=1}^{n} \sqrt{1 + (\frac{\Delta y_{i}}{\Delta x_{i}})^2}(\Delta x_{i})$

Because $f'(x)$ exists for each $x$ in $(x_{i-1}, x_i)$, the Mean Value Theorem guarantees the existence of $c_i$ in $(x_{i-1}, x_i)$ such that $f(x_i) - f(x_{i-1}) = f'(c_i)(x_i - x_{i-1})$

$\frac{\Delta y_i}{\Delta x_i} = f'(c_i)$

Because $f'$ is continuous on $[a, b]$, it follows that $\sqrt{1 + [f'(x)]^2}$ is also continuous and therefore integrable on $[a, b]$, which implies that $S = \displaystyle \lim_{||\Delta|| \to 0} \displaystyle \sum_{i=1}^{n} \sqrt{1 + [f'(c_i)]^2}(\Delta x_i)$                                                                               = $\int_{a}^{b} \sqrt{1 + [f'(x)]^2} dx$