Let \(f(x)\) be a differentiable function defined over the interval \([a, b]\) with its derivative \(f'(x)\) being continous. Such functions are also called smooth. Our goal is to calculate the arc length of that function from point \((a,f(a))\) to point \((b,f(b))\).
We start by approximating the curve using line segments. To do so, we create a regular partition \(X=\{a=x_0, x_1, ... b=x_n\}\). Now for each point \(P_i=(x_i, f(x_i))\) we construct line segments from point \(P_i\) to \(P_{i+1}\).
To find the length \(|P_i - P_{i-1}|\) of each line segment, we need to look at the horizontal and vertical change over the distance of each interval \([x_{i-1}, x_i]\). Since we use a regular partition, the horizontal change is \(\Delta x = x_i - x_{i-1} = \frac{b-a}{n}\). The vertical change in distance depends on the function value, so \(\Delta y_i=f(x_i)-f(x_{i-1})\).
The length of the line segment can be calculated using Pythagorean Theorem:
\[|P_i - P_{i-1}| = \sqrt{(\Delta x)^2 + (\Delta y_i)^2}\]
By Mean Value Theorem we know there is a point \(x^*_i\in[x_{i-1}, x_i]\) such that
\[\begin{array}{rl} f'(x_i^*) &= \frac{f(x_i) - f(x_{i-1})}{x_i-x_{i-1}}\\ &= \frac{\Delta y_i}{\Delta x} \end{array}\]
Therefore the length of each line segment can be expressed as
\[\begin{array}{rl} |P_i - P_{i-1}| &= \sqrt{(\Delta x)^2 + (\Delta y_i)^2}\\ &= \sqrt{(\Delta x)^2 + (\Delta x)^2\left(\frac{\Delta y_i}{\Delta x}\right)^2}\\ &= \sqrt{(\Delta x)^2 \left[ 1+\left(\frac{\Delta y_i}{\Delta x}\right)^2\right]}\\ &= \Delta x\sqrt{1+\left(\frac{\Delta y_i}{\Delta x}\right)^2}\\ &= \Delta x\sqrt{1+\left[f'(x_i^*)\right]^2}\\ \end{array}\]
Summing over the lengths of the line segments approximates the arc length of the curve already
\[\begin{array}{rl} L &\approx \sum\limits_{i=1}^n |P_i - P_{i-1}|\\ &= \sum\limits_{i=1}^n \sqrt{1+\left[f'(x_i^*)\right]^2}\Delta x \end{array}\]
The approximation gets better and better the larger \(n\) becomes. Taking the limit as \(n\to\infty\) we get a Riemann sum that can be translated to a definite integral, which allows us to calculate the exact arc length of the curve
\[\begin{array}{rl} L &= \lim\limits_{n\to\infty}\sum\limits_{i=1}^n \sqrt{1+\left[f'(x_i^*)\right]^2} \Delta x\\ &= \int_a^b \sqrt{1+\left[f'(x)\right]^2} dx \end{array} \]