# Determining the Length of a Section of a Curve $L(f, a, b) = \int \limits_a^b \sqrt{f'(x)^2+1}\,dx$ ## Proof The distance between any two points on any curve can be approximated using Pythagoras' theorem: $D = \sqrt {\Delta x^2 + \Delta y^2}$. This approximation is more and more accurate as the two points are closer and closer to each other. Because of this, for an infinitesimally small change of $x$, we can express the change of the total distance we're interested in $L(f, a, b)$ as $dL = \sqrt {dx^2+dy^2}$ This expression can then be manipulated as follows: $ \begin{align} dL &= \sqrt {dx^2+dy^2} \\ &= \sqrt {dx^2+\left(\frac {dy} {dx}dx\right)^2} \\ &= \sqrt {dx^2+\left(\frac {dy} {dx}\right)^2dx^2} \\ &= \sqrt {\left(1+\left(\frac {dy} {dx}\right)^2\right)dx^2} \\ dL &= \sqrt {1+\left(\frac {dy} {dx}\right)^2}dx \end{align} $ We just evaluated the derivative of $L$, meaning we can take the antiderivative to get $L$: $L(f,a,b) = \int \limits_a^b \sqrt {1+\left(\frac {dy} {dx}\right)^2}dx$