Let $y=f(x)$ be a function such that $\deriv{f}{x}$ is continuous on $\ba{a,b}$. The length of the curve $y=f(x)$ from $x = a$ to $x=b$ is: $\huge L = \int_{a}^b \sqrt{1+\pa{f'(x)}^2}\d x $ >[!info]- Why the [[Derivative]] needs to be [[Continuous]] >The reason why the [[Derivative]] needs to be [[Continuous]] is to prevent calculating the arc length of [[Infinity|Infinite]] length. >[!example] Find the length of the [[Curve]] $y= \ln(\sec x)$ from $x=0$ to $x=\frac{\pi}{4}$. >>[!check]- Solution >>$\huge \begin{align} y' &= \frac{1}{\sec x} \sec( x) \tan(x) \\&=\tan (x) \\ \\ y'^2 &=\tan^2{x}\\ \\ L&=\int_{0}^{\pi/4} \sqrt{1+\tan^2 x} \d x \\ &=\int_{0}^{\pi/4} \sqrt{\sec^2(x)} \d x \\ &= \int_{0}^{\frac{\pi}{4}} \sec(x)\d x \\ &= \ln \left| \sec x + \tan x \right| \biggr \vert^{\frac{\pi}{4}}_{0} \\ & \dots\\ &= \ln \left| \sqrt{2} + 1 \right|\op{units} \end{align} >>$