Suppose $f:J\to\mathbb R$ is differentiable and $f'$ is continuous on the domain. Then if $a,b\in J$ and $a<b$, the length of the curve of the function on interval $[a,b]$ is \begin{equation} l(f)=\displaystyle \int ^a_b \sqrt{1+f'(x)^2} dx \end{equation} This is my proof:
Since the arc length is defined as $l(f,P)=\displaystyle \sum_{i=1}^n \sqrt{1+\dfrac{f(t_i)-f(t_{i-1})}{t_i-t_{i-1}}}\Delta t_i$
$f'(x)$ is the result from using MVT one each interval $[t_{i-1},t_i]$ created by the partition P
then the definition of arc length can be turned to be
\begin{equation} l(f,P)=\displaystyle \sum_{i=1}^n \sqrt{1+f'(x_i)^2} \Delta t_i \end{equation}
Then since $f'$ is continuous on $[a,b]$,so $\sqrt{1+f'(x)^2}$ is continuous on $[a,b]$. Hence it is also bounded on $[a,b]$. Thus $\sqrt{1+f'(x)^2}$ is integrable on $[a,b]$
For any partition P=$\{t_0,...,t_n\}$, and on any interval $[t_{i-1},t_i]$, suppose $M_i$ is the supremum of $\sqrt{1+f'(x_i)^2}$ and $m_i$ is the infimum of $\sqrt{1+f'(x_i)^2}$
Then it's clear that $\displaystyle \sum m_i\Delta t_i\leq \sum \sqrt{1+f'(x_i)^2}\Delta t_i\leq\displaystyle \sum M_i\Delta t_i$
Then take $\displaystyle \lim_{|P|\to 0}$ to this inequality
Since $\sqrt{1+f'(x)^2}$ is integrable on $[a,b]$
$\displaystyle \lim_{|P|\to 0} \displaystyle \sum m_i\Delta t_i=\displaystyle \lim_{|P|\to 0} \displaystyle \sum M_i\Delta t_i=\int^a_b \sqrt{1+f'(x)^2}dx$
Thus $\displaystyle \lim_{|P|\to 0} \sum_{i=1}^n \sqrt{1+f'(x_i)^2} \Delta t_i=\int^a_b \sqrt{1+f'(x)^2}dx=l(f)$