Let $f:[a,b] \rightarrow \mathbb R$ be $C^1$. Is the length of $\{f(x): x \in [a,b] \}$ and the total variation of $f$ the same thing ? The definition are extremely similar to each others:

* The total variation of a real-valued (or more generally complex-valued) function $f$, defined on an interval $[a, b] \subset \mathbb{R}$ is the quantity
$$
V\_{b}^{a}(f)=\sup \_{\mathcal{P}} \sum\_{i=0}^{n\_{P}-1}\left|f\left(x\_{i+1}\right)-f\left(x\_{i}\right)\right|
$$
where the supremum runs over the set of all partitions $\mathcal{P}=\left\{P=\left\{x\_{0}, \ldots, x\_{n\_{P}}\right\} \mid P\right.$ is a partition of $\left.[a, b]\right\}$ of the given interval.
* Let $f:[a, b] \rightarrow \mathbb{R}^{n}$ be a continuously differentiable function. The length of the curve defined by $f$ can be defined as the limit of the sum of line segment lengths for a regular partition of $[a, b]$ as the number of segments approaches infinity. This means
$$
L(f)=\lim \_{N \rightarrow \infty} \sum\_{i=1}^{N}\left|f\left(t\_{i}\right)-f\left(t\_{i-1}\right)\right|
$$
where $t\_{i}=a+i(b-a) / N=a+i \Delta t$ for $i=0,1, \ldots, N .$ This defintiton is equivalent to the standard definition of arc length as an integral:
$$
\lim \_{N \rightarrow \infty} \sum\_{i=1}^{N}\left|f\left(t\_{i}\right)-f\left(t\_{i-1}\right)\right|=\lim \_{N \rightarrow \infty} \sum\_{i=1}^{N}\left|\frac{f\left(t\_{i}\right)-f\left(t\_{i-1}\right)}{\Delta t}\right| \Delta t=\int\_{a}^{b}\left|f^{\prime}(t)\right| d t
$$