Suppose $f$ is integrable on $[a,b]$ and $F(x)=\int_a^x f(t)dt$. Prove that
$$ V_{[a,b]}(F) = \int_a^b |f(t)|dt, $$ where $V_{[a,b]}(F)$ denotes the variation of $F$ on $[a,b]$.
Attempt Let $\mathcal{P}=\{x_0,x_1,\dots,x_k\}$ be a partition of $[a,b]$. We have
\begin{align*} V_{[a,b]}(F,\mathcal{P}) &= \sum_{i=1}^k \left| F(x_i) - F(x_{i-1}) \right|\\ & = \sum_{i=1}^k \left| \int_a^{x_i} f(t)dt - \int_a^{x_{i-1}} f(t)dt \right| \\ &=\sum_{i=1}^k \left|\int_{x_{i-1}}^{x_i} f(t)dt \right|\\ \end{align*}
Not sure how to get the absolute value inside the integral then the rest follows obviously.
