I need to prove that every polynomial $p(x)$ is of boundary variation in the interval $[a,b]$ and $V_a^b p \leq \int_a^b \vert p'(s) ds \vert$ where $V_a^b p = \sup \{V(p,P) \; : \; P \; \text{is a partition of} \; [a,b] \}$. To show the first I take any partition of $[a, b]$, $P=\{ t_0=a, t_1, \dots, t_n=b \}$, and we know that $p(x)$ is continuous and $p'(x)$ exists then by the mean value theorem for every $k$ there is $x_k \in (t_{k-1}, t_k)$ $$p(t_k) - p(t_{k-1}) = p'(x_k)(t_k - t_{k-1}) $$ $p'(x)$ is another polynomial then $\vert p'(x) \vert$ is a continuous function and by the extreme value theorem we know it has a maximum $A$, then $$V(p,P) = \sum_{i=1}^n \vert p(t_k) - p(t_{k-1}) \vert = \sum_{i=1}^n \vert p'(x_k) \vert (t_k - t_{k-1}) = \sum_{i=1}^n A (t_k - t_{k-1}) =A (b -a) $$
Then every for every partition the variation is bounded by $A (b -a)$.
The problem is in the second point $V_a^b p \leq \int_a^b \vert p'(s) ds \vert$, the solution of the first point seems to be related, for the sum $\sum_{i=1}^n \vert p'(x_k) \vert (t_k - t_{k-1})$, but I don't know how to conclude.
The idea here is that the sum can be rewritten as a sum of integrals using the fact that polynomials are smooth functions: $$ \sum |p(t_k)-p(t_{k-1})| = \sum\bigg|\int_{t_{k-1}}^{t_k}p'(s)\,ds\bigg| \le\sum\int_{t_{k-1}}^{t_k}|p'(s)|\,ds = \int_a^b|p'(s)|\,ds. $$ Since this inequality holds for arbitrary partitions, taking the sup over all partitions of $[a,b]$ gives $V_a^bp \le \int_a^b |p'(s)|\,ds$.