Show that $\vert\int_{-1}^1 \omega(t) dt \vert \leq 2^n \int_{-1}^1\vert \omega^{(n)}(t)\vert dt$ I am stuck with the following problem: 

With $\omega: [-1,1]\rightarrow \mathbb{R}$, $\omega\in C^n(-1,1)$. Suppose that $\omega$ has a finite number of zeroes $t_1<t_2<\cdots <t_n$ (i.e. $\omega(t_i)=0,\forall i$) on $[-1,1]$. Prove that $$\left\vert\int_{-1}^1 \omega(t) dt \right\vert \leq 2^n \int_{-1}^1\vert \omega^{(n)}(t)\vert dt$$

I think I should show it inductively, but I can figure out how to do it. If someone could give me some hints that would be greatly appreciated.

Applying Rolle's theorem an awful lot of times, you will find that all derivatives of $\omega$ of orders $\le n-1$ have at least one zero in $[-1,1]$. This allows you to run the following, outrageously wasteful, chain of inequalities for $k=0,1,\dots, n-1 $: 
$$\int_{-1}^1 |\omega^{(k)}(x)|\,dx \le 2 \sup_{[-1,1]} |\omega^{(k)}(x)| \le 2\int_{-1}^1 |\omega^{(k+1)}(x)|\,dx$$ 
where the last step uses the fundamental theorem of calculus: $\omega^{(k)}(x)$ is given by the integral of $\omega^{(k+1)} $ from some zero of $\omega^{(k)}$ to $x$.
