Proof for an integral inequality For a function $f:\mathbb{R}\to\mathbb{R}$, $|f|\leq M$, $f(s)\equiv 0$ for $s<0$, define
\begin{equation}
I^n(t)= n \int_{t-1/n}^{t} f(s)ds.
\end{equation}
In my book they conclude that
\begin{equation}\label{eq}
|I^n(t')-I^n(t)|\leq 2n|t'-t|M.
\end{equation}
How can I proof this inequality? I only see that
\begin{equation}
|I^n(t')-I^n(t)|\leq |n[(t'-t'+1/n)M-(t-t+1/n)M]|=0.
\end{equation} 

hint
Put $u=t-s $.
then
$$I^n (t)=\int_0^\frac 1nf (t-u)du $$
and
$$I^n (t)-I^n (t')=\int_0^\frac 1n (f (t-u)-f (t'-u))du $$
