Is the function $G(x)=\int_0^x \left| \frac{d}{dt}\int_{t-1}^tf(s)ds \right|dt$ bounded? Let $f:\mathbb{R}\to \mathbb{R}$ be a continously differentiable bounded function. Then the function $F$ defined for each $x>0$ by
$$F(x)=\int_0^x \left( \frac{d}{dt}\int_{t-1}^tf(s)ds \right)dt$$ is bounded since
$$\left| F(x)\right| =\left| \int_{x-1}^xf(s)ds-\int_{-1}^0f(s)ds\right| \leq 2\sup_{t\in \mathbb{R}} \left|f(t) \right|. $$
What about the function $G$ defined for each $x>0$ by
$$G(x)=\int_0^x \left| \frac{d}{dt}\int_{t-1}^tf(s)ds \right|dt$$
is it also bounded ?

Let $f(s)=\cos(\pi s)$ so that $f(s)=-f(s-1)$. Then 
$$
\int_{t-1}^tf(s)ds=\int_{t-1}^t\cos(\pi s)ds=\frac{1}{\pi}(\sin(\pi t)-\sin(\pi(t-1))=\frac{2}{\pi}\sin(\pi t). 
$$
Then 
$$
\frac{d}{dt}\int_{t-1}^tf(s)ds=\frac{d}{dt}\Big[\frac{2}{\pi}\sin(\pi t)\Big]=2\cos(\pi t)
$$
so that 
$$
G(x)=\int_0^x|2\cos(\pi t)|dt\to\infty
$$
