A step function built from the floor

Step function. On each open interval $]k,k+1[$ with $k\in\{0,1,2,3\}$, the integer part equals $k$, so $f$ is constant there with value
$$c_k=(-1)^k\,k.$$
Hence $f$ is a step function associated with the subdivision $\{0,1,2,3,4\}$. The values at the subdivision points do not affect the integral.

Values on each subinterval.
- $]0,1[$: $c_0=(-1)^0\cdot 0=0$,
- $]1,2[$: $c_1=(-1)^1\cdot 1=-1$,
- $]2,3[$: $c_2=(-1)^2\cdot 2=2$,
- $]3,4[$: $c_3=(-1)^3\cdot 3=-3$.

Integral.
$$\int_0^4 f=\sum_{k=0}^{3}c_k\cdot 1=0+(-1)+2+(-3)=-2.$$

Conclusion. $f$ is a step function on $[0,4]$ and
$$\int_0^4(-1)^{\lfloor x\rfloor}\,\lfloor x\rfloor\,dx=-2.$$