Laplace Transformations and Piecewise Functions I am trying to understand why it is that Laplace transformations can simply be "added together" when finding the transform of a piecewise function. My professor has quite extensively talked about the linearity property of these transforms and their inverses, however, I still have yet to understand why they can just be added together like integrals of the original piecewise function. For example, here is a basic problem I'm working on:
\begin{align}
\mathcal{L}\left\{f\left(t\right)\right\},\:\text{such that}\:f\left(t\right) =\begin{cases}
-1, & 0\leq t<1, \\[.3ex]
1, & t \geq 1.\end{cases}\tag{1}
\end{align}
If I were to find the basic integral of this function, I would add the integrals together as such:
\begin{align}
-\int_0^{1-}dt+\int_1^{P}dt,\:\text{such that}\:P\in\left[1,\infty\right).\tag{2}
\end{align}
But since the transform has an extra term inside the integral, why is it that I can still add the two pieces together? Is it because I can write the integral as
\begin{align}
-\int_0^{1-} e^{-st}\:dt+\int_1^{\infty}e^{-st}\:dt\implies -\int_0^{1-}\:d\left(\frac{-e^{-st}}{s}\right)+\int_1^\infty\:d\left(\frac{-e^{-st}}{s}\right)?\tag{3}
\end{align}
Or, more generally:
\begin{align}
\int_0^{P_1}f\left(t\right)\:d\left(\frac{-e^{-st}}{s}\right)+\int_{P_1}^{P_2}f\left(t\right)\:d\left(\frac{-e^{-st}}{s}\right),\:\text{such that}\:P_1,P_2>0\:\text{and}\:P_1<P_2
\end{align}
Or does this not make sense...
I guess that in (3) I'm thinking of the integral as more of an operator.
My apologies by the way if any of you feel this is a duplicate, I have looked to see if I could find a duplicate to save some time and didn't find anything other than posts that just stated the definition instead of really investigating/stating the reason why that definition holds.

This is just basic properties of the integral.
$$
\int_a^bh(t)\,dt+\int_b^ch(t)\,dt=\int_a^ch(t)\,dt.
$$
In your case
$$
h(t)=f(t)\,e^{-st}, \quad a=0,\quad b=1,\quad c=\infty.
$$
