Find the Laplace Transform of
$$f : (0, +\infty) \rightarrow \mathbb{R}, \quad f(t) = \left\{\begin{array}{lr} \sin(t), & \text{for } 0 < t < 2\\ 1+2t^3, & \text{for } t \geq 2\\ \end{array}\right\}$$
Usually the laplace transforms on piecewise functions are only really defined on one interval or zero on all other intervals, but if it's defined on multiple intervals that means there are two different transforms with two unique answers respective to their intervals, right?
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If a function $f$ is absolutely integrable on $[0, \infty)$, then the integral $$\int_0^\infty e^{-st} f(t)\;dt$$ converges for any $s \in [0, \infty)$. That means you can define a function $\mathcal{L}f$ on the interval $[0, \infty)$ by setting $$\mathcal{L}f(s) := \int_0^\infty e^{-st} f(t)\;dt.$$ It doesn't matter whether $f$ is defined using a single formula, or defined piecewise, or defined as the solution of some weird differential equation. As long as you can compute the integrals above, you can define the function $\mathcal{L}f$ on the whole interval $[0, \infty)$.
Per the definition, you can write this:
$$\int_0^\infty e^{-st} f(t) dt = \int_0^2 e^{-st} \sin(t) dt + \int_2^\infty e^{-st} (1+2t^3) dt.$$
A more common way of actually doing the calculation of the Laplace transform of a piecewise function, is to write it as a sum of functions multiplied by step functions. In your example, your function starts out as $\sin(t)$ and switches to $1+2t^3$ at $t=2$. You can write this in terms of step functions with
$$f(t)=\sin(t)+(1+2t^3-\sin(t))u(t-2)$$
where $u$ is the unit step function at zero. (You can multiply the first $\sin(t)$ by $u(t)$ if you want, it doesn't change the Laplace transform.) This is more useful when a Laplace transform table is available.