$$g(t)=\int_0^t f(\tau)d\tau$$ If we can automatically assume $F(s)$ and $G(s)$ to be Laplace transforms of $f$ and $g$, how can I show that $$G(s) =\dfrac{F(s)}{s}$$
The only conclusion I can draw from this question is that g' = f. But, I'm not sure how to apply this information to prove the Laplace equivalence as asserted.
Hint: With definition $${\cal L}(f)=\int_0^\infty e^{-st}f(t)dt=\int_0^\infty e^{-st}g'(t)dt$$ Now use integration by parts.
Edit: Let $u=e^{−st}$ and $dv=g′(t)dt$ then apply integration by parts \begin{align} {\cal L}(f) &= \int_0^\infty e^{-st}f(t)dt\\ &= \int_0^\infty e^{-st}g'(t)dt\\ &= e^{−st}g(t)\Big|_0^\infty+s\int_0^\infty e^{-st}g(t)dt\\ &= s{\cal L}(g) \end{align} with $g(0)=0$.