Use Definition 7.1.1. DEFINITION 7.1.1 Laplace Transform Let f be a function defined for t ≥ 0. Then the integral ℒ{f(t)} = ∞ e−stf(t) dt 0 is said to be the Laplace transform of f, provided that the integral converges. Find ℒ{f(t)}. (Write your answer as a function of s.)

Answer:[tex]F(s)=\frac{1}{s}[/tex]Step-by-step explanation:Let [tex]f(t)[/tex] be defined for [tex]t\ge0[/tex].The Laplace transform of [tex]f(t)[/tex] is an integral transform given by the Laplace integral:[tex]\ell(f(t))=F(s)=\int\limits^{\infty}_0 {e^{-st}}\cdot f(t) \, dt[/tex], provided that this improper integral exists.If we let [tex]f(t)=1[/tex], then[tex]\ell(f(t))=F(s)=\int\limits^{\infty}_0 {e^{-st}}\cdot1 \, dt[/tex][tex]\ell(f(t))=F(s)=\lim_{a\to \infty}\int\limits^{a}_0 {e^{-st}} dt[/tex][tex]F(s)=\lim_{a\to \infty}-\frac{1}{s} {e^{-st}}|_{0}^{a}[/tex][tex]F(s)=-\frac{1}{s} {\lim_{a\to \infty}e^{-as}}--\frac{1}{s} {\lim_{a\to \infty}e^{-0\times s}}[/tex]Recall that: [tex]\lim_{x\to \infty}e^{-x}=0[/tex][tex]\implies F(s)=0+\frac{1}{s}[/tex][tex]\implies F(s)=\frac{1}{s}[/tex]