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)} = [infinity] 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.)

[tex]\int\limits^{\infty}_{0} {e^{-s\cdot t}\cdot f(t)} \, dt = -\frac{1}{s}\cdot e^{-s\cdot t}\cdot f(t)|_{0}^{\infty}-\int\limits^{\infty}_{0} {e^{-s\cdot t}\cdot f(t)} \, dx[/tex]

[tex]\int\limits^{\infty}_{0} {e^{-s\cdot t}\cdot f(t)} \, dt = -\frac{1}{2\cdot s}\cdot e^{-s\cdot t}\cdot f(t)|_{0}^{\infty}[/tex]

[tex]\int\limits^{\infty}_{0} {e^{-s\cdot t}\cdot f(t)} \, dt = -\frac{1}{2\cdot s}\cdot e^{-s\cdot t}\cdot f(\infty)+\frac{1}{2\cdot s}\cdot f(0)[/tex]

[tex]\int\limits^{\infty}_{0} {e^{-s\cdot t}\cdot f(t)} \, dt = 0+\frac{1}{2\cdot s}\cdot f(0)[/tex]

[tex]\int\limits^{\infty}_{0} {e^{-s\cdot t}\cdot f(t)} \, dt = \frac{1}{2\cdot s}\cdot f(0)[/tex]

[tex]\mathcal{L}\{f(t)\} = \frac{f(0)}{2\cdot s}[/tex]