Question

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.)
f(t) = −1, 0 ≤ t < 1 1, t ≥ 1.

Answer 1

Answer:

The Laplace transformation of the function would be

    $F(s) = -\frac{1-2e^{-s}}{s}$

Step-by-step explanation:

According to the information of your problem

$F(s) = L\{f(t)\} = \int\limits_{0}^{\infty} e^{-st} f(t) dt$

And the function given is

$f(t) = \left \{ {{-1 \,\,\,\,\,\,\,0 \leq t < 1} \atop {1 \,\,\,\,\,\,\, t \geq 1}} \right.$

Therefore

$F(s) = L\{f(t)\} = \int\limits_{0}^{1} -e^{-st} dt + \int\limits_{1}^{\infty} e^{-st} dt$

And when you compute those integrals you get that

$F(s) = -\frac{1-2e^{-s}}{s}$