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

The Laplace transformation of the function would be

`F(s) = -(1-2e^(-s))/(s)`

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) = -(1-2e^(-s))/(s)`