Question

(1 point) If f(t) is continuous for t≥0, the {\it Laplace transform} of f is the function F defined by F(s)=∫[infinity]0f(t)e−stdt and the domain of F is the set consisting of all number s for which the integral converges. (a) Find the Laplace transform of f(t)=1. (Make sure you can state the domain of F if we ask for it later!) F(s)=

Answer 1

Answer:

The Laplace transform of f(t) = 1 is given by

F(s) = (1/s) for all s>0

Step-by-step explanation:

Laplace transform of a function f(t) is given as

F(s) = ∫∞₀ f(t) e⁻ˢᵗ dt

Find the Laplace transform for when f(t) = 1

F(s) = ∫∞₀ 1.e⁻ˢᵗ dt

F(s) = ∫∞₀ e⁻ˢᵗ dt = (1/s) [-e⁻ˢᵗ]∞₀

= -(1/s) [1/eˢᵗ]∞₀

Note that e^(∞) = ∞

F(s) = -(1/s) [(1/∞) - (1/e⁰)]

Note that (1/∞) = 0

F(s) = -(1/s) [0 - 1] = -(1/s) (-1) = (1/s)

Hope this Helps!!!

Answer 2

In this exercise we have to use the knowledge of the Laplace transform to calculate the total value of the given function, thus we will find that:

$F(s) = (1/s) \\for \ all\ s>0$

So we have that the Laplace transform can be recognized as:

$F(s) = \int\limits^\infty _0 { f(t) e^{-st} \, dt$

Find the Laplace transform for when f(t) = 1, we have that:

$F(s) = \int\limits^\infty _0 { f(t) e^{-st} \, dt \\\\ F(s) = \int\limits^\infty _0 { 1 e^{-st} \, dt$

$F(s) = \int\limits^\infty _0 { e^{-st} \, dt = (1/s) [-e^{-st}]$

$F(s) = -(1/s) [(1/\infty ) - (1/e^0)] \\F(s) = -(1/s) [0 - 1] = -(1/s) (-1) = (1/s)$

See more about Laplace transform at brainly.com/question/2088771