Question

Let f0(t) = 1 − e−λt, where λ> 0.

(a) show that sx(t) = e−λt.
(b) show that μx = λ.
(c) show that ex = (eλ − 1)−1.
(d) what conclusions do yo

Answer 1

To show that sx(t) = e^-λt, we need to verify that the Laplace transform of f0(t) is equal to the Laplace transform of e^-λt.

Step-by-step explanation:

To show that sx(t) = e-λt, we need to verify that the Laplace transform of f0(t) is equal to the Laplace transform of e-λt.

(a) Taking the Laplace transform of f0(t), we have: L[f0(t)] = ∫[0,∞] f0(t)e-st dt = ∫[0,∞] (1-e-λt)e-st dt

Using integration by parts, we can simplify this expression to: L[f0(t)] = (1/(s+λ))-((λ/(s+λ))e-(s+λ))

To find the Laplace transform of e-λt, we can directly calculate: L[e-λt] = 1/(s+λ)

Since the Laplace transforms of f0(t) and e-λt are equal, we can conclude that sx(t) = e-λt.