Laplace Transform of
(1-e^-t)u(t-3)

Question

Laplace Transform of 
(1-e^-t)u(t-3)

anonymous · Answer

$$ (1-e^{-t})u(t-3)$$

anonymous · Answer

@Loser66 Lol, I was going to ask you.

loser66 · Answer

I forgot all the stuff. ha!!

loser66 · Answer

woah... you post the whole book ??

anonymous · Answer

How are you in high school!

accessdenied · Answer

I'm not remembering the specific identity with multiplying by the heaviside step function, but by definition u(t - 3) is neutralizing any values below t < 3.
So the Laplace transform of (1 - e^(-t)) u(t - 3) is equal to:
  $ \displaystyle \int_{0}^{\infty} (1 - e^{-t} ) u(t - 3) e^{-st} \ dt $
= $ \displaystyle \int_{\color{blue}{3}}^{\infty} (1 - e^{-t} ) e^{-st} \ dt $
We would need to substitute u = t - 3   so that the adjustment brings back our lower boundary of 0, which is the definition of the Laplace transform of our function after substitution