Laplace transform with abs value in exponential

Hi everyone,
I have this problem and just need someone to check or correct:
[tex] f(t) = -4e^{-3|t|}(u(t + 3) - u(t-1))[/tex]
where u(t) is step function: u(t) = 1 for t >= 1 and 0 elsewhere;
so, i guess I need to break abs value into 2 cases and have 2 different equations? anyway, here's what I have if someone would be so kind and check my work (at first I applied linearity property and distributed [tex]-4e^{-3|t|}[/tex]:
1. for t > 0:
[tex] L[f(t)] = \frac{-4}{s+3} + \frac{4e^{-(s+3)}}{s+3} [/tex]

2. for t < 0:
[tex] L[f(t)] = \frac{-4}{s-3} + \frac{4e^{-(s-3)}}{s-3} [/tex]

thanks in advance!

ps: i guess one thing I should explain is that by definition of unilateral laplace transform, even though first part of signal starts at -3 we do not consider it, what we are doing is one-sided Laplace transform, so I started integrating from 0^- the first part of the expression.

edit: to (hopefully) increase chances that someone looks at this here's the Lapl. trnsf. that are relevant:
[tex] L[u(t-k)] = \frac{e^{-sk}}{s}[/tex]

[tex]L[e^{at}f(t)] = F(s-a)[/tex]
but you probably know this anyway :shy:

 

ooops, sorry about the typo, u(t) = 1 for t >= 0, you're right.
so I would have to integrate from 0 to 1 (for the one-sided laplace trnsfm)...
thanks!