Find laplace transform of step function

Hello good MHF folk. I'm getting very frustrated with a few exercises in Stroud's Advanced Engineering Mathematics 5th ed. I just can't get to the answers provided, I'm hoping someone may be able to help me (Hi)

First one is this. Sketch the graph then find the laplace transform of:

$\displaystyle f(t)\left\{\begin{matrix}3^{-2t} & 0\leq t < 2\\ 0 & 3 \leq t \end{matrix}\right.$

I can draw the graph ok, I understand that. And my answer is almost the same but there is an

$\displaystyle e^{-6}$

in the given answer that I just don't understand... Here is their answer:

$\displaystyle F(S) = \frac{1}{s+2} (1 - e^{-6}e^{-3s})$

Re: Find laplace transform of step function

Aha! Nevermind! I got there.

I was taking the laplace transform of

$\displaystyle e^{-2t}-e^{-2t}u(t-3)$

I wasn't writing the second $\displaystyle e^{-2t}$ in terms of $\displaystyle (t-3)$. So I did a quick check:

$\displaystyle e^{-2(t-3)} = e^{6-2t}$

so

$\displaystyle e^{-2t} = e^{-2(t-3)}e^{-6}$

made that change and ended up with

$\displaystyle e^{-2t}-e^{-2(t-3)}e^{-6}u(t-3)$

and all suddenly became clear! It then easily works out to the provided answer above...

I hate the frustrating times when studying math, when nothing seems to work or make sense. But then it's so satisfying to figure it out in the end! Sometimes even writing the question out on here helps to shift things about in your mind a bit. (Rock)