Second Shifting theorem to find Laplace transform

[helpinghand]

Can some one help me to understand this:

I need to use the second shifting theorem to get the Laplace transform, given:

f(t) = { 4 - t² , t < 2 ...... 1
{ 0 , t >= 2


They got this:

f(t) = (4 - t²)(1 - u(t - 2)) .... 2


I know that the second shifting theroem says that L(g(t-k)u(t-k)) = e^-skG(s)

But the thing is I don't know how they get from 1 to 2, can someone please explain this?

Cheers

 

[Too Easy]

Mark's notes leave a bit to be desired

you have to use the heaviside step function u(t-a)
I think of it as a function that can be switched on at any time t and remains on forever
knowing that the function equals zero for t<a and equals 1 for t>a.
We want the function to be equal to 4-t^2 for t less than 2. but u(t-2)=0 for t<2
so we must add the expression 4-t^2 for t<2, then take it away for t>2

hence:
f(t)=(4-t^2)-(4-t^2)u(t-2)

As far as i can see there is no "proper" way to do this, certainly not in the lecture notes anyway