00:01
So in this question we have a function f of t, which is equal to e to the t if t is between zero and two, and it's zero elsewhere.
00:17
So let's write this in terms of step functions.
00:19
So let theta of t be equal to zero for t less than zero, and one for t greater than or equal to zero.
00:30
This is our step function.
00:32
Then f of t is equal to e to the t times theta of t.
00:37
That cuts it off when t is less than zero.
00:41
And then we need a function one minus theta of t minus two, because then when t goes to two, this step function kicks in, and that's going to make the function vanish when t is greater than two.
01:02
Now let's make the laplace transform of f.
01:05
L of f, which is a function of s, is the integral from zero minus to infinity f of t e to the minus s t dt.
01:20
But since we have these step functions cutting off our f of t, we only have to integrate from zero to two e to the t e to the minus s t dt, which is the integral from zero to two of e to the one minus s t dt...