 Let's do this, the Laplace transform of this function f of t equals e to the power negative 3t. I remind you that the Laplace transform of any function, of any function f of t equals the improper integral of e to the power negative t times the f of t dt. That is what we have to do. If we do the Laplace transform, it's got to be this curse of L, which I can never do well, of e to the power negative 3t. Well, it's this improper integral of e to the power negative s t times e to the power negative 3t dt. I suppose I can write this a bit better. It'll be e to the power, and it'll be a negative s plus 3t dt. I suppose I could write it like that. And what would be the integral of that? Remember it's the limit as b goes to infinity of the integral of going from 0 to b. Remember all of that. But this is going to be, there should at least be a 1 over s plus 3, s plus 3. Let's see, there's got to be an e to the power negative s plus 3t. But if I integrate this, if I would integrate this now, I'm going to be left with a negative in front there. So there's got to be a negative there as well, going from 0 to infinity. Now just think about it a bit. We want this to be, land up in the denominator. The only way that this is going to land up in the denominator, if all of this is larger than 0. So here we have the fact that s plus 3 must be larger than 0, or s must be larger than negative 3. If x is negative 4, we're going to have negative 1 times negative, that's positive e to the power t. And that's not going to work for us. We want this to remain negative, and the only way, before that was easy, we just had s larger than 0. But here the s plus 3 must be larger than 0, so s must be larger than negative 3. So let's do that, we're going to have negative 1 over s plus 3, and we're going to have 1 over e to the power s plus 3 t, going from 0 to infinity. There's going to be negative 1 over s plus 3. There's going to be a negative 1 over s plus 3. This is our look. And then we're going to have the fact that we need to do this bit. There's going to be 1 over e to the power s plus 3 times infinity, minus 1 over e to the power s plus 3 times 0. That equals negative 1 over s plus 3. And what is this going to be? Well, this becomes 0. I have this infinity in the denominator. e to the power 0 is 1, so that becomes negative 1. So our end result here, you can see there, negative 1. So this becomes 0. This becomes 1. 0 minus 1 is negative 1. Multiply out there as 1 over s plus 3. Very awkward in the beginning these Laplace transforms, but there you have it. The Laplace transform of e to the power negative 3 t, so we have a few in our arsenal now how to do some of these. So if that was negative 4 t, this would have been 1 over s plus 4. I think you could see where that is going as well. The only provisor that we have here is that s must be larger than negative 3 for this to converge, for this improper integral to converge. And we saw that because we needed this negative s plus 3 t to remain a negative for the values that we choose so that we could put it in the denominator so that when we do take these bounds that we do end up with this kind of scenario. So yeah, s plus 3 had to be larger than 0, 2 for this to remain a negative.