If f(t) is continuous for t ≥ 0, the Laplace Transform of f...

If $f(t)$ is continuous for $t \ge 0$, the Laplace Transform of $f$ is the function $F$ defined by $F(s) = \int_0^\infty f(t)e^{-st} dt$ and the domain of $F$ is the set consisting of all numbers $s$ for which the integral converges. Find the Laplace transforms of the following functions. (Enter domains as inequalities.) (a) $f(t) = 38$ $F(s) = \text{________}$ with domain ${s \text{________}}$ (b) $f(t) = 22e^t$ $F(s) = \text{________}$ with domain ${s \text{________}}$ (c) $f(t) = 21t$ $F(s) = \text{________}$ with domain ${s \text{________}}$

Transcript

00:01 All right, for part a, we are integrating from 0 to infinity the expression e to the negative std.

00:14 Now, it's important to note that if we want this integral to converge, we are going to need s to be greater than 0.

00:24 If s was equal to zero the power here would be one no the power here would be zero and the entire expression would be one the integral there is not going to converge and if s was less than zero excuse me if s was less than zero the power would be positive and then the integral wouldn't converge either and by the way we know for sure that that's true because we know for sure that t is from zero to infinity.

01:03 It's not negative, so t is not changing the sign of the exponent.

01:09 All right, anyway, with this in mind, the antiderivative of e to the negative s t here is itself divided by the coefficient of t in the powers, so we divide by negative s, so we multiply by negative 1 over us and we evaluate from 0 to infinity as t goes off to infinity the exponent goes off to negative infinity so e to the negative s t goes to zero and we get zero we subtract when we plug in 0 negative s t equals zero so e to the negative s t is one and we get negative one over s so for part a we get a laplace transform of one over s part b oh i meant to write that in black uh part b part b is actually extremely similar to part a what we are integrating here e to to the t times e to the negative s t d t well this is equal to the integral from zero to infinity of e to the negative s minus one t d t right negative s plus t that's a negative s minus one t okay so notice this integral here e to the negative s minus 1 t d t is exactly the same as the integral from part a, except we have s minus 1 plugged in for s.

03:19 Since in part a we got a laplace transform of 1 over s, here we should get a laplace transform of 1 over s minus 1.

03:29 But let's actually work that out for real.

03:33 The process is similar.

03:37 We first note that if we want the integral to converge, we need s minus 1 to be greater than 0.

03:45 Next part, the anti -derivative here.

03:49 Well, this is negative 1 over s minus 1, just dividing by this coefficient of t.

03:57 And we multiply by itself.

04:04 So this is the anti -derivative here, and we're evaluating from zero to infinity.

04:11 And as t goes off to infinity, the power goes off to negative infinity, so e to the negative s minus 1t goes to zero, and we get zero.

04:24 And when we plug in zero, the power equals zero, so e to the negative s minus 1t becomes 1, and we get negative 1 over s minus 1.

04:37 So, as expected, we get a laplace transform of 1 over s minus 1 for part b.

04:51 Now, for part c, what we want to integrate is from 0 to infinity.

05:00 What we want to integrate is t, e to the negative stdt.

05:08 Now this one is an integration by parts problem.

05:13 So let's integrate by parts...

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