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Determine the Laplace transform of the given function.$$f(t)=\left\{\begin{array}{rl}1, & 0 \leq t<1 \\-1, & 1 \leq t \leq 2,\end{array}, \text { where } f(t+2)=f(t)\right.$$

$\frac{1}{1 \cdot e^{-3 s}} \cdot\left(\frac{e^{-1}-1}{3}\right)^{2}$

Calculus 3

Chapter 10

The Laplace Transform and Some Elementary Applications

Section 3

Periodic Functions and the Laplace Transform

Second-Order Differential Equations

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So this is problem. Seven were given this periodic function. That's one 14 this range and minus one for T in this range. And we've got that f of t plus two is equal to half of tea. So the period T is equal to. So to find the LaPlace transform, we used their, um 10.3 point three in the book. That's going to be 1/1 minus E to the minus period times s terms, the integral from zero to the period of f of tea, times E to the minus s T d t. Okay, now, this is peace wise defined, so we're gonna have to do the integral like this, split it up into an integral from 0 to 1, and then from 1 to 2 can This is multiplying this entire thing. And so this is just the integral of one times this experiential and then minus this. Okay. And so, evaluating the integral this is going to give us minus one over arrests. Eat the minus. Ste. Evaluated between zero and one. And then this is minus. This integral evaluates between one and two. Okay, so this term right here, what is that going to be? That's going to be negative. One over arrests times E to the minus s and then plus one of rest times one. And then this is minus on. Don't forget, everything's in brackets. Negative. One over us eats the minus two s. Then that's minus. And then therefore, that becomes a plus, one of rest times E to the minus s. Okay, so let me rewrite this, okay? So rewritten it becomes this. And so let's just simplify. So we're going to get with. So let's expand and simplify. So these a role s is, by the way, there's no fives in here, okay? And so get it. All under this common denominator, this becomes so it's one and then minus to the minus s, and then plus heats the minus to us. So that is our answer for the LaPlace transform. What about

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