the-laplace-transform-of-a-continuous-function-over-the-interval-0-infty-is-defined-by-fsint_0inft-4

Question

The Laplace transform of a continuous function over the interval $[0, \infty)$ is defined by $F(s)=\int_{0}^{\infty} e^{-s x} f(x) d x$ (see the Student Project). This definition is used to solve some important initial-value problems in differential equations, as discussed later. The domain of $F$ is the set of all real numbers $s$ such that the improper integral converges. Find the Laplace transform $F$ of each of the following functions and give the domain of $F .$
$$
f(x)=e^{a x}
$$

K B · Accepted Answer

So in this question, you know that the loss transformer continuous function can be rewritten. As this is the interval from zero to infinity, including syrup is gonna be eating the power of negative at Saks. And then here we put our function, which, in our case, is eating the power of a axe. Not yet. So now we&#x27;re gonna be right. This is the limit, which is? The limit is t goes to infinity of and then this is gonna be t and your infinity hair T is going to infinity Sweep a t here instead now And you The power, eh? Minus ass times Acts and and d s, which can now be recognized one over. We&#x27;re reading this worse. You keeping the limit? One over a minus. Fast times either power. But they might just ask. That&#x27;s you&#x27;re okay. So now we can rewrite again, keeping this limited. Steve goes to infinity, will rewrite this portion. It&#x27;s one over. A minus ass tries. You do the power being minus past acts. You&#x27;re a minus s t now minus one over a minus s, which evaluate. Now we&#x27;re violating the limit. It&#x27;s well, it&#x27;s just gonna be won over asked me to say so. This is actually our function after us. It&#x27;s gonna be one of her ass minus a. And our domain is gonna be a to infinity. So these right here are finally answers to this question.

K B · Answer

So when this question we know that we cannot we write our function half of us for this boss transformation with a continuous function over the intervals your infinity as this integral infinity zero. And then here we have either the power of negative at Saks. This is from the formula. And then here we put effort backs, which in our case, is just gonna be acts. And then dx the box is given. And the question is, now we can rewrite. This is the limit. The limit is two goes to infinity of our interval t zero, actually the power of negative at Saks Steeples. Over. Wish we cannot write. Rewrite. This portion is negative. One over a squared, even power of negative ass axe one boss ass Fax this entire thing. Use your So now this portion again, we&#x27;re keeping this limited t goes to infinity. We can make this into one minus either the power of negative Esty Times one plus Ask Steve And then we find now that F Vass is just gonna be equal and for this portion right here. But actually, we&#x27;re gonna want to put the limited&#x27;s t goes to infinity. So it&#x27;s just this limit. And then Juan over Escort is on the outside in this case. So when we evaluate this limit, you realize that after bus, and, um, in the end, it&#x27;s just gonna be one of grass squared. And the domain about the best is gonna be zero to infinity, not including zero. So these are our final answers to this question.

Kira Schwander · Answer

So in the student project of your textbook, your textbook list listed the LaPlace transform over the range zero to Infinity as having this function given by this integral. So we&#x27;re currently looking at the function f of X equals one. So we&#x27;re going to plug in one whenever we see f of X in our integral. So we&#x27;re going to rewrite this as f of s equals are integral from zero to infinity of exponential, negative s, times X. Let&#x27;s write that times one with respect to X. So now we&#x27;re going to have to rewrite this as a limit because we can&#x27;t integrate all the way up to infinity. So we&#x27;re gonna have our limit as teas approaching infinity of our integral from zero to t of. I&#x27;m going to rewrite this a bit. Um, all right, now just exponential negative s times X And I&#x27;m not going to bother writing that one, cause it&#x27;s just a constant with respect to X. So now we can solve this anti derivative, so we&#x27;re gonna have our limit as t approaches infinity of our anti derivatives exponential negative s Times X. And now we&#x27;re going to divide by the derivative of negative s X. So it&#x27;s going to be divided by negative s. We&#x27;re evaluating this from zero to t. So now we&#x27;re plugging in tea and zero. So we&#x27;ll have negative or the limit as T approaches. Zero charity approaches. Infinity of negative exponential negative s t over s minus. Negative. Exponential. Negative s time. 0/0 over s. So now we&#x27;re going. Teoh, rewrite this so that we don&#x27;t have our negative exponents so that we can see what&#x27;s actually going on here so that we&#x27;ll have our limit as t approaches infinity of negative one over Exponential s t times s plus one over s times exponential zero. So for this first term, as T gets closer and closer to infinity, this denominators getting larger and larger, causing the fraction itself to get smaller and to converge all the way down to zero as the fraction gets smaller and smaller and the second term is going to be one over one over s plus times one. So our function of s is going to be equal to one over s. So now we need to find the domain of this function. So this is going to exist any time. That s is not that the denominator is not equal to zero. So this is going to be equal. So over the domain of zero to infinity, exclusive on both ends of zero and infinity.

Kira Schwander · Answer

the student project in your textbook defined the LaPlace transform over the region zero to infinity by this integral. So when we&#x27;re given a function of eggs such as co sign of two X, we were able to plug in this function for f of X. So that will make it capital. F of s equals the integral from zero to infinity off exponential negative S X Times co sign of two X. So we need to keep in mind that we can&#x27;t go all the way to infinity here. So we are going to have to use a limit so we&#x27;ll have the limit as t approaches infinity of the integral from zero to t of exponential negative S X Times Co sign of two X with respect to X So now ignoring bounds. For now, we&#x27;re going to look at just the integral of exponential negative s X tonnes co sign of two X with respect to X and then we can see what the bounds are later. So to solve this, we have to separate X terms being multiplied together, which is going to be difficult to integrate on their own so we can use integration by parts which you&#x27;ll remember is the integral of you times DV mine equals u times v minus the integral of V times to you. So say you equal to exponential negative s ex and D V equal to co sign of two X. You could also reverse them. I&#x27;m just choosing to set you equal to exponential negative s eggs so that derivative is equal to negative s times. Exponential negative s ex and the anti derivative of co Sign of two X is equal to sign of two X divided by two. So now we&#x27;re going to plug this in to our formula. So we have you ties V so 1/2 times one over Exponential s ex times Sign of two X minus the integral of v times Do you? So 1/2 time sign of two eggs, times negative s times, exponential, negative at sex. So we&#x27;ll notice that we have two constants in our into girl. So I&#x27;m going to pull these out so that they don&#x27;t get us confused here. So we&#x27;re going to end up with one. We&#x27;re gonna end up with sign of two eggs over two times Exponential s ex just simplifying some things on the outside plus s over two times the integral of sine of two x times Exponential negative s X, with respect to X. So we&#x27;re going to use integration pipe parts again. So we&#x27;ll set you equal to exponential Negative s ex again and Devi equal to sign of two x with respect to X so that the derivative of you to the negative SX is going to be negative s times Exponential, negative SX and the anti derivative of sign of two X is negative co sign of two X divided by two So now we complete this into our formula. So we have sign of two X over two times Exponential s ex plus s over two times u times v so negative 1/2 co sign of two x times one over exponential s ex minus s over to times the integral of you three times do you so 1/2 times co sign of two x times exponential negative s X and I ran out of room almost so again we&#x27;ll notice that we have. These two isn&#x27;t s here. We have these two constants on the inside of our integral. So we&#x27;re gonna pull them out again so that we can have sign of two X over two times Exponential s ex plus rather minus s turns the coastline of two eggs over 24 times Exponential s ex combining things in that second term minus s squared over four times the integral of exponential negative S X Times co sign of two X So you&#x27;ll notice that this is the same as theory Jinling Integral that we&#x27;ve that we had. So all of this is equal to the integral of exponential Negative SX Times Co sign of two X with respect to X. So I&#x27;m going to scroll down so that we can have more space to evaluate this. So if we add as squared fourths times this integral to both sides, we&#x27;re going to end up with sign of two X over two times exponential s X minus as times co sign of two X over four exponential s ex equals one plus science s squared over four times the integral of exponential negative s ex Times co sign of two x with respect to X. So if we divide this term from both sides will end up with 1/1 plus as squared over four times. Signed of two x over two times Exponential SX minus sign s Times Co sign of two X over four times exponential SX as equal to this original Integral. So because our A rented original Integral had the bounds zero to t, we can now evaluate this from zero to t so plugging in tea and zero will have the limit as T is approaching Infinity of sign of two to t over two times Exponential of s T minus s Times Co sign of two t over four times Exponential ste times 1/1 plus X squared over four minus sign of two time 0/2 tens. Exponential s time zero minus s temps Co sign of two times 0/4 times Exponential s time zero times 1/1 plus as squared over four. So this is going to be equal to as we evaluate this as T is getting closer to infinity, this denominator of distraction gets larger and larger, bringing the whole fraction to converge to zero. So we&#x27;ll have zero minus. Similarly, in the second fraction we have t is approaching infinity, causing the denominator to approach infinity, meaning the the fraction itself is approaching zero. So we have zero minus zero turns 1/1 plus as squared over four. So that&#x27;s just gonna be zero minus sign of zero is equal to zero. So we have zero minus s times. Co sign of zero is over four times. Exponential zero is going to be s times 1/4 times exponential zero, which is also equal toe one. So s times 1/4 times, one times 1/1 plus as squared over four. So if we multiply these together, this is gonna be positive. Will have s over four plus s squared as our solution, So f of s is going to be equal to s over four plus as squared. And now we need to find domain for us. So one thing to notice is that we do have in s in this in the denominator of our fraction, however, we&#x27;re adding a square of that s so over our original domain of zero to infinity, there will never be an s for which this fraction is not does not exist

Wen Zheng · Answer

if FT is continuous for tea is greater than zero. The lap last transform off. Off. It&#x27;s a function, captain. Half defined by bucks. Yes, he got to interrupt from zero to twenty after house later. Negative as stated. And is that the way off? Half is a such considering off our numbers for which the integral converters I find that the low class transforms the foreign functions A after Izzy or wine Hey of us. Yes, we caught you. Name it. A cost to infinity. Into girl from zero to I in connective Castillo. Thank you. This it is. He caught you, Lim. A costume affinity. Negative one over us. You too. Negative asked house, eh? Minus one. This&#x27;s the culture. One over us. Years at the Wayne is US authorities are zero. How to be lefty is you got to eat the power off. Yes, It&#x27;s the conscious that limit a cost to your vanity. Integral from zero to a You two one one of us. Yeah. This is about you claim it. A cost to vanity won over one of us terms. I need to make you one. It&#x27;s you one minus asked. House, eh? Minus one This is equal to one over one negative one over one. Dewayne is as great is on one part of C. After you see on TV, then we have a CE That&#x27;s the limit. A cost three affinity Integral Found zero to a he has connective. As to you Here we use integration. My parts you hear the result is lim a cost to infinity. A hams, Negative one over us into a negative times. A minus one over us square halves. Each connective asked house, eh? Minus one If it is a good time. Zero we have limit of this part is equal to zero. And this this part is you got zero. So the answer is one over us square. Here&#x27;s what the mind is as the secret is on zero.

Alex Johnson · Answer

can bring use the plus transformation of, ah, function FX. And in this case, we&#x27;re gonna let FX be a constant value, see? And we&#x27;re gonna take a look at what we get if s is gonna be greater than zero. So let&#x27;s set up our integral We have a constant C. I&#x27;m gonna pull up to the front of the inner girl of the Integral from zero to infinity of G to the negative s x d x, and we&#x27;re gonna turn this into ah limits. We have C times limit as our goes to infinity of the anti derivative, which is e to the negative s X divided by negative s evaluated from zero to our This gives us see times the limit as our goes to infinity of e toothy Negative s are divided by negative s plus if one divided by s. And if our goes to infinity e to the negative s arm, go to zero And so what we get is C over s so l of f of s is equal to see over s if FX is C

Problem

Use the formula for arc length to show that the c…

Problem 404 Hard Difficulty

Answer

$F(s)=\frac{1}{s-a}$
Domain of $F$ is $(a, \infty)$

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Gilbert Strang,

Calculus

Grace H.

Catherine R.

Anna Marie V.

Samuel H.

Integration Review - Intro

Definite vs Indefinite

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$F(s)=\frac{1}{s-a}$Domain of $F$ is $(a, \infty)$

Calculus

Grace H.

Catherine R.

Anna Marie V.

Samuel H.

Integration Review - Intro

Definite vs Indefinite

Gilbert Strang,Calculus

Grace H.

Catherine R.

Anna Marie V.

Samuel H.

$F(s)=\frac{1}{s-a}$
Domain of $F$ is $(a, \infty)$

Gilbert Strang,

Calculus