integrate-each-of-the-given-functions-int_02-frace2-t-d-tsqrte2-t4-3

Question

Integrate each of the given functions.
$$\int_{0}^{2} \frac{e^{2 t} d t}{\sqrt{e^{2 t}+4}}$$

Tyler Moulton · Accepted Answer

Yeah, we want to integrate the given expression that is the integral. Or rather the definite integral from 0 to 2 of each of the two T. D. T divided by the square root of each of the two T. Plus. For this section has been challenging our ability to use exponential integration which states that the integral of E d u d U equals W plus C. However, the given definite integral instead of using simply on each of the U. D. U. Form has to eat to be used in it. It has an eating at U. T. In the numerator as well as in the nominator plus four. So it must be that we have to use a different method of integration. We&#x27;re going to use the power rule for which we identify you D U N N. You see the two T plus four. Do you see the beauty and its negative one half which is you know the end. Thus by the power rule we have that are integral, evaluates to two times each of the two T plus forward. The positive one half, 10 0-2 Or root mean squared plus 4 -5. No, this is all this problem. We use the powerful formula here, which states the integral of U. To the n. D. U. Is one over N plus one U. To the N plus one plus. See where we don&#x27;t use C. For the constant of integration because we have a definite integral.

Aishwarya Krishnakumar · Answer

problem. Nine were asked to find the definite in true. And to do this, we are going to first take out the constant. So in this case, the concept is too, so you can take that out and rewrite it, and then it makes it easier for us to just solve for the integral. So if you add one wide to over two, then it would be t three Hobbs times to over three minus. If you add one, it would be he squared over two. And this whole thing, since it&#x27;s a definite integral, would be evaluated from 0 to 4. So now all you have to do is plug in four and then minus the value you get when you plug in zero for tea. Well, let&#x27;s go ahead and do that. It would be 2/3 four since you&#x27;re bugging and four for the T values minus four squared over too. This minus 2/3 times zero to the three halves. Power minus zero squared over, too. And if you compute this, you will get to crimes negative. Eight over three, which equals negative. 16 over three

Xiaomeng Zhang · Answer

so we can say those one dressy close to two times into grow off the square Rules off T right now two times in Q grow off t and see what does one that echoes who true, comes to over three. He to the power offs trio were true in the Rangers here in full for those one grotty co two t Squire over to zero. Well, so that he comes to 4/3 through the power of history. That is a to minus zero. And for those one were how ah, just he Esquire he calls two of 16. See here we how circuit to over three minutes. Ah 16 that IHS 48/3. So on how 16/3 miners

Aishwarya Krishnakumar · Answer

So for problem 29. Where assets? All of this definite intergroup. And we&#x27;re going to use u substitution since we have a function inside for functions. So we&#x27;re gonna let you equal three plus eat to the two teams. And if you take the derivative of both sides, you would get two need to the to D Katie and we&#x27;re gonna go ahead and move this to to the other side sober. We have 1/2 do you and then the two goes away. And as you can see, we have a this We have it right up here in the original. You couldn&#x27;t. So that&#x27;s gonna help us when we write the actual integral. But before we do that, we have to change the bounds. Since these bounds are related to X and we have to do a u substitution. So we have to change these bounds. So that is related to you. So to do that, we just have to plug in both of the bounds. Oh, my bad. These air actually related to t not ex. There are no exes in this problem. A little zero because t and one he calls and we&#x27;re gonna change those teas into use. And to do that, you have to plug it into this equation. So we&#x27;re gonna say three plus e to the two times one. Since we&#x27;re plugging in the upper bound into this tea and then to find a lower band, we do three waas e to the two times zero. So if we do that, we get the lower bound to be four in the upper bound to be three plus knee squared, and then we can go ahead and do the actual u substitution so it would be won over. We plugged with you into the original equation and we will get you squared. And then for the green circles, that would be 1/2. Do you 1/2 to you so we can move the 1/2 to the front sense of the constant. We have the same in a girl. And to make things easier, you can also move the U to this second power to the numerator. So it would be you to the negative to power. And now all you have to do is take the undergrowth until you add one and there would be you to the minus one times a minus sign. And don&#x27;t forget this 1/2 and you would find this with respect to four and three plus U squared. So multiply the negatives, turn out and then it would be one over you plug in three plus c squared into you will be three close e squared And what I did was moved the move you to the denominator and then minus won over Plug in the for into the U And if you plug this into a calculator, you get zero point zero seven. I think eight.

Xiaomeng Zhang · Answer

we all know that the relative off each other power off to T plus three just equals two times. But we&#x27;re sorry to times I need to the power off to key. So we are uses up a Cugini in those question So sorry blocks e to the power of two key Squire and the the relative off beat the power of two people. Three Here we need to comes one tryst here because it&#x27;s too here and those are We need to change off our range status he to the power off to He loves three need the power off kooky plus three It&#x27;s a new indie grow You call to you to the problem of to a plough straight those wankers before run over Oh, over you, Squire You So those when you go to u to the power off miners Well, over minus one my nose. One christ The ghost approximately e counts. You pawned zero horned 0768

Mike Gaerlan · Answer

In order to find the integral of this, we need to find the integral of each individual piece here. So let&#x27;s start with the integral of this here bein a girl of t e to the t d T that&#x27;s going to require on integration by parts. I&#x27;m gonna let you equal t and d t equal e to the t d t. So I have the u equals d t. Ah, sorry. This should be Devi Some of this and then v is equal to e to the t. So according to integration by parts, I&#x27;m gonna have you ve which is t e to the T minus the integral of V, which is either the tee times do you is your just et so then are integral. Um, let me write out our t d t. This is equal to, And then the first part is going to be equal to t e to the T minus. And then we&#x27;re gonna have e to the t and then plus a C one and is in the eye hot direction. Okay, let me move that a little bit over. Was working anymore Space Now in order, find the integral of the same component here. We&#x27;re going to use a U substitution you. Is he going to pee? There&#x27;s you is going to be equal to T Square. So do you. It&#x27;s gonna be due to t d t so this integral now becomes when we do our substitution, it&#x27;s gonna be vulnerable, equal to the integral of 1/2 of sign of you, the U. So that is going to have the integral remember derivative or integral of Sinus negative co signs. We&#x27;re gonna have negative 1/2 co sign of you, and then you is just he squared. Listen, the J hat and then lastly here, the integral of negative to t divided by t squared minus four DT again, we&#x27;re going to take, um where we&#x27;re going to use U substitution here, where we&#x27;re gonna let u equal t squared minus four. The U is equal to two team DT. So when we do the substitution, we&#x27;re gonna have negative d you over square root of you. So remember, this is the same as you to the negative 1/2. So then we&#x27;re gonna add one. So becomes you to the 1/2 and then divided by 1/2 which means a time to on the front, which we really We have a minus two. So becomes minus two. You to the 1/2 and then remember, you is equal to this, so it&#x27;s gonna become minus to times square root of T squared, minus four. And then we have a posse. Three here. A swell. Okay, So really, we have a plus, uh, here in front, negative to this plus C three k hat, and then I&#x27;ll just go ahead and move this actually done. So the bottom here. So Okay. And then I forgot to add a C two here, so that should be a C two here. So this is going to be our indefinite integral here, you said Plus here. Yeah.

Problem

Integrate each of the given functions. $$\int \fr…

Problem 26 Easy Difficulty

Integrate each of the given functions.
$$\int_{0}^{2} \frac{e^{2 t} d t}{\sqrt{e^{2 t}+4}}$$

Answer

$\sqrt{e^{4}+4}-\sqrt{5}$

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Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus

Caleb E.

Kristen K.

Samuel H.

Michael J.

Precalculus Review - Intro

Functions on the Real Line - Overview

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$\sqrt{e^{4}+4}-\sqrt{5}$

Basic Technical Mathematics with Calculus

Caleb E.

Kristen K.

Samuel H.

Michael J.

Precalculus Review - Intro

Functions on the Real Line - Overview

Allyn J. Washington, Richard S. EvansBasic Technical Mathematics with Calculus

Caleb E.

Kristen K.

Samuel H.

Michael J.

Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus