integrate-each-of-the-functions-int-fracleft13-e-2-xright4-d-x6-e2-x-3

Question

Integrate each of the functions.
$$\int \frac{\left(1+3 e^{-2 x}\right)^{4} d x}{6 e^{2 x}}$$

Tyler Moulton · Accepted Answer

Yeah. We want to evaluate the following the integral from one plus three E. The negative to actually four D. X. Divided by six. Each of the two X. This question is challenging our understanding of methods of integration. In particular. It&#x27;s challenging our understanding of our recently learned method called the power Rule for integration. The powerful states that the integral of U. To the N. D. U. Is U. To the endless one over endless ones. Let&#x27;s see. So if we can identify you, do you And and in this problem we can solve since one plus three equals negative two X. To the fourth is in parentheses raising power four and it has derivative negative 62 X. Or rather than negative two X. We must have you as one plus three. Either negative two X. Do you as negative six negative two X. The X and n equals four. In this integral we have an extra factor of negative 36. So we must have that are integral is negative 1/36 times 1/5 plus one times one plus three negative two X plus C. Or negative one over 100 and 81 plus three E. To the negative two extra 50 plus C.

Kelly Maluccio · Answer

So today we will be solving the integral from 2 to 6 of either the two x plus three y d X. Now the first thing we do when we look at this is we we can figure out howto integrate e to the X. Something simpler. So we want to use what&#x27;s called u substitution. And to do that, we have to pick what&#x27;s Are you gonna be So are you in this case is going to equal to X plus three. Why? And that means that our d&#x27;you has to equal, um, two times DX because three wise like a constant and so the derivative of that in terms of X will be zero. So now I have my you and might do so I can rewrite my integral here, and I&#x27;m going to write just a definite interval and then we&#x27;ll bring back the two in the six. So this will be the mineral of E to the u times D&#x27;You And don&#x27;t forget about this two over here, we have to make ah, 1/2 and I&#x27;m gonna bring this in the front. So since it&#x27;s a constant, we can take it out, and it could be 1/2 times this interval of E to the U D U. So now we know how to integrate this because the inter role of either the U is just eat of the U. Um, if you don&#x27;t remember this, it&#x27;s also the interval of or it&#x27;s e to the you and then over natural log of your the base. But the natural log of ease Just one. And so that&#x27;s why he is a special function where when we integrate this, we just get 1/2 e to the U. Normally, if we were just solving the definite and a role we would put plus C at the end. But we don&#x27;t need that here because we&#x27;re gonna end up plugging in this, too. In this six sow, the seed would cancel. So I&#x27;m gonna write 1/2 e to the you, which I&#x27;m gonna replace my you back with two X plus three. Why? So say two X plus three. Why? And I&#x27;m gonna solve this from 2 to 6 or integrate this rate from from 2 to 6. So first, what we&#x27;re gonna do is plug in the six for X. Okay, so we&#x27;re gonna get 1/2 times E to the to 10 6 which is 12 plus three. Why and then minus. And now I plug in the two. So I have 1/2 again times E to the two times two, which is four plus three wives. And now we can write this a little bit nicer just by writing the 1/2 once. And you could have done this to start. But now we&#x27;ll have 1/2 and then we&#x27;ll do e to the 12 plus three. Why minus e to the four plus 31 And this will be our answer for our indefinite integral above right here.

Matt Just · Answer

I don&#x27;t know. We have the definite integral zero two four of minus X to the three halves plus x to the one half d x. I could start by finding an anti derivative. So we have so this minus Khun stay are here. And then the entire derivative is so we add one to the exponents We get X to the five hands over five hands plus x to the three halves over three. House evaluated from zero to four. Okay. And so four to the five halves We&#x27;LL take the square root and then take the fifth power. So it&#x27;s two to the fifth, which he be thirty two over five halves plus okay. Squared two cubes of eight over three halves and then minus this evaluated your over that Syria. Okay, so, uh, this is minus can multiply by their sim prickles of both the numerator is get doubles at sixty for over five plus sixteen over three and so we need to get a common denominator. I mean, we could leave it like this, but if we do get the common denominator of fifteen, you should get negative. Six fifty six. Divided by fifteen is the final answer

Xiaomeng Zhang · Answer

So those one dressy coast through my nose three times acts to power off, Serie over to and miners into grow off access to the power off house. Okay, so for our 1st 1 that a cow shoe True over five times, actually. Power are 5/2 40 and my noticed us to over three times acts to the power off. 3/2, full zero. So for the 1st 1 that they go to six. So our five I&#x27;m too through the power off five that is 32. And for those one, we have to over three times, two to the power off street so we can see that he has to six times or be too. And that&#x27;s why he goes to 16/3. So we&#x27;re have the answer. You kill two minor 65 6/15

Tyler Moulton · Answer

Yeah. We want to evaluate the given definite integral integral negative 1 to 1 even to experts through I. D. Y. This question is challenging understanding of integration in particular. At first glance, it seems like it&#x27;s testing our understanding of multi variable calculus really. What is testing us though is non X integral. So this is a single very eligible for which the differential dy is not with respect to X. So what we&#x27;re going to do to solve this problem is integrate with respect to Y. Where we treat the term X simply other constant. That&#x27;s the solution will be a functional form X. So for this it&#x27;s real. We can use the integral to be ubuntu. It was either you evaluation it&#x27;d be this is using the fundamental theorem. The country that&#x27;s combined with exponential integration for you is to expose three Y. Therefore D u three Y. Thus we have integral negative 1 to 1 of the two experts. Dy dy one third. Either to expose three Y evaluate the negative 1 to 1, which gives us the to expose three. You know, the two explains 3/3

Amy Jiang · Answer

we have a phone in your girl on This is what Exes? Dustin. So look, me and our inner low for you was this Looks like, um, go number three. And in this case, since we have less than a one of six that&#x27;s gonna be in terms of or a solution will be in terms of hyperbolic tangent. Inverse. Let&#x27;s first note that we have this each of the X selling me to to Carol. That is, uh why don&#x27;t we use use up? Why don&#x27;t we take you? We&#x27;re not used. Why don&#x27;t we just take you to the equal Teoh each of that and then we know that. Do you? Is you go to eat it X yet. Okay, so let&#x27;s replace that we get the in a row. Uh, if your ex the ex, this is equal to do you and then we have over 36 minus use. What? And then, in this case we have, are you Value is less than so. You know, X is equal to other six. We have used to go to E l one of six. That gives us six. Okay, so now using equation dream, we get this is equal to 36 is 60. Politician so yet one over R six hyperbolic tensions and worst of X over six and then plus C. So I had been you and we should replace. Are you with our ex who said that X or you is equal to eat it x So this is equal to 1/6 of about tensions in verse of each other X over six, both feet.

Problem

Integrate each of the functions. $$\int\left(1+\s…

Problem 24 Easy Difficulty

Integrate each of the functions.
$$\int \frac{\left(1+3 e^{-2 x}\right)^{4} d x}{6 e^{2 x}}$$

Answer

$-\frac{1}{180}\left(1+3 e^{-2 x}\right)^{5}+C$

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Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus

Grace H.

Catherine R.

Kayleah T.

Michael J.

Precalculus Review - Intro

Functions on the Real Line - Overview

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$-\frac{1}{180}\left(1+3 e^{-2 x}\right)^{5}+C$

Basic Technical Mathematics with Calculus

Grace H.

Catherine R.

Kayleah T.

Michael J.

Precalculus Review - Intro

Functions on the Real Line - Overview

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

Grace H.

Catherine R.

Kayleah T.

Michael J.

Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus