evaluate-the-integrals-int-2-t-e-t2-d-t-3

Question

Evaluate the integrals.
$$
\int 2 t e^{-t^{2}} d t
$$

Bobby Barnes · Accepted Answer

we want to evaluate this indefinite Integral here. Now, they tell us in this chapter that if we integrate into the you should get you to you plus C. And maybe we can make it look like that for us to interpret. So this right here is what we&#x27;re going to take to be used to use it with a negative T square. Take the derivative of this We&#x27;d get to you is even too negative two t DT or differential. And if we we&#x27;re to multiply each side by negative negative too. But negative one, we&#x27;ll get to t d t is able to negative to you. And that is exactly what we have here. Which means we can Well, I&#x27;m gonna factor that negative out, and we&#x27;ll have do you and then e to the U Now integrate beat you is just eating you. So we have negative e to the U plus C and plugging in you. We&#x27;ll give negative e to the negative t squared plus C four hour indefinite integral

Bobby Barnes · Answer

we want to evaluate the indefinite integral to t e to the negative t squared e t. Now you might notice that we have this to t here and composed inside of e to the t we have negative t squared and the derivative of negative t squared is negative to t, which looks really close to just two t bite. So So you might have some inspiration Say, Well, let&#x27;s let you even negative t squared. Let&#x27;s find the differential So we get you is a negative to t Bt. And then if we multiply that negative over, we get negative. D&#x27;You is equal to to t e t And so now that to t d t is what we have on the right hand side here. And so we could first go ahead and re invite this as so First is gonna be e t u and then d t to t d t becomes negative. T you Let&#x27;s go ahead and move the negative out front. So we get negative interval of e to you, do you? And in this chapter, we learned that interval from e to you, Justin&#x27;s of being e u. And then since this is indefinitely at our constant integration. And the last thing we need to do is plug. Are you back in and doing that? Gives us negative e to be negative t squared, plus c. So this is what we get for evaluating this indefinite integral.

Kyler Gray · Answer

for a problem. 16. We&#x27;re gonna use integration by parts in the tabular method to solve this integral. So we&#x27;re gonna use t squared, as are you. So we&#x27;re gonna list out its derivative so t squared will become to t 20 and then we&#x27;re gonna use eat the 40 as our devi. So eat the 40 and we now need to differentiate or sorry, integrate that. So that will become 1/4 e to the 14 and then 1/16 each. The 40 and then one 64th you need to the 14. So now if we connect all these dots, we go back to connect effect. We can alternate the science plus minus plus. And now we can go ahead and write out our solution. So are integral. Will be 1/4 eats the 14 times t squared, minus 1/8 a to the 14. Sorry, there&#x27;s also a team there. And then plus one 32nd e to the 40. So we could go ahead and factor our 1/4 eat of the 40. So 1/4 e to the 40 and then that will give us only factor it out here. We&#x27;re just left with t squared factor out here and we&#x27;re left with minus one half t. Then factor here and we get +18 And then, of course, the plus c. So there we go. That is our solution.

Amit Srivastava · Answer

so suppose ah question. Like there the square of a three d DT. Now, the only way to solve this question is using by apart. So first part as a fabric johnson and other part is a public 32 So, as you know, uh, UV application of you be by parts. So let&#x27;s I equal to Yeah, we are using by Park Esperance a will be last part. So p square beauty at the first fence in dirty minus different station off this T square, it will be to t And in Dickerson, I&#x27;ll be 30 by three. Ready? So handle part will be here. He power treaty by three and we can again used by parts here T e 30 dirty 40 square feet. Public 30 by T minus two. Ready? So here it is. Teach the U and E treaties. Tapi 30 will go outside integration off it. Pretty pretty. Minus deposition of stage one. So you 30 by 3 30 Ah, now you were poorly utilised. Eat power three t by three minus two by three be it ready by three minus e three TV night. So this is the very dancer off this question

Patrick Vaughn · Answer

in this problem, we want to evaluate the integral of each of the four T plus two e to the to T minus eat of the tea over a piece of the to T plus one t. So we have a fraction, but it&#x27;s not in terms of polynomial owns, so we&#x27;re gonna have to manipulate it in some way to get that form we want. And so whenever we do that, we&#x27;re gonna doing a change of pace. So let&#x27;s go ahead and try you equal to eat of the tea because we look at the new Maria, you see, each term contains a factor. Eat the TS. Let&#x27;s go ahead and just rewrite that, Um, so we can say this is equal to put our sites will have me to the three t looks to eat the to T minus one times eat of the TVT over E to the to T plus one. So remember, if we do a new substitution, you equal to eat of the tea and do you with the either the t d t, which is really nice because we have that right here that&#x27;s do you. And then these would become like this would be you. Cute. This would be to you squared. So to get us into a forum where we can start using some of the things we learned in the section. So this becomes you Cubes Plus to use where it minus one over you squared plus one, Do you okay, so we can almost break this up into is your things. But first we&#x27;re gonna need to dio as we&#x27;re gonna need to use long division. Teoh, divide this denominator into the numerator. So let&#x27;s see, we have you squared plus zero you plus one divided into you Cubed plus two years squared zero u minus one. Something tells me I messed up here, so let&#x27;s see. Yeah, this term here, as you to the to t So this term should be Just eat of the tea, See where else that screwed us up over here. Just be you. This should just be you. And so we got to rewrite this. So we have plus zero you squared. Plus two u minus one. Okay, looks a little better. So first we&#x27;re gonna be multiplying by you, so that will be you. Cubed plus zero. You squared. Plus, you which the track that would get +00 U minus one. So this term here is already lower power than that. So we&#x27;re done. That&#x27;s gonna be our remainder. So are equal to U minus one. Okay, so we re write the fraction, um, as u plus U minus one over. You swear to plus one, do you? Okay, so in order, Teoh, evaluate this. We&#x27;re gonna need to split it up. So we&#x27;re gonna have you over a U squared plus one minus one over you squared plus one. And the reason we did that IHS we can use when we integrate, this will get Tanja inverse of you the winner. When we integrate, this will have to use the change of variable. But it works really well, because when we different to your bottom with respect to you will have to you do you. And since there&#x27;s a yuan talking, why would do you and the integral for care that&#x27;s gonna cancel really nicely? Okay, so that we have a you hear? Uh, you Okay, so let&#x27;s put this up and do things where we need to use a substitution and things where we don&#x27;t. So I have u minus one Overuse squared plus one. Do you, plus integral of you over you squared plus one. Do you? So let&#x27;s let be equal to you Squared plus one DVD will equal two years eu So we integrate. These will have 1/2 you squared minus tangent, inverse of you plus a constant. And then we&#x27;ll have this interval of 1/2 1 over be dp So that&#x27;s gonna be plus 1/2. Um, Helen, Absolute value V. We&#x27;ll put the constant over here. So have 1.5 years squared. Minus tangent. Inverse of year plus 1/2 Ellen of you policy. All right, So what&#x27;s Viv iwas You squared? Plus one. So we get 1/2 you square minus tangent. Inverse of you plus 1/2 Ellen. So you squared Post one is always positive. So we&#x27;ll just have Ellen of, uh, you squared plus one. Close the constants. Okay. And what was you you was equal to eat of the t. So last but not least, we have 1/2 each of the two t minus tangent. Inverse of either the tea plus 1/2 Ln of he to the to T plus one plus a constant. And that finishes our problem. You

Joshua Silva · Answer

All right. So we have another integration by parks problem here. And the end of grand that we have is t squared eat of fourty d t now just to give us a quick reminder of the formula for integration by parts integration. But parts tells us that when we have an integral of the form you TV, we can rewrite that equivalently as being you times V minus the integral of the do you. And also just to give us a bit of a visual intuition for why integration by parts works the way that it does, we have here a little graphic that can help you to see how the different pieces fit together to give us this formula here, in this case, for the definite in a girl version of integration by parts. But it also works for the indefinite and girl version as well. No, onto are integral. We have t squared eight of the 14 duty and we want to find some part of this expression here to be Are you and some other part to be our devi? Now we want to pick for Are you something that differentiates down to something simpler so picking each of the 40 for instance, wouldn&#x27;t do anything because we would just get back another exponential function that wouldn&#x27;t get us anywhere. So we want to pick Are you to be the T Square bit? And so we write down, let you equal t square, then do you equals to t DT? Then our Devi will wind up being eat the 14th duty and then V by immigration Well, equally to the four t over four. All right, so that you have this all the pieces we need to get this integration by parts of problem rolling. Now we need you times V So Ust Square and V is either 40 over four. So multiply it all together we get t squared over four times eat of the four T minus. The integral of E d U and V again is each of the 40 over four and d&#x27;you is to t d t so multiplying all that together. We&#x27;re going to get tea over to times each of the 40 d t and this is gonna wind up being another integration by parts problem. So we have to pick a U and we have to pick a V in this case, are you is gonna be tea over to let you equal t over to then do you equals one over to D. T. Devi In this case is again eat of the fort E d t. And then V is going to be the integral of this, which is again each of the 14 over four. All right, so bringing this all back down, we have you times V where USC over two and V is either the four key over four multiplying this all together. We&#x27;re gonna get tea over eight times E to the 40 minus the integral of video where V is either the 40 before and d&#x27;you is 1/2 a d t. Putting that together, we&#x27;re going to get e to the 14 over eight DT And this case. Fortunately, we don&#x27;t need integration by parts anymore. This is just a U substitution. So that u substitution is gonna be e to the 40 over four times eight, which is 32. So eat the 40 over 32. Finally, we&#x27;re gonna add a constant here, but also this is a plus because the minus is gonna distribute over that plus there when we, uh, just affected that a lot. And then we&#x27;re going to write down all these other terms as well. T over eight, Heat of the 40 and T squared over four. Eat before t Now, technically, we could be done at this point, but we can simplify things a bit further by factoring out this one forthe first of all and then also factoring out all these t terms. So that&#x27;s 1/4 time&#x27;s a T squared minus t over to plus one over eight because we&#x27;re pulling the four out of the 32 and then that all gets multiplied by each of the 14. And we had a constant and now we have our solution.

Problem

Evaluate the integrals. $$\int \frac{\ln x d x}{x…

Problem 17 Easy Difficulty

Evaluate the integrals.
$$
\int 2 t e^{-t^{2}} d t
$$

Answer

$ - e^{-t^2} +C $

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Joel Hass, Christopher Heil, Przemyslaw Bogacki

University Calculus: Early Transcendentals

Grace H.

Kristen K.

Samuel H.

Michael J.

Integration Review - Intro

Definite vs Indefinite

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$ - e^{-t^2} +C $

University Calculus: Early Transcendentals

Grace H.

Kristen K.

Samuel H.

Michael J.

Integration Review - Intro

Definite vs Indefinite

Joel Hass, Christopher Heil, Przemyslaw BogackiUniversity Calculus: Early Transcendentals

Grace H.

Kristen K.

Samuel H.

Michael J.

Joel Hass, Christopher Heil, Przemyslaw Bogacki

University Calculus: Early Transcendentals