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1 1 1 15-28 1 1 In Exercises 33 33 L Integral '(shuoripuoj 1 1i [ 1 1 1 3 1 Indutinita 1 Antiderivative :7 1 2j 1 that satisfies the inltial 1 1 H [ Finding Fi...

Question

1 1 1 15-28 1 1 In Exercises 33 33 L Integral '(shuoripuoj 1 1i [ 1 1 1 3 1 Indutinita 1 Antiderivative :7 1 2j 1 that satisfies the inltial 1 1 H [ Finding Finding r W) 1 Funcrion 5 sketttd L Vector-Valued 1 3 1 Vector-Valued Aeetate L Functinn 5 8 In Exerciss 17 and [8, find the limit: 1 1 31 Graprprey Graph 1 8012 1 1 1 Qjuds JuI Joj dorzunJutd ~iJ+ 1 1 1 { E Limit M 3 23 Representing Representing 4 Finding & 15 'z"s) #

1 1 1 15-28 1 1 In Exercises 33 33 L Integral '(shuoripuoj 1 1i [ 1 1 1 3 1 Indutinita 1 Antiderivative :7 1 2j 1 that satisfies the inltial 1 1 H [ Finding Finding r W) 1 Funcrion 5 sketttd L Vector-Valued 1 3 1 Vector-Valued Aeetate L Functinn 5 8 In Exerciss 17 and [8, find the limit: 1 1 31 Graprprey Graph 1 8012 1 1 1 Qjuds JuI Joj dorzunJutd ~iJ+ 1 1 1 { E Limit M 3 23 Representing Representing 4 Finding & 15 'z"s) #



Answers

Evaluate the integrals in Exercises $7-20$
$$
\int_{1}^{e} \int_{1}^{e^{2}} \int_{1}^{e^{3}} \frac{1}{x y z} d x d y d z
$$

Hi there. In this problem, we were asked to evaluate this triple integral. So as always, let's start from the inside Out. And let's just focus on this inner integral with respect to X. So as always. Just so we don't forget where we are. Well copy these outer to integral. But we're really just focusing on the inner integral in this step. So we need the anti derivative of why times sine Z with respect to X. Luckily for us. Uh, since we're with respect to X, Y and Z are both those constants. So you really can pull those out. In other words. Think of this whole thing as why? Uh, why Sandy Times 1? Um, so the anti derivative with respect to X is just X. Right? And then times this constant and we pulled out. Okay. Um, the limits of integration then should go right here. As always. This is just just like calderon. And we'll evaluate by plugging in Hi in zero for X. And that's attracting. Okay, so the Y signs the those will stay there since they're outside the brackets. And if we plug in pie for X. And subtract, I'll write it out. We plug in pi for X. And then subtract the results of plugging in zero for X. Well, clearly this will just end up as pie right here. I'll give you a couple of seconds to look at that. But I'm pretty sure at this point in Calif three you can look at this and realize, well, everything here just becomes pie. In other words, we just get why sci fi times pi and we can pull that pie out to the very outside if we want. So why don't we do that at this step? Just so it looks a little easier. So, there's times pi here since pie is a constant, we can just pull that to the outside. All right. So we went from three into girls down to two. That is going in the right direction. Now again, we'll start from the inside. And now this is with respect to the variable. Why? So, we are going to need the anti derivative of all this with respect to Y. Well, again, sign Z can be pulled out since the Z is a constant. We're only uh finding anti derivative with respect to why and the anti derivative of the Y. Part. We know what that is. That's why I squared over two. And again, the pie in the zero as the limits of integration will put those there. And let's see what we get. So, we get signs E. Okay? So, if we plug in pi for why we get pi squared over two, then if we plug in At zero for why? Zero squared over two, which of course becomes zero. All right, So let's see where we are copying the outside. You get signs E. In this hole Heart here is just pi squared over two. So again those are all constants. Let's pull those out to the outside. Just makes your life easier. So pi times pi squared is pi the third all over two. So that's our constant on the very outside. And the good news is we are down to one variable just dizzy. So this should feel exactly like a calico an integral because it is Uh huh. Anti derivative of sine Z. Is minus. Cosine Z. Yeah. And we're plugging in one and 0. So let's see if we get minus co sign of one minus minus will make that a big plus Co sign of zero. So we're basically done. Let's just simplify what we can here, Coastline of one. We really can't simplify but coastline of zero we can co sign of zero is just the number one. So this certainly could be a final answer or if you prefer to change the order. Either way we get part of the 3rd over two times 1 minus the co sign of one. And we are done. And hopefully this was helpful.

Here were to find the integral from one. The negative one of the quantity are plus one quantity squared. We are now. There's lots of ways you could do this problem. One is to just simply expand our post one quantity squared. U u r squared plus two are plus one g r, which is going to give you 1/3 are cubed plus R squared plus are evaluated from one to negative one, which is gonna get you a negative 1/3 plus a one minus one minus quantity. 1/3 plus one plus one look, well, thes cancel and you're left with essentially negative 1/3 minus 1/3 minus one minus one, which is going to get you negative 8/3 for your answer. Another way you could do this problem is if at the beginning, you let visited and color that you equal are plus one. So then therefore do you equals d armed this emptiness institute except the bounce. Okay, so this has become a U squared. Do you question however you need to plug in your bound so one plus one would be to and negative one plus one would be zero. This gets us 1/3 you cubed evaluated from 20 And if you plug in a zero first, you get zero minus 1/3 time's two cubed you negative 8/3 which matches our bottom answer there.

Is what we have you then together, like this 11 Well, plus, like Texas, where you go with the others. Yeah, they were beautiful. They were doing this wishing, using option. Frustrated the greats. Alexis, you therefore fixes the suit. Why be postal? One first will go on my six. What if you didn't work with this social disconnection? You might like this one or this fire. The expression is X squared. Why? When this one was five will be closed. So that designation. So this is a question off. So good. Explain the normal Su minus one. So the graph of this week's up near this? No, the same music video. Also, this is the same time separates of this movie. Well, this one this moving what so committed This. So this farm on the substance to be who anywhere you can do anything to do park So there's just one. They did that You know that Then you will be the basis stools and I results going through one. The SPD off this seven circles radio on December 7 by do they suddenly people. So my right toe When the spy who since you did you mean this part Planes funds of Orvis fishing they less and that

Here we have a choice of X is equal to the integral from 12 square root X Z square, divided by Z to the fourth plus one DZ. Now let's go ahead and calculate the derivative using the fundamental theorem of calculus. So we're taking the derivative with respect to X. That means that we're taking the immigrant and we're evaluating it that square root X and then what we're going to do is calculate the derivative of square root X So what we get as X divided by X squared, plus one times one half x to the negative one half and let me just go ahead and rewrite this so it's slightly simpler and that's it.


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