Integration by Parts, Complex Exponentials and Volumes

I think I hope you're still going to, or you still you had your beautiful dinner, and Elaine, let's actually begin with your question. Elaine has some issues with certain things we did when we did integral in the previous session, and Elaine, spell it out for us, what was troubling you. I understood that the antiderivatives would have a sign and.Alright, the quickest way to do it. I'm going to actually do that again. If you generically, you begin with the cosine of the ax now. I'm going entirely generically, and e of the bx dx now. We're trying to find the antiderivative. And first of all, is there anybody who does own the process and would like to share it?If you can, I don't have to. I would be over glee. I can see that Ivy did it in your homework, although maybe you did it in the conventional way of dealing with the sine cosine using integration by part instead of. At least I didn't see a comprehensive solution using what I promoted as the eyeball method because it's eyeball. Sometimes I can't directly see in the in your calculational steps here. So Ivy, do you know?How to do it, sort of just at one fell swoop directly write out the answer. Taking the complex number approach. Hmm. Um. We could, write. We could also write the cosine ax, as like. Um.We could like add the sign i sine x, so it would we would write it as e um b x times e i a x. Yeah, so this could be actually generalized into that. It's not, it's the real part thereof. Very good. So I would say.We're visualizing it as really just an expanded coefficient. Instead of just have a b, we have the b plus a i. X. This is anti-different. We're finding. I might representing your idea. Faithful. Yeah. Okay. Beautiful. And then when we take the.Antiderivative of this, we do e to the b plus ai times x, and then we divide by the b plus ai. Correct. That's all you do. So in fact, later you could really just compress all the steps into. This is antiderivative now. There's only one more step. By the way, doing this, you're getting the antiderivative of two things. You haven't not just in integrate sorry integrate.Integrated this one. You have also integrated. It's country, or rather, it's missing part here. You also get the solution for that. We're getting basically the two birds shot at one stone, so the answer is bundled up in that. They are correspondingly the real and imaginary part. Okay, beautiful. So how do I translate it back into a real function? Yeah, very good. We're going to multiply. Ah, yes. Divide by the conjugate on the denominator, so that we're going to actually produce this shared coversion. No matter what, that part of the coversion is shared, and on the numerator we're multiplying by the b minus a i. That's exactly right. So we're multiplying by the same quantity on both the top and the bottom. And so, what would be the antiderivative of the cosine ax b to the bx e to the bxAt this moment, here Elaine, can you take over and tell me what it is. Is it b e to the b plus a i x minus a i times e to the b plus a i x. You're just multiplying the bound to get the real component. Right. So you.You would actually use the real of that one multiplied by the real, which leads to the b under the cosine of the bx. Oh, sorry, cosine of the ax. Sorry. And all of these will be multiplied onto. In fact, why don't I pull out this first? Eventually, every single term has that component, and inside will actually be multiplied by the real, which is cosine of the ax. And then plus the imaginary.But a bad imaginary, because this is minus AI. So what are we getting? For the other source of the real component. Um, the a squared.Um, sign ax. So we're actually using the negative AI multiplied by the sign of the ax now. So that gives you plus. But you, you didn't say plus or minus, or did you? Um, I just no, I just said a square. Ah. Wait, which part is the sign?We're looking for the real component. Oh, oh. Forget the base. Yeah, it's a a sign ax. So this is anti derivative. Now we're done. Elaine, did you quite follow this time? Yeah. Okay. Oh, then Elaine, using the same principle, can you tell me directly what would be the the other anti derivative, which is this.That one equal to that's not. What is that equal to? Is it the same thing because it's a conjugate? You can't just say it's the same thing. Yeah, they're part of the same solution, but you have to spell it out for me.Certainly not equal to each other, right? Is it like a negative version of the antiderivative for cos ax e to the bx? What? Say it clearly. What do we have? Your answer should be different because this is actually starting from the different. Integer. So be careful. Just say it clearly. What is your entire derivative, please.Can I speak it?```中文嗯。``````Elena, share with us your mental process. We're waiting for you. Oh, I'm doing the same thing, but we are we are exactly doing the same thing. Can you just spell out what terms you're dealing with? Ah, right now I have sine x e to the v x times.B plus A I over V squared plus A squared. So we get this. Okay, and what's in the parentheses? ```呃，b sine ax minus a。Cosine ax, b of the sine ax. Very good, and then minus or plus. I got minus. Excellent, done. I'm judging from your conclusion whether you fully understood it or.You simply expanded out that's cosine ax, right? You simply expanded out the multiplication multiplication of the two terms. Did you not? Yeah, that's it. That's it. Very good, Eddie. Are you getting the same? Is that the same as what you are trying to say? Yes. Brilliant, then. I think you guys are formidable. Well, then why don't I actually just do a quick check on this one? If I actually give you seven sine four x now.On the e of the negative two x, Elaine, can you give me the answer? What is antiderivative of this guy? E to the negative two x over sixteen plus four times, which is the twenty.Okay, on the denominator. What happened to the coefficient seven? The seven was, um. Ah, we can put the seven like it's seven e to the negative two x. Sorry, I failed to understand. You don't just copy the coefficient. Yeah, that's what I meant. Okay.Good, good. Thank you. Then yeah, you did it correctly. And then what else inside? Uh, negative two sine, four x minus four cosine four x. Addy, are you getting?I am not. I don't think there's a negative. Wait, uh. The four is imaginary part. This four here. If you go for the sign, that's the imaginary part. But here we're looking at the imaginary, so we want to combine imaginary with real. So this part is perfectly done. I agree, and there is a negative here, no doubt about it. Because the real part has a negative too. What about the imaginary part? That must be, I mean, the other component now, which is using the real, which is going to be actually the cosine for x, to be combined with the imaginary. Was that negative four or positive four?I'm getting a negative four. Yes, because it's counter. Very well done, you know. Any issues anymore? I will watch critically about what's going on. Yeah, I think I got it. Excellent, I believe. Well, I believe you got it actually, even from your homework. So this is integration by part, but apply the nimble way because sometimes you don't necessarily have to go for the formalism of integration by part. You can just counter.Up, guess and check, and and and find it backward here. Um, two more examples here, and then we're gonna look ah move on to more application within the word problems. Meaning we're not only doing the mathematical calculation, we're actually see what's the meaning thereof. Okay, so if I actually give you this natural log of the x squared, um, x cubed, and I want to find the antiderivative of this guy. What do we do?There are several ways to go about it that you could guess and conjure. A, if you're savvy, very often I don't bother doing the integration part. I just, I just basically go for write out one term and cancel the unwanted one, etc. , etc. But you can take it anywhere you want.Let me do integration my part. Alright, let me see how you do it.嗯，嗯，我们 choose the X, um the X Q does the function, like easy to. Yeah, and I do review. Good. Ah, this is important comment Ivy is making. She's identifying the natural log part as hard to integrate. It is, but.And it's decently easy to differentiate. I assume that it's easy to do either. You can find antiderivative or derivative. Then polynomials are always going to be easy. So, and I is going to actually integrate that x cubed first and leave the natural log of x on the outside to do a swap. Go ahead, walk me through it. X to the fourth over four.Orion times a natural log x squared. Okay, so we need to take away the part we accidentally produced, which is. Entire derivative. Of um. The. The. Derivative of natural log x squared is um a, is it one over x squared? One over x squared. Ah, you are taking the derivative of natural log like.Right.I got two natural log x over x. Yeah. Over x, yeah, but when you combine the over of x cancel with that one half of the x to the fourth. And then what you're getting is we're subtracting by the antiderivative of this guy now, aren't we?好，We. Um, yes. So, now we're going to actually repeat the process again. And Idee, eh, you demonstrated this process very well. And then Idee, do you want to pick up from here. We posit the first term in antiderivative.Now we need to make up for this. In addition to what, I mean, this additional integrant that we inadvertently introduced, we need to remove it. So we need to integrate that. When you launch such a chain, you're always trying to confirm. Oh, no matter what, we're getting better. Meaning this integrant is getting easier. That means we're moving the right direction. Okay, if next time when you end up with actually natural log of x cubed, that works, we're moving the wrong direction. You need to actually go the other way.So, go 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four over four times two log n at log. This is actually four over what? Over four. Is that the antiderivative of this term? No. Right, a slight modification. Over, over eight, isn't it? Over eight, yes, good. And you just copy that. Correct, good, and now what else do we do? When I do this, and I again use the product rule, I definitely would get this term here to cancel out that term we produced earlier, but we also introduce another term, which is.So now we need to cancel that out, but because this is already subtracted, so that's the integral we further produce. We need to cancel it, so we need to add back in that term here, which is going to be actually this now. Correct. Be sure to double check very carefully and give me a.Confirmation or disconfirmation. Hmm, I got the same thing. Beautiful, and you're seeing. In fact, we overall we have an actual force everywhere, and then you're looking at some kind of combination of a different powers of natural logs. This will work even if I give you a very high power of natural log of x.我们看，不吧？ yeah, beautiful. Okay, so that's the integration by part. Uh, what's the graphical meaning of the integration by part? Meaning, if you're multiplying the two functions together, this is what I said earlier. If you look at the differential of the FG, now you can represent this as the f, the other as the g. Now they don't have to be equal to each other.A differential of the FG are these two parts. This is the FDG and the DGF. Ah, that's sorry. This is FDG, and let's say the upper is the G now, and the height is the F, and this is the GDF. When you add them together, you get the total change of the differential of the FG. FG would be the antiderivative, but if you have only one slice, then you not only have the FG, you need to also subtract.By the antiderivative of the GDF. Hopefully that piece is decently easy to integrate. This is our goal. Are we happy? Clear, huh? About the picture. Can you like repeat what you said about the shaded part? I'm actually saying this.If you're integrating the FDG now, and we know from the shaded from the shades here, the sum of the two slices here give you the total differential of the DFG. So when you actually do the DFG now, what you're getting is FDG, and then plus the GDF. So that this is going to be the total differential of a DFG minus the F, sorry, minus the GDF. Meaning it's the entire.Increment of the rectangle minus this horizontal slice that gives you the vertical slice, which is what you begin with in the integral. Makes sense. Yeah. Okay, for this one, for this one, the first part of the integral is done because the integral of the differential you just end up the fg, and then again we end up integrating its complementary part here, and hopefully its form will be easier. Somehow you'll be able to do it. So that's the whole mindset. I will clear. If we are now, we're gonna actually look at more applications. Yeah. Okay, let's actually look at the following. This is strictly application under this one major part on the calculus AB. So we're gonna actually look at the.Volume of a 3D region, but before I give you what it is now, I'll tell you how to build it. You build it from actually ground up, meaning I begin with actually a two-dimensional figure. For example, I'll give you, let's say, a parabola. This parabola is simply going to be the y equal to the x squared. No need to go fancy at the moment. At the same time, I'm also giving you a sinusoidal function. Let's say this is one to one, and meanwhile you have this.This is not very well drawn. Something like this. The other part of the function is y equal to sine x, and then they're gonna actually intersect at this small region. Of course, you need to find out what would be the intersection, but these are calculational details, and very often you would resort to the calculator because this is a transcendental function. You can't solve it by hand. But we just know it happens before the function reaches one one. You look at this little leaf.And what do we do here? We're gonna actually lay it out on the ground like this now, and that's my parabola. You have to do three three D visualization, okay? And that's your sinusoidal function. And for easy construction here, I'm gonna actually expand it out. This is a region we're looking at, and that's my x direction, and that's my y direction.And I'm getting ready to construct some architecture from it. Do you follow me so far? Yeah. Alright, now we're gonna actually look at every single crop section. It's perpendicular to the y-axis, meaning if I'm gonna actually cut it using a segment perpendicular to the y-axis, right? And I'm just going to use that one as the base of an equilateral triangle.So I'm gonna build an equilateral triangle on top of that. This is a erecting in the three D in the vertical direction. Maybe I'll do it that like this. That's an equilateral triangle. And now I'll choose a different way to cut it, and then that's another equilateral triangle. These equilateral triangles are not of the same size because it really depends on what is the base segment. By the time you're getting to very close to the boundary, boundary, they're gonna get very small. So you get this pile of the equilateral triangle.You write it in space, and when you connect them together, what you're getting would be the dome-like shape. So that you can see on the side here, that's what you're getting. On the other side, it's almost like a kind of a three D, a nugget of an ice cream. Meaning on the bottom you have this now, and on the top you have.Yes. Could you visualize what it is? Yeah. Okay, well then you're great. I don't think I did a very good drawing, but if you cut it using a slice here, do a cross section perpendicular to the y, what you're getting is always going to be an equilateral triangle. You don't really get a smooth shape, okay? It's not a smooth body of this body, but instead it just a skinny equilateral. It's straight on all sides, and I'm asking you to find the volume of this weird nugget. How do we do it? For the time being, I don't want to. I don't want you to spend time finding that intersection again. This is a calculator job. So for simplicity, I'll just call that the coordinate a and b. Coordinates a and b now. And I will actually give you a minute or two to really think about.How is this related to an integral at all? Calculus is what divide and conquer. You differentiate, you're chopping things up into infinitesimal chunks, and now integration is to assemble them back together.`````````````````` 嗯。We like think of it as finding the area under the top curve. The top curve. What do you mean by the area of the curve? I I'm trying to be, harisplitting in terms of the terminology, and for for the sake because you.To actually write a perfect medical exposé on AP. And I think you're on the right track, but try to try to say it more accurately. I'm talking about like the curve that connects the tops of the equilateral equilateral triangles. Like.You're talking about this curve here. Yeah, right. Ah, there will be actually there's a trajectory of the location of the top vertices. They definitely give vital information about the height of this dome that we're calculating the volume thereof. And what do you mean by the area? Do you mean the length or do you mean the height, or do you mean something else?嗯，嗯嗯。嗯。I was thinking. Like, maybe both the length and the height.Aside being the part of the equilateral triangle, would always be maintaining certain ratio with the base segment length. Before we tackle the entire problems here, or plunge into the calculation of detail, we want to have a roadmap. Meaning, what are the necessary parts of the information we need to finish the task? Right. Well, I would say if you want to look at the entire volume.Since we're in the mindset of assembling, integrating, synthesis, not break down on the conquer, but rather put them back together. So I actually want to write, I want to find the total volume, which is the integral of individual pieces of the volumes, a infinitesimal chunk of the volume, and that chunk of the volume is made up of this triangular slice. And let me actually expand it out with a certain thickness. So if I draw it on the side now, what do we mean by a.It's a chunk of the volume here. Hey, come on. It's made up of between the two curves. And I'm given this is actually with infinitesimal difference or thickness in the y. That's actually increment of the dy here, which is gonna be actually this, which we call the thickness dy. Which actually identify this little slice of the area between the two curves. Now, upon which we're constructing this. Equilateral, triangle that form a slice of this prism. Correct, and we want to write what is going to be that d volume. And later, when we assemble all of these together, then we'll be able to know what is the total volume.So far, so good. Yeah. Beautiful. Well then, how do we write our TV specifically? Well, since we're looking at a.Which is equal lateral triangle with a little thickness. Then we do base times the height, right? We do the area base area of the equal lateral triangle, which is simply going to be if I call the base length the a, then basically we have the a and that's half of the a and that's root three over the two a. And if you do half of the base times the height, then the area really just equal to the root three over four a squared. This is the base area, and you multiply by the height, which.Which is a dy, so this by definition is our dv, isn't it? Hmm. Yeah. Alright, well then we almost have a have the question reduced to finding the antiderivative, although this is in the future.You're going to be a definite integral because you're given the lower bound and upper bound. So we know we're integrating from zero to whatever this y max, which is our b now. Oh, oh, oh, I'm using this as coordinate that's a a, so that's a bad move. I need, I was, yeah, inadvertently using the double notation here. That should be called s. So this is going to be the s squared. There's only one thing at the moment: s is not explicitly written.In terms of the functional y, yet to find the antiderivative, we need to have a function in terms of the dummy variable y, but it can it can be written as a functional y, can it not? We just need to do so. Now back to this graph here, we're really just saying, and I'm resorting to this graph because it's the clearest. So, if we're actually looking at it with a small increment of the y here, I'm cutting it right here.And this is my dy. I don't want to find out how long is this little s. This is segment s. Now we're using to build the equilateral triangle, isn't it? Yeah. How may we do so.Eddie and Elaine, are you on the same page? Do you know what we're trying to do at the moment? Uh, yeah. And Eddie, you. Yeah. And could you restate it? What are we trying to find? Oh, we're trying to find S in terms of Y, so we can integrate it. Yes, we're trying to actually find the S now by this. Specification on the graph. It is interesting to the two curve at the height of the y. Now you'll be able to find out how long that segment between the two curves, and then that will be appear as a base length, so that it would actually replace the integral as an integral. And then after that, you're looking at the classical calculus problem, just finding the antiderivative. Well, let's find out how the s is related to the y.中文嗯。嗯。中文嗯。Eddie, you want to share your process.Are you lost? Update us where you were.好的。Feel free to say it. Hey, please do talk, and otherwise we're about to wrap up the session. Andy, can you narrow down to the part that you don't know how to do.You know, you're just doggedly keeping your silence. It's verging on not very decent because the whole classroom is waiting for you. I can't move on. I don't even know what's going on.Can you talk? Well, I'm afraid I have to call the day. I have other students here.And I need to call him. What so? Can we take its homework? Try to finish it. Find out the the x dependence on the y. It won't be too probably easy. Okay, after you even write the integrand, you might have trouble doing it. And we might actually need integration by parts. But I wish you the best luck. Next session, we're gonna actually start right from here. And Eddie, no matter how you're, maybe you lost the track of what's going on. Listen to the recording and recover it, would you.Okay, alright. Take care. Enjoy your Thanksgiving evening. Bye, especially Ivy. Sorry for getting into your dinner.

Integration by Parts, Complex Exponentials and Volumes

Integration by Parts, Complex Exponentials and Volumes

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