中值定理、积分技巧与体积

Conceptual point. In mediu theorem, finally we're talking about the averages of functions. But I begin with my usual question: any question from previous sessions or homework? We were doing integrals of a volume of a rotational solid. Let me quickly recap homework. It actually did stimulate quite some discussion.Got it. Hold on, where's my pen? Thank you. Ah, shoot. One second. Where's my pen? Hmm. Sorry, forgot. Alright, let me quickly recap where we are. And you guys didn't answer my question. Do I need to go?Go through homework problem or not. Yeah, I think so. Good. Let's go through it. I give you a region to begin with, which is a square. This is y equal to x squared. Well, the other one is y equal to sine x. And we're creating a rotational solid by rotating around the x-axis.The way, no, no, no, no, we're not creating rotating solid. Okay, different chapters. We're only creating a solid by using this as the bottom crowd section. So fundamentally, we're actually looking at one crowd section at a time, perpendicular to the x direction. No, no, I give you this. Did I not? We're creating a shape using that one, and as a basically equilateral triangle. That's how we're creating it. Ethernet. Can I get a confirm?Information, please.Into infinitesimal volume, and this infinitesimal volume element, as we could easily identify from the graph here, it's a little slab which is made up of, ah, this is a dy that's in the y direction that's the thickness, right? So we write down the dy first. The minute I'm writing down the dy, the integral is written in that variable. So we want to write the integral solely in terms of the y. But if you want to find out what's the area of each slice of the triangle.It would equal to the root three over four times the base a squared. This base here is a difference, it's a horizontal difference between the two, um, x coordinate on the two curves here, and given the same y. So absolutely, we would have the square root of the y because observably this is bigger, right? And the minus the arc sine of the y, and that one would be smaller. The angle is smaller than this horizontal coordinate. This is a root y, sorry, root y.And the others are sign y. So far, very well done. What the remaining part is, we're going to square it because the whole area scaled by that squared, an under root three over four dy. This is conceptually speaking the total volume. If you reach this far, you're getting eighty percent of the points on AP. The remaining part usually just take one point. And that's calculation part. This is suppose without calculator. We're getting the old game of having to integrate a function which is y plus the arcsine y squared, and then minus twice the y of the arcsine y. And I would keep the scaling factor on the outside. The only job is to integrate this guy, right? I'm admitting this is definitely beyond the difficulty level on your AP now. Usually, you're not going to see such a problem, but when you learn, you have to learn just with more power so that you.Could perform well on the on the test one under test condition, whatever they give you, would be actually easier than what you're used to. Okay, we're not using any other technique than integration by part we have just learned. It's not a big deal. So let's find the antiderivative of the integral. Well, the first component is trivial; it's just one half of y squared. There's no need to really talk about it. Now let's integrate this first because that's more identifiable integration by part format.So knowing that y, it's very easy to integrate. Meanwhile, arcsine resist integration, so it stays on the outside. I'm actually getting a minus. I'm including even the negative here. Okay, so I'm getting a minus y squared arcsine y. This is going to give me what I want. Correct, but at the same time, when you take the derivative, you have also you have also created minus the y squared. The square root one minus y squared dy.Correct.Because fundamentally, I'm saying, later the theta be this arcsine of the y. Well, if you do that, you might as well just have done that earlier. But it's good, it's fine. If we do it here, there's no loss. Okay. By the way, you probably want to make the same move by doing the substitution of the variable even there, because this is not a very good integration by part. We don't even have a any part that's integrable. Alright, so later foundation now. If I do that substitution, what would become aYou know, guys, why don't we start from earlier? This would work out, but it would be even easier if we just do that substitution or all around because we're gonna need it to resolve the second integral. So if I do so, I'm gonna actually set a new variable theta. You know, this is also a cheap way to get get rid of whatever that's really blocking your way to integration, right? Because I don't know how to handle that outside. Let me just replace it with the angle. The minute you're doing that substitution.Be sure to substitute all the parts: integral, differential, and boundary. But we don't have any boundary here because we're not doing definite integral. So we have only two parts to handle. So arcsine becomes actually theta. This is theta squared minus twice of the this y here because theta is arcsine. So y equal to sine theta. So that's the sine theta, and then times the theta itself. Now, like I said.The differential must be replaced too, so you're gonna actually write your y equal to the sine theta, therefore the dy equal to the cosine theta d theta. Now I'm gonna bring that in. What you end up with is theta squared, then the cosine theta d theta. So far, are we good? Now we're looking at another. It's getting a lot easier now because we're only looking at the combination between polynomial, very low power polynomial for. That matter, and sine cosine. We can do this to find the antiderivative. I do basically my integration by part in my head now. Okay, if you have the theta squared cosine theta, look at that part first. There are several ways to go about it. You can look at cosine as part of the complex differential, a complex exponential, but you don't have to. We can just guess and check. Then I'm actually guessing one third of the theta cubed cosine.Cosine theta to begin with. Now, if I guess that, I definitely get what we want, but I also, oh, stupid move. Why do I want to make that higher? I certainly want to make that lower. So my first guess must be the theta squared the sine theta to begin with because this one when you take the derivative, you'll surely get the theta squared cosine theta, right? But it also gives you something else. It gives you a two theta sine theta. We not only not.And that we actually have negative two theta sine theta, you can see how how nice it is. Had I given you that one, you know, then you're done. Ah, but that's not we're not so lucky. So that actually means I need to minus four times of the antiderivative of this kind. Agree. Hmm. So, please.How you got the theta squared sine theta? Oh, I really want to look at this component here. My antiderivative of the integral is theta squared cosine theta. And what is a cheap way to get it? I'm just conjuring my antiderivative of theta squared sine theta because when you perform integration by parts, you fix the theta squared first, you derive sine theta, don't you get exactly what you need, which this term checks out.You follow me. Yeah, but on top of that, you're also adding two theta sine theta, right? That's also part of the derivative, which is not there. Not only we shouldn't have that, we actually should have negative two theta sine theta. So, don't I need to minus four times of this antiderivative to get that? Oh, should no, that's not.No, no, no, that's actually two theta sine theta cosine theta. My bad, they're different issues. Okay, well we just use cancel this out separately because that term here. Let me rewrite here. What you're getting is cosine theta. Sorry, theta squared cosine theta minus the theta of the sine two theta by using the double angle formula, so that we could absorb twice of the sine theta cosine theta back into the sine two theta, and this is not a bad data. Better now.Ivy, if you identified my mistake, that's why you were asking. Then thank you, you are very good. So we're only working on the first term, okay? I'm not even working on the second term yet. But now my goal is to minus this quantity. I need to cancel the antiderivative of this guy. Right. Right, you. Uh huh, very good.And to do that, I need to add onto and a two theta cosine theta. Because when you take the derivative, you use a product rule. You're fixing the two theta part, and you're deriving the cosine theta. Voila, you're getting actually this negative two theta sine theta. Good, so that term is successfully cancelled out. But now we we generate another term when we take the derivative, which is going to be twice of the cosine theta. So we. We need to subtract the antiderivative of that, but that part is decently easy. You just need to add onto, oh no, no, we need to get rid of it. So we just need to subtract by twice of the sine theta. So this is using integration by part repeatedly, by canceling out the unwanted term. This, when you take the derivative, you are supposed to get just that. Go ahead and take the derivative in case.There was a small step you couldn't follow me. Now here is your chance to follow. And after you're done, give me a signal so that we move on. For the screw, y minus arcsine y, why do you square it? Like here. Because we're integrating by the area, the area with the base of the a now of an equilateral triangle. It's a root of a root three over four a.Squared. Oh, after all, integrating volume, right? The volume must be made up of meter cubed, and dy, although very small, but it does have the unit of meter. It's just going to be a small thickness in the y direction. So this is the one that mentioned that's meter. And the within the parentheses, what you're getting is a length that's only meter. And this squared here gives you the meter squared, so that you would end up with an area times a height, which gives.有的保留。Please do confirm it once you double check. That is the.我哭，我好难过。In case you finish early, you can continue to find what would be the antiderivative of the remaining part in exactly the same manner. It's a process to review of an integration part.I checked, and it's correct. You saw that Elaine. Why didn't you go forward? Use the same idea. Look at the two theta as a single angle. It's a lot easier that way. Look at how do we find anti derivative for the second parts. 你到底在 mean that?中文嗯。Yeah, I um I also checked it. Excellent. And then go forward. Once you're done with the previous three terms, ah, go ahead. I mean the antiderivatives of the three, and then move on to take the antiderivative of this. I do I do need your confirmation that you understand where they come from. These are the antiderivative of that.One single term, by repeated. Yeah, I understand. Oh, good, cool.中文嗯。中文嗯。```中文``` ``Ànd I were getting the same answer for the second part. I'm still working on it.中文```嗯，Can I talk about what I got up to so far? Yes. Um. So.I did a negative theta squared over two times sine two theta. Oh, if if you want to do integration by part by looking at the theta squared, then you're getting something of a higher power. It wouldn't be as convenient as actually integrate the other component. Because this way you lower the power of theta. What you're doing is all wrong.It just said, "We're not moving in the right direction." So I do sine to theta first. Yeah, you see, we did that earlier. I fix the theta squared that integrated the cosine theta first, to guarantee my powers of the theta are always decreasing. Right. Yeah.`````` 嗯。嗯，I integrated the sign to theta first.And I got um. Negative cosine two theta over two. Yes, except for there's that negative to begin with, so we enough adding, correct. Yeah. Very good. But it also generates a its own derivative, which is actually one half of the cosine.Two theta, right. So we need to remove that. We need to remove ah the antiderivative thereof. Yeah. Very good. And what would that be. Um. Sign two theta over four. Yeah, there you have it.I got the same thing. Beautiful, Eddie. You mute the Eddie. Doctor, could you repeat the question again? I'm asking you, what are you getting.Oh, um, I'm still doing it. I'm not doing it. Uh, okay, continue. Now we're done because finally we finish the integrating. We got all the components of every single part now, except for they're not written in terms of our y. Remember, we actually did the substitution of the variable. At this moment, if you're doing the definite integral, then there's no need to go back to the y.Actually, Eddie, I'm going to wait for you because I need you to be following this. In itself, it's a new knowledge point. But for Elaine and Ivy, if you know, we found the antiderivative now, which consists of five terms. These are actually all the terms that we found, correct? And plus, don't forget that we'll have the wise. Oh, really beautiful. Okay, now kids, and we eventually our goal is to find the volume. This is only a.Intermediate step for us to realize what's the antiderivative because we did substitution. Now the antiderivative is written in terms of the. This is weird notation. That's a theta now. If you're doing the definite integral, you can really just plug in the boundary for the theta, but you have to translate it from the boundary of the y. Now, there are two alternative ways. Or you can remember how you defined the y and then just plug back in, replace the theta by the y.I'm gonna replace the data by the y, just so that we can finish the antiderivative of this guy rigorously. Okay, remember we were in the middle of finding what's the antiderivative here, and we still need to have a one half of the y squared that comes out of the first term, and this whole bundle here created plus two theta. But the minute you're seeing the theta, you got to replace it back by the arc sine of the y now. But what about the cosine theta? We know the sine theta is y, therefore the cosine theta. The must be the square root of one minus y, right? So this term actually become twice of the arc sine y, and then times square root of one minus y squared, and then plus the theta squared, which is really arc sine y squared sine theta. That is the y itself, the minus twice of the y, that's twice of the sine theta, and now we're adding one half of the cosine two theta.Which is simply twice of the cosine theta, cosine two theta, actually, call twice of the cosine theta squared minus one. I hope you still remember your trig identities, double angle formula. So we're really getting the one minus two y squared, and then that's the cosine two theta part, and there's actually one half of the theta, which is the arc sine of the y. Then find.And again, we're using double angle formula. Sin two theta equal twice of the sine theta. So sine theta, that's two theta. So when you copy, what you get is one half of the y. All of these are. Ah, please do your pencil work very carefully. You do need to be able to do this much on the AP.Then finally, we'll be able to plug in our boundary conditions.``````呃， yeah, I got the same thing. And Elaine, you. I'm still working on it.Alright, got the same thing too. Excellent. Okay, and this is how we do finding the antiderivative, carrying out the integral part, and then we plug in the boundary. What would be the boundaries? Now you have to keep in mind what we were doing. We're finding the volume. Zero and B. Yeah, what is a yes? We're doing the from zero to B, but what is that? So for the B. Ah, it's like the y. Coordinate of the intersection point between the two equations. Correct. Where the x squared minus x, you're not required to solve that. You count. You have to use your calculator. And I keep emphasizing, this is a high time that you guys get familiar with using the TI instruments. Could somebody show me or quickly solve it for us? Use your calculator. You're not permitted to do it with Desmos or, a online. Actually, maybe that's.Changing, but do it. Oh, you don't have a calculator. Bravo, you know I commend your mom for it. But I'm afraid it's about time that you get a calculator because part of the AP curriculum actually requires a TI calculator. And they have different versions from the Starter TI eighty four, these are graphing calculators, all the way to the TI Inspire, which is a a.You could actually program it. It has algebra manipulation functions, and all of these are permitted except for if you do carry and programmable. And then actually, they're going to require you to erase all your programs when you take it into the AP test. I mean, you personally programmed. Which one should we get? I don't care. I would say get a cheap one. Okay, you could go to eBay. This is this kind of a you know, it's about one hundred fifty.$50. I don't think it's worth it, but nevertheless, it has requires it. I actually got for one of my foster home students here for only forty five dollars, almost a brand new calculator on eBay. So it is totally functioning. Okay, let's come back. Oh, thank you for being waiting to take the trouble. Well, if that's the case, go to Desmos and find it out. And because there's still a bit of nuance.After this, because those are the values that, ah, well, basically you're setting them up. Ah, you okay, just get me the number and tell me what to do afterwards.嗯，my odd zero point seven six nine。 Okay. And how do we plug it into the?Why here, I equal to zero point is, let's say it again, seven six nine, seven six nine. Thank you. So how do we use that number? Our job is to plot. You know what, from zero to that number. So what's our upper bound? Our upper bound is the y coordinate. Yes, very good.And how do we achieve that? How do we find it? We can. The wife coordinate was zero point seven six nine. Oh, you have already done that. Meaning you have already squared your x now, huh? Ah, thank you, thank you. So I V is anticipating what we have to do. This is why the wife coordinate. That means the x you're getting is about zero point eight eight eight. Uh. Two point eight, seven six, or eight seven. Yeah, that's a pretty close. So when you square it, you're getting that. So this is our upper bound, and then you plug it in, and it's going to give you the answer. Although you don't really need to manually plug in each one of those, and if your calculator would take the antiderivative, do the plug in, and show you the graphical sweep of the area for you, for you to do some sanity check. Alright, so that's the whole process.Includes. If you're done with the technical detail, make sure that you have complete notes now. Now I'm gonna actually give you a summary of the takeaways. Of the entire procedure of doing this kind of problem. But you need to give me a confirmation that you have you have carefully worked out all the details.``Ì think I worked at everything. Beautiful. So may I start a conceptual summary? Yeah, this is really important because problem. I would put you in the driver's seat and make you take initiative. You have to work out something on your own now. So if you're dealing with a bulk figure, for example, a volume now, in order to utilize Integro.And what you want to do is to break and conquer, meaning you want to break it down into integral of the d volume, which way you are going to slice or break the volume, it's up to you, it's up to what's convenient, okay, don't get hung on one kind of a slicing, so you break it down into infinitesimal volume, you really want to visualize it. For example, what we did would be this this whole figure, and you slice it in a certain direction each time, you end up with a slice of the volume. But next time, if you're looking at a.Cone, maybe you want to slice it in this manner, right? Every single time you're looking at the, on this kind of a slice. So the minute you look at your DV now, you write down all the three dimensions. Usually, it's consisting of a thickness, which is our choice of the variable. That's our integrating variable. Earlier, we had a the dy, but now in general, I'll call that the dh because it's referring to the thickness. And then there's me area that gives you the volume. This area, depending on the shape, would.It depends on the two dimensions in a specific manner. For example, if the circle, the only part you need to know would be the radius, and there would be really pi radius squared. But if it's equal, well, if it's a rectangle, you have to find all the two dimensions, and and after that, you need to change this conceptually correct decomposition of the volume into d volume now into a favorable form that's able that we're able to integrate. Meaning, what integral must.Be written in terms of that, there's a differential, the same variable now, and that usually involves drawing the information from the original description of the area or the volume, meaning you have to go to the equations that actually confine them. So at this moment, our model is done. The second step will be actually really technically finding the antiderivative. You start your calculation now, finding the antiderivative, plug in the boundary, you. Whatever the necessary technique, including integration by part or substitution, or sometimes there are other, it would such some other skills. And after you find the antiderivative, theoretically speaking, you really have a the complete information about how the volume depends on the variables. Now, then the final step is to simply plug in the numbers. And not to mention, there's always going to be sanity check, even after you get the number, you have to use.Your intelligence to see whether that's reasonable. Our good Sophia. Yeah. Okay, then let's actually prove such a famous formula. And you might you you may not even think that there's any need to prove it anymore. Meaning let's actually prove the spherical volume. Volume of a sphere with a radius of r just equal to the volume of.Sphere would equal to the four pi r cubed over three. It's a well known formula. You have to visualize a sphere and start slicing it. The equation of the sphere, it's very easy. So there's some convenience immediately, right? And I'm trying to draw the three D now, and natural coordinate system. I'll give you a minute to think about it.This is definitely easier than what we did before. But do not skip any step.中文嗯。```We may like separate it into like a bunch of infinitesimal like.Circum, like disk, I guess, just like this segment, and then like here, that we like. Oh, what a wonderful idea! Yeah, and Eddie is going for a simple but very powerful way to slice it off into each, a circular disk at a time. And we're visualizing it up. Sorry, sorry. I'm truly at the height of a. Let's call it the one here. We're. Licensing off a layer. And you have to visualize that that's a certain thickness now, so that it would have volume, right? And Eddie, would you write for us that what would be our DV now, referring to this disk? This a this very flat, thin chunk of the DV volume.中文嗯。嗯，嗯，对。Absolutely. The, pi r squared. Then, D E. Uh huh, very good. Ah, now I've been noticing it's a disk. So the area of the disk is actually pi r squared. But I want you to realize that.That's not the big r. It's not radius of the sphere, right? To be safe, like with that little r. We do know depending on the location your joint. That little r definitely is getting smaller. So the pi little r squared, and then the dy. You don't have to put that dy. You can put that dr, but we're actually deciding where the location at the y coordinate. So this is what I call the oh dz is better because that's the x and that's why this is the z coordinate. Let me write this.Ivy, am I doing what you intended? Yeah, beautiful. So in the way we have already found our volume now, except for we have moved beyond model setting, and we just need to make sure that R can be written as a function of z. Otherwise, you can't perform your integral, right? Yeah.```中文嗯。``` 中文嗯。中文`````````嗯，show me your mental process. Maybe we could like label the thickness as D Z and try to find the.Like, you want to notice, does the geometry tell you any relationship between R and Z? Again, it's not a function of DZ, it's a function of Z, right? Because you're looking at the integral. Yeah. And do they obviously have some relationship? Yeah, definitely they do. So in fact, remember this is R and that's the Z. Now, the higher the latitude.The smaller the R, the bigger the Z, right? Oh yeah. How do you connect them in the functional relationship? This is like doing the relative rates. The whole point is to relate the variables so that they're all written in terms of the Z. Now. I was also thinking of the big R because when the Z is zero, like little R is equal to the little R is equal to the big R. So. Oh yeah, that's correct.Absolutely. Like, make a triangle. Hmm, beautiful. Ah, I, I surmise that you're constructing this radius now. Use that as a point use, right? Draw on my board. Uh huh, that's an important move.Then the partners, be the big or very good, excellent.The Louise Pigeon theorem. Yes, very good indeed. Ah. Oh, the little r squared. Is c squared equals the big r squared. Ah.So very good, and the I V is finding a relationship we could take advantage of. Now little r squared plus the z squared would equal the big r squared, so we could immediately just change it into the big r squared minus the z squared, right? There's no need to really find the little r by taking any square root. All we're interested in is to replace the integral. Done. This is a highly integrable. Go ahead and finish it. Be careful with what is the boundary.You're well on your way to success. Fully finding the volume.Our squared would just be like a constant, right? Oh yes, absolutely.``Àre the boundaries negative r and r? Yes, brilliant. And you notice actually, we need to go all the way from the negative r. Excellent. This is from negative r to positive r. Although I would say, if you notice the symmetry after all, these are even functions, right? And also, this up. But sine is here, definitely it's equal to the lower, and you can just double it only integrate from zero to r. Definitely integrating from zero has its advantages because usually the functions are easier if you plug in zero. However, it's not harder if you just do it that way. Go ahead and finish it. To convince yourself, you are indeed getting that standard formula.``````中文嗯。所以，你。We don't read the announcements. We're starting ten minutes later. Today, for once, only for today. That was announced on WeChat. Ah, there has been more people that one who don't read.Nick, Nick, I send out the announcement. We're starting ten minutes later because I am just my every single class is late today. Okay, I'll be back in ten minutes. Yeah, thank you. Anybody finishing?但我还必须工作。 Don't skip anything because this is this is absolutely IP level, meaning it's your stuff. You try that.```Yeah, I think I got four power.Cube or three, excellent. It's just the anti-derivative would be made made up of. Let me actually say this is the two pi. I'm gonna go go for the second one. Anti-derivative simply the r squared z, and then minus one third the z cube, right? And you plug in the boundary from zero to r. Yeah.Beautifully done. Are we comfortable? And Adian and Elaine, where are you guys? Oh Toby, we're starting ten minutes late. We send out the announcement in the group chat. Oh, sorry, sorry. You don't read the announcements. Okay, sorry. Yes, you in a bit. Ah, yeah, I got four pi r. Ah, cube number three.嗯，对，嗯，对，beautiful，and it's very typical of AP, especially the rotational solids. Why don't we try out one more? And if actually in this case, I'm not giving you a standard geometric object now, but instead I'm going to actually give you this parabola. No, that's too easy. Let me give you a exponential function. Why equal to the e x Okay, and the graph would look like this. And well, the other.Inside is not part of the graph, but rather we're creating it by rotating this around your y-axis. I'm rotating it around y-axis, and then you would generate a open volume. However, I'm telling you we're only integrating actually this part here, so that that quantity here is. No, I'm giving you the x now. The x here is a two, so this is a zero one, and you work out basically just e squared. That's.The pair of coordinates here, and I want you to find out what is going to be the volume of this cup, inside of this thing. You think about it this time, including the interior, and then what do we do? We can still break it down into like really small circles, right? Because we're spinning it like circular.嗯。So, it could be like, right, the volume. That's the, the, integral of pi r r squared t. Excellent. So we we write the volume as actually the dd and the dd, sorry, the dd here. It's according to your.Suggestion: We're cutting one slice like that, right? And I'll go ahead. I interrupted you, just to follow up with you. Hi, R Square. Uh huh, and however, though, what is the R? You mean that's the R, right? Yeah, the reading. Good. Uh huh, and then D. Why? Correct.So again, we're facing the need to translate R as a function of y. I think you are not wrong. This certainly can be called R now, but I have a suggestion. Because you are given a graph, this horizontal segment has a meaning in terms of the given x y coordinate. So it's not also just an x. What you call R is actually the x coordinate. So why don't we write it as x now? At least immediately, these variables are connected by functions. Do you see?Yeah. Excellent. Now let's go ahead and finish that.So, coming right as x equals the natural log of y. We're actually looking at first of all, and Ivy is making the right move. He's saying we need to write everything in terms of the variable y instead of variable x. Now that's a little inconvenient, so she changes it into natural log of y now.Beautiful. So this is oh sorry, and that integral just equal to the pi comes out, and we're integrating the natural log of the y squared dy. Once again, we're the need of integrating my part, but it's okay. And what are the boundaries? Knowing that these are the boundaries for the y, one and e squared. 嗯，yes，from a one all the way to the squared, you got it. Alright, finally, we're getting to the last two points on the AP now. Up to this point, you probably win three points. Ah, no, no, no, no, two only two, and that's only because it's easy. There's no need to really analyze something, and you have only one curve. Now, what about the anti derivative of the natural log y squared?Still in the spirit of trying to do something, just to get that to begin with.I even wonder this is worth leaving as homework because it's sort of a trivial now. No, no, no, it is homework. Be sure to work through the steps. Okay, to get you started, I would say immediately you could eyeball one way. Don't make the natural log go to higher powers. So I would say this is the easiest starting place.And you begin to subtract and add, etc., etc. And good luck. Makes sense. Okay, take care. Bye.

中值定理、积分技巧与体积

中值定理、积分技巧与体积

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