 I have to say it's a free body diagram, it's like something that you could get like one of those holistic health policies. You know, you go in and they give you a free body diagram, which aren't going to come for you. Okay, so, yeah, I didn't say anything about that, yes. So, yes. So, maybe you're not giving the story quite right. And if you do everything, you do 75% no. Or 80% correct? If you can't know 80% of the basic stuff, you don't deserve the message. But, wait, wait, wait, wait, wait, wait, wait, wait, wait, wait, wait. So, here's how the final will be structured. Part one, easy problems. Really easy problems like, okay? Easy problems. They won't all be this, right? Problems along the lines of the level of the problems on the placement exam. So, a lot of you maybe didn't take the placement exam because you had 80 credits, but they're straightforward, easy problems. First, they cover everything. So, you know, like write the Taylor series for easy to text or something like that. But covering all the material. But easy straightforward problems that if you know how to do the material in this class, you should be able to get through it in half an hour. And if you do 80% correctly on that, then you get a C for sure. Okay? So, if you do that 80% correct, you get a C. And then the rest is to determine whether you get an A or a B or a C. We have to negotiate. I don't know. I'm hoping that won't happen to any of you. So, you should focus on the first part until it's done. Because the first part is supposed to be a measurement of the minimal skills in this class. Demonstrate mastery. Demonstrate the minimal level of understanding. Then you should pass the class. So that's what the first part is for. And the second part is to decide what this class would be C or A. Fold it down a little bit so I can get here again. Part is going to be a minimum hurdle. If you clear the minimum hurdle, then you're guaranteed to pass. Beyond that, you'll look at the rest. So, I mean, you can think the first part is worth, let's say, 50 points. And in order to get a C, you need a 50 on the test. Other items, so there's that issue about the final being in two parts. There's a basic easy part that everyone who deserves to pass should be able to do relatively quickly. And then there's the rest, which will be the harder problems that will be like on the previous midterms. Right on the previous midterms, there were a couple of easy problems gathered among harder problems. Not really. Well, there should have been a couple of easy problems gathered among harder problems. Certainly, that was true in the first midterms. On the second midterms, more of the problems were middling difficult. There was no crazy hard problem, but there was no trivial problem. The other thing is, sorry, I don't remember whether it's jazz 111 or 110. It's one of them has a, I don't know, it holds 103 pieces. I think that's 111, right? I think so. So the room, the biggest room that I could get was that one. So they gave me jazz 111, Friday at 2.15, and I will just do whatever problems people want. Also, I have my office hours on Monday, but so if you have questions that you want to ask specific questions during the week, you can make an appointment. I'm around. You can make an appointment and come in. So I'm not holding schedule hours other than my usual hours on Monday next week, nor is Professor Donna Fann. But both of us, if you say, I need help. I really don't understand partial fractions. When can I see you? And then we'll say, okay, how's Tuesday at 2? Or, no, I'm giving the final answer, but I won't say how's Thursday at 2? Or whatever, okay? Any other questions about where the final is, what it is, how it is, why it is? Okay, so let's move on to doing problems, reviewing. We left off in techniques of integration. Somebody asked me to do a partial fractions problem. So let me briefly review partial fractions. So you use partial fractions when you have an integral line. Maybe I would do, well, okay. And you have an integral which has a bottom that you can factor. And then that splits it up into something like, let me just already leave it factored. So say I have an integral like this, and one, one, okay, I don't care. Say I have that integral. I can put something more complicated on the top, but make it more complicated, okay? So what the idea here is that we want to split this up into two easy integrals that look like, but we can't quite. So it's just algebra to try and split this up. So we just forget about integrals, and we just do algebra. So we say if there are A's and B's so that this is true, what are they? Note that it's important. A couple of things that are important. We must have the degree of the top smaller than the degree of the bottom. If the degree of the top is larger, you have to do long division or some other way to reduce the degree of the top. You have to split out all of the stuff that is higher degree than the bottom. Let me hold the special case for a minute. So we just do this. So it's just an algebra problem. So we cross multiply. We need to find A and B, so the 2x plus 1 equals A times x minus 1 plus B times 3x plus 2. And either you can now multiply this out and equate coefficients. Let me do it that way. Or you can choose the distance choices of x to kill off the factors. I like that way better in most cases, but sometimes the other one's easier. So let me do that way first. So if, because this has to be true for all values of x. So that means that if we choose special values of x, it still has to be true. So if x is 1, we have 2 times 1 plus 1. 1 is 3, this is 0, and this is 5B. So that tells us right away that B is 3 over 5. What? No? So if x is 1, if x is 1, 2 times 1 plus 1 is 3. A times 1 minus 1 is 0. 3 times 1 plus 2 is 5 times B is 5B. So if x is 1, 5B is 3. So I know right away that B is 3 fifths. And if x is negative 2 thirds, then that tells me, well these numbers are a little grosser, but okay. Negative 2 thirds, so that's negative 4 thirds plus 1 is negative 1 third. And this is negative 2 thirds minus 1 is negative 5 thirds. And this is 0. So that means that A is a fifth. Okay? Do you want me to go through the other way? Yo. Yeah. Yeah, no. Okay. So the other way, which is completely equivalent, but sometimes a little easier to use, I guess I'm going to give up on my free body diagram. Oh well. That one's just stuck. Okay. The other way, which is equivalent, is we multiply out this thing. 2x plus 1 equals Ax minus A plus B3Bx plus 2B. So that means that A plus 3B is 2 because I'm matching the coefficients of x and 1 is minus A plus 2B. And then you solve these equations simultaneously. Here you would add them together, get 5B equals 3, and then substitute back and get A equals a fifth. Okay? So it should be the same. Okay? And then does anyone need me to do this integral? I'll just do it. It's quick. So the integral 2x plus 1, 3x plus 2, x minus 1 dx is equal to the integral A is a fifth, 3x plus 2 plus the integral 3 fifths over x minus 1. So this is one-fifth the log of 3x plus 2 because we make the substitution u equals 3x plus 2 except this is a 3x. So that means that u equals 3x and then du is 3dx. So divide by 3. So am I saying that too fast? So this is one-fifteenth log 3x plus 2. And then this one du is dx. So this is just 3 fifths log x minus 1 plus the constant. Did I do this too quickly? Everybody's okay with this, right? Okay. So one caveat with partial fractions is sometimes you can't reduce the factors. You're left with a quadratic factor or you have a repeated factor. So if I have something like, let me just put a 1 here over x cubed x square plus 1. Suppose I want to use partial fractions for this. Then I would split this up. Maybe I want, let's put a x minus 1 here too. I'm not going to actually do it. I'm just going to set it up. I would split this up, read things. And what I need is that the degree of the polynomial on top with the a, b, c, and q's in it is one less than the one on the bottom. So here I would have a x squared plus bx plus c. So I would have to find three variables in order to understand what's on top of the x squared. Here I would have what I need to be e x plus f and here I would have the g. So I'm not going to do this one because it's quite complicated. But I just want to point out that you split it up into each of the factors and the degree of the thing that goes on top is one less than a denominator. So this one would certainly not be on the first part and if it's on the second part I would expect most people to blow up. Alright, so let me leave that first we have to cover here. I think that's it for the techniques of integration because I did most of them last time. Any other techniques of integration I need to review? No. Okay. So what else was I would... So the next topic is volumes. So now we're into chapter six I guess. Oh no, no, no, I missed improper integrals. Sorry. So improper integrals would be something like the integral from one to infinity of one over, doesn't matter, x squared dx. So this is saying if we have this curve one over x squared and we look at this area is it a finite number or is it not? And the only thing with improper integrals is that, well this doesn't really make explicit sense unless we extend the definition of integration because we can't put infinitely many little rectangles under here. So we just remember, we just extend this to mean this is the limit as m goes to infinity of the integral from one to m of one over x squared dx. So we just do this integral which is very easy and then we take a limit. So the integral of one over x squared is one over x, negative, and we evaluate it from one to m. So this is the limit as m goes to infinity of, lost my train of thought here, negative one plus negative one over m. Yeah, wait. So, no, no, the derivative of negative, what am I doing? Isn't this right? Yeah, the derivative of x to the minus one is minus x to the minus two. So the derivative of this is this. So, yeah, so this is right. And so, but something's wrong here. Oh, I did this backwards. Sorry. It's minus one over m plus one. That's better. Okay, and then if we take the limit, this term goes to zero so I get one. So this guy converges to one. So we would say that this converges. Okay, so this is pretty straightforward and I know when you first thought it seemed very confusing because we had this infinity in there and all sorts of things, but this is very similar to the kind of stuff we're doing that we did with the infinite series. So maybe after having to deal with the infinite series, this makes a little more sense where at least you have a little more practice. Of course it will diverge if I replace, if I change the power around, I can get one that diverges or converges or whatever. The idea is very similar. The other situation would be if the thing blows up somewhere inside. So for example, the integral from zero to, I don't know, zero to four over one over x minus two, dx over x minus two. So this one would diverge. It blows up at two. We have to check. So this we have to write as the integral from zero to two and the integral from two to four and then take the limit, this is the limit as x goes to two from below and this is the limit as x goes to two from above, but this blows up, right? So this is the log of x minus two evaluated from zero to two log x minus two evaluated from two to four but now I have to take the limit here, here and here. Yeah. If one of them blows up, you don't need to bother with the other one. So this two from below of log x minus two, so this diverges. Of course I could change the problem and make something that converges but if you put a square root there it's going to converge. So these are the two variations on improper integrals. I feel like I'm forgetting something else. Oh, no, it's not it because we also have citizens rule and stuff like that. So any questions on improper integrals? So the other topic in chapter five that I just forgot about and now I'm just remembering is the citizens rule. So the idea here is to just think carefully not even that carefully think a little bit about what the integral means. So if I have some integral like that that I can't do because there's no nice formula for the rule of sin x over x although we can do it by power series there's no closed form for this. What we do I want to integrate from one to two and my function, I don't know looks something like that it doesn't really matter too much for what we're doing. What I do is I chop this up into some number of pieces and I just put little rectangles or little shapes underneath it to try and calculate the area here. So we have three methods that we tend to favor so the most efficient one well, so the easy to remember would be say should I do four or doesn't matter. The easy to remember would be say the trapezoid rule which so I take so we choose some number n and this is how many plus or minus one how many points we're going to evaluate our integral at and so maybe I will give you in probably I will give you in we choose n and for the trapezoid rule we calculate the area of that figure so the trapezoid is the area of the trapezoid is the base times the height the average of the two heights the right height and the left height so here I've chosen n in this case I'm taking n equals four so I divide this up into four pieces so that means that this point is at one this one is at one and a quarter this one is at let's call that five fourths this one is at six fourths this one is at seven fourths and this one is at eight fourths which is also known as two I know that six fourths is also known as three halfs and so to use the trapezoid rule what we will do is find the area of this so we're going to take the width of the base so here the width of the base is a quarter every time and so we're going to take the width of the base times and now we just add up the heights at each of these points but remember that the ones in the middle get double counted because we're going to average them so I want to average f of one which is the sine of one divided by one plus and now here at five fourths from this one I'm going to count it once and from this one I'm going to count it once so this will be twice the sine of five fourths divided by five fourths twice the sine of six fourths divided by six fourths twice the sine of seven fourths divided by seven fourths and then here I've only got one side so I only count it once and I'm averaging these numbers so I have to divide by two so this is the trapezoid approximation for the symmetry another way we can do the same thing we get through this am I another way we can do the same thing is take the midpoints and just add them up do I need to review that no nobody wants to hear it so another one we can do is the midpoint rule which is the same idea we just add up the heights of the rectangles that goes through the midpoints so that would be this picture take this plus this plus this so I have to evaluate these points so I only have four numbers and then I also have Simpson's rule which is more complicated because I'm actually fitting a parabola here so we just draw the picture again so if I'm doing Simpson's rule I'm actually with n equals four means that I'm taking two rectangles and I'm putting little parabolas that fit best the pattern on each one of these is I take one of these and one of these and three of those but since this one gets double counted it actually goes one, three, two, three, one so it would be these same numbers actually so here I take three of these guys two of those guys two of those guys one of those guys so it's one, three, two, three, one four four oh yeah, it's four, isn't it? sorry that's four yes, because one, four, two, four, one and since I'm averaging them three at a time but I'm also averaging I divide by six oh, wait a minute three yes and a half no, four is for never and there are error formulas here that tell you how much how many things you need to know if I ask a question on the error formula like I say how many terms of sine x over x would we need to get the error to get the integral within three places or 10 to the minus 50 or whatever it is there's a formula and you just solve it and the thing that's tricky is you have to remember that the formula depends on in this case the maximum of the second derivative over the interval or in this case the maximum of the fourth derivative over the interval yeah that would not be a first part kind of question okay so that's okay so now we should move on to volumes I guess so the basic idea of all of the volumes whether they're volumes of revolution or volumes with a given cross section or whatever is that the volume of the solid is the integral of the area of the cross section whatever the x or the y so if I have some weekly object here with an easily understood cross section then I can find its volume just by slicing it I'm integrating the volumes of the slice okay so if I have so maybe I should do an example of this but so here the area maybe I should do this and all of the problems that we do even the ones where you can look out a formula like the volumes of revolution with the washers or the discs or whatever it is it's really just this everything is really just this think about how to slice it up the slices have a nice description and then you just add up the area of the slices so for example suppose we have suppose we have so suppose we have some shape let's say it has so I have something that has a square cross section I don't know if you can see let's cut the end off so suppose I have a shape that looks like that so the base looks like so this is looking at it a little bit weird this is the x axis this is the y axis but I laid it down and put the thing usually we make x come out of the board let's make this be the x axis this is the y axis and so the shape of this thing the base looks like this and let's say just for argument's sake this is the line y equals 3 minus x and it goes from well here is 3 and here let's put it at 2 at 1, 2 and the cross section the cross section is perpendicular to the y axis or squares does everyone know how to do this yes some people know so what this problem is saying these slices if I slice it this way I see squares I can also slice it the other way but it's a much more complicated thing to do so let me do it this way so this means that to do this we want to set this up so tell me how I would set this up so ok what would I integrate one of the bounds on the edge 0 to 2 and I'm integrating dx or dy dy right so 0 to 2 dy tells me that I'm looking at slices that go like that starting off from y equals 0 to y equals 2 rewrite this as x equals ok so x x equals 3 minus y that was easy and so now what do I integrate that square yeah because I have sitting over here a square so if I look at y here some arbitrary height I have a square that is 3 minus yy by 3 minus y high dy fit I'm just adding up all those little slices of cheese so I have 3 minus y square and that's the one if you want me to do that square it out we'll make the substitution and you're equal to 3 minus y and so dy ok are we ready ok with this we can of course complicated where the cross section is the disk so here obviously this one I can't have the cross section but I could have a similar setup where I take something like let's just take the same object and do things too in various ways oh I'm sorry I forgot to mention one other thing in terms of so this is a pause I am putting up it's not quite finished but it will be finished probably tomorrow no it's extra credit so you don't have to do it but you can do it if you want ok so I'm adding another web of sign which is optional which has somewhere on the order of 60 to 70 problems because if you wish each one counts for half a point this adds on to your homework problem but if you have more than 100% on your homework average I don't care now I'm ready for battle ok so I don't know where it is so I have to leave it so there will be a web of sign I'm going up over the weekend which is optional Monday the final so do by the final but it's not do just once you do as much as you want so then do it that's to study so what is this web of sign all I did is I did problems from the beginning to the end some are straight forward like part one some are harder like part two and I put a bunch of problems so there's a problem like this there's a problem like ok so let me come back to this problem this side's like this side, this side, there now I really look ready for battle you know I can give myself like a thing like I forgot his name I can give it to heroes not Mike Tyson ok so let's just continue this a little bit suppose suppose I took this shape and I revolved it around the axis to get piece of the cone then the process would be similar here this is still the line x equals 3 minus y I revolved it around the axis it goes from 0 to 2 the cross section is no longer a square now it's a circle so this would be the integral from 0 to 2 of pi r squared dy if I revolved it instead of around the line y equals 2 some other line well I'll get some other thing if I revolved it around the x axis I revolved it this way so that I would get something that looks like a cylinder a point on it then it's more complicated I could either do this by slicing it this way in which case I have circles for part of it and I have two functions that I have to integrate so I could integrate this dx as the integral from well, if I'm doing it this way this is 1 this is 3 so I could say I could integrate from 0 to 1 of no, the height is 2 0.5 that's one of these circles in this bit and then I have to integrate from 1 to 3 of this line which is 3 minus x dx or I could slice it the other way and have a watcher okay so does everybody understand where this came from? yes okay, so I have this thing that looks like a very short pencil it has this elliptical bit like that and then on the end it has something that comes to a point okay, so I've exaggerated this width that's 3 so this part it's just a circle of constant radius what is the radius? radius is 2 so what is the cross-sectional area of a circle with radius what is the area of a circle with radius 2? pi r squared r is 2 so that's for pi so I can write that as 2 squared the radius is constant so I have to integrate one function to get this part it's a length 1 so it's really just for pi and then this part is a little more complicated it looks like that okay, so I could also if I were I don't know why I would do it this way but I could do it this way I could think I take this shape it's dy and I revolve them around the axis so what I'm doing is I'm decomposing this thing into a bunch of concentric shells right, so I have a bunch of little tubes and I want to find the surface area of each of these tubes times its thickness so this would be related by shells sometimes you have a choice of whether you want to do it this way and sometimes it's really the easiest way to do it this way so if I wanted to do this for some reason so some people and some problems say do this problem by shells I won't say that, if you want to do it by shells do it if you want to do it by washers do it by whatever method works so if I wanted to do this by shells then that means that what I have to integrate is the area is that the area of one of these little cylinders so these cylinders go from if I take a typical one it looks like this the radius here is y the width here is it goes from zero to three minus x and if I cut it open so this distance this is really if I were to cut it open it's a rectangle which is three minus xy and two pi y a ramp because the circumference of a circle is pi d or two pi r two pi y those three minus y three minus y so I want to integrate two pi y times three minus y d y as y goes from zero to two so those two intervals if you do that should be the same I recommend that you not memorize the formula for volumes by discs or volume by washers understand them and re-derive them on an asmetic basis because you will not remember these formulas in six months but if you remember the idea that you slice it and you find the area of a slice that's something that should stick with you anybody want me to do more volume problems? yes do you have one in mind? I mean they're all just like this so they all come down to I only have two minutes so I don't have time to do another volume problem so if you have more volume problems than you want done let me know so let me move to one last topic that we can do in two minutes before I get my work hanged off which would be something like I don't know average value or earthquake you want it on the test it has to be on the test how can I not put average value on the test? that's your birthday present so we have some function like this not I was going to do average value but I think you're arguing I'll write both down so for average value all that means is I want to find where the area the area if I swap off the top and stick it on the bottom the area of this rectangle is the same as the area under the curve so this is very straightforward if I'm going from A to B well the area is just the integral from A to B of f of x dx and so the average value is just like that like you might say so this is why she loves it so much because it is so easy so you just calculate the area and you divide oh I forgot to do an area between curves for the arc length it's I mean for all of these things it's sort of similar the arc length is we want to figure out by adding up lengths of a bunch of little segments well these little segments if you look at a small scale are hypotenuses of triangles where they're dx, y and they're dy, high and so by the Pythagorean theorem this is the square root dx squared plus dy squared but we could rewrite this as 1 plus as prime of x square dx and so that means I don't have time to do an example but I have time to write down the formula that means that the formula for arc length is we integrate that yeah I'm going to write it down again so I integrate from the beginning to the end the square root of 1 plus the derivative square dx because all I'm doing again is I'm adding up the lengths of these little hypotenuses so there's really not time to do an example so let's stop there I'll see you on Monday