| Start | Well, we'll just do an example.
Um, was 28 assigned? There were a couple here that I hadn't covered those types because I didn't think we were going to. I don't have them with me. Twenty four 24 you should be able to do. What else? |
| 0:31 | 21, 24, what else did you get?
31 Oh, 31 you have to do division. And 30? Ok. So there's a couple of things. So you have to look at the factors in the denominator. So, suppose you have This was 24. Those are the factors in the denominator, right? Right. |
| 1:06 | But we don't care about the dx for the moment, ok, we wanna figure out how to break that up.
So when you take the x^3-x you can factor into this. So you have 3 terms. So you're gonna need A/x + B/(x+1) + C/(x-1). Ok, so you're gonna end up with 3 equations and 3 variables. |
| 1:30 | I won't do all the steps for you but A is gonna get multiplied by these two.
So A is gonna be times (x+1)(x-1). B is gonna get multiplied by x and x-1. And C is gonna get multiplied by x and x+1. And that's gonna equal x^2+2x-1. |
| 2:08 | We love these, right?
Eh, don't love the algebra? So, skipping a step you should get that. |
| 2:32 | You guys saw I put up a whole bunch of problems over vacation?
There was one small type I didn't do, which was those trig substitutions. I'll do some of those this week. So you can practice. The exam's in a couple weeks, right? I don't know how much more we're gonna get to before the exam, my guess would be areas and volumes. And then we'll have some time for review. I'm looking to see if I can get a classroom to do an extra review session, but it's tricky. Because I could get like, review for say 12 of you, and then I could pick the ones I like best, which usually you demonstrate with chocolate. |
| 3:05 | Um, or of course gold. Gold works.
But to get a room for say, 250 of you, sometimes takes a little maneuvering. Alright, so now we have Ax^2 and Bx^2 and Cx^2. We have Bx and Cx. And we have -x. |
| 3:33 | And that's gotta equal x^2+2x-1.
So you're gonna have A+B+C has to equal 1. 1x^2. -B+C will have to equal 2. And you know what A is. You know -A=-1 so you know A=1. |
| 4:01 | So you should be able to solve from there.
Ok, and then you're gonna get something over x, something over x+1 and then something over x-1. Not gonna do every step. No, there was a long division one. 30 or 31? 30. You win. I'm gonna show you a cheat. |
| 4:39 | Now, I'm gonna show you a trick first, which most of you probably won't really pick up on, but that's ok.
You take the numerator and you subtract 16 and you add 16. And you go why would I do that? Well, you'll see. You could break that into r^2-16/(r+4) + 16/(r+4). |
| 5:14 | Now, does r^2-16 factor?
(r+4)(r-4) I hate making r's. They're very hard to do in chalk. |
| 5:32 | And then those cancel and this integral becomes r-4 +(16/(r+4)), which is a very simple integral, right?
r^2/2, 4r, 16 times the log of r+4. Now you say to me, but I would never know that trick. Right, you guys see what I did? So I subtracted 16 and I added 16, and that enabled me to write two integrals, two fractions, and the first I can simplify. |
| 6:06 | Now many of you would say I would never think of doing that. That's ok.
The other way is you do polynomial, you do division. Ok, or you could use synthetic division for those of you who remember synthetic division, which is more fun. Ok? So r goes into r^2 r times. |
| 6:31 | Because r times r is r^2.
You multiply them together. And subtract. And you get -4r. Did I do that right? Yeah. And there's no, because um, sorry, right. |
| 7:06 | Ok you bring down the 0. That's a 0.
And now -4 times r is -4r, so this becomes -4r-16. Subtract, and you get 16. Since r+4 doesn't go into 16, you get that. So you can end up in the same place. So we math geeks know a lot of shortcuts that you guys probably don't know. |
| 7:32 | But that's ok.
And this, as I said, this you should be able to integrate very simply. I hope so. So again, if you wanna do the polynomial division, you say how many times does r go into r^2? Well r times r gives you r^2. So you multiply it out and you get r^2 +4r. You subtract and you get -4r+0. And then -4 times r gives you -4r, so you multiply the -4 by both terms. And you subtract. |
| 8:01 | Ok? That's the polynomial division.
And if you had x^2 +4 and x^3 +4, that's a nasty one. I never teach these. They're hard. But you should be able to do them once you get the hang of them. So if you wanna do |
| 8:36 | There's other clever tricks but let's just do the division.
So you wanna do x^2 +4 goes into x^3+0x^2+0x+4. You have to have a 0 for all those missing powers of x. So x^2 times what gives you x^3? x^2 times x. |
| 9:01 | So this is x^3+4x
and subtract
and you get -4x+4, now x^2 doesn't go into -4x.
So this is -4x+4/(x^2+4). |
| 9:33 | So this integral would become integral of x+ (-4x+4/(x^2+4)) dx.
Everybody following me? You guys are all being very quiet. Hi. Ok, so this would now become x- (4x/(x^2+4)) + (4/(x^2+4)) dx. |
| 10:11 | So this is easy, this is just x^2/2.
This you could do with substitution. That's inverse tan. Ok? I'm leaving that part up to you guys. How we doing on these? Eh? They're hard. |
| 10:32 | Ok, we'll do a couple more. I'll do some of the Appendix G ones.
I forget which ones were assigned from Appendix G. |
| 11:54 | Ok, what do we do with the first one?
Well this, we didn't do this type yet. So we should learn this. |
| 12:00 | The bottom here factors into x^2 and x+2.
So we have x^2, it's slightly different cause you have what's called a repeated, a repeated term in the integrand. Or in other words in the denominator. So now you would have to make this A/x + B/(x^2) + C/(x+2). So when you have a term and it's been squared, you do once over the term and then again over the term squared. |
| 12:33 | So if it was cubed, you'd have A/x, B/x^2, and C/x^3.
Got the idea? So if you have the denominator and it has like a (x+1)^2 in it, then you would do an A/(x+1) and you'd do a B/(x+1)^2. Ok? My handwriting's a little off today. I'll make that look prettier. |
| 13:05 | Ok?
Yes. Well, it has to do with the way you can put the rational expression together. So when you have a repeated factor in the denominator, a factor that's raised to a power, you have to take one letter over the factor and another letter over the factor squared. If it was cubed you'd have to have a third letter over the factor cubed. Ok, so as I said, so if you had something over (x+1)^2, you'd have to have a letter over x+1 and a letter over (x+1)^2. |
| 13:34 | So that's the way this would break up.
So you would have 5x^2 +3x-2 over x^2(x+2) equals A/x + B/x^2 + C/(x+2). |
| 14:01 | And that takes a little algebra.
The A gets multiplied by x^2 and x+2. Let's see, you have 5x^2 + 3x-2 And then you get B, I'm sorry A only needs to get multiplied by x. Sorry. |
| 14:32 | Cause when you multiply through by x^2 the x^2 cancels with the x.
Ok, B just gets multiplied by x+2 and C gets multiplied by x^2. That's what happens when you multiply through by x^2 times x+2. You clear the denominator. Ok, so again, so the A gets multiplied by x and x+2. B by x+2 and C by x^2. And then everything has the denominator of x^2 and x+2. |
| 15:03 | Yeah?
Lost? Yes. We're lost. Alright, I'll write the intermediate step. Ok? So I have (A/x)(x^2)(x+2) + (B/x^2)(x^2)(x+2) + (C/x+2)(x^2)(x+2). |
| 15:37 | Ok, and in the first one, one of the x's cancels.
So you get 5x^2 +3x-2=A(x)(x+2) +B, the x^2's just cancel is times x+2 and C, the x+2's cancel, so C is just multiplied by x^2. |
| 16:09 | Well you want a common denominator, so what would a common denominator be?
It'll be x^2(x+2). So B really only needs to get multiplied by x+2 to make a common denominator. Right, it's already got x^2 in the denominator. Right? Common denominator is this, right? |
| 16:34 | Right.
Well no, B gets multiplied, everything can get multiplied through by the same thing. If you multiply through by the common denominator, which is x^2(x+2), and then you do the canceling, you end up here. Ok. I know, you're not totally satisfied. The common denominator is x^2(x+2) right? |
| 17:00 | You already have B/(x^2).
So you really just have to multiply top and bottom by x+2. Right? Then everything would have the denominator of x^2(x+2). So B only needs to get multiplied by x+2. Right. You're multiplying through by the denominator of the left side to clear the denominator. So then there's no fractions. So, you have 5x^2 plus- I find these problems exhausting. |
| 17:31 | Ax^2 +2Ax + Bx +2B +Cx^2.
So that's A+C has to equal 5. 2A+B=3. And |
| 18:02 | 2B=-2.
So we know B=-1. That makes A equal to 2 and C equal to 3. I'll let you check that in your spare time. |
| 18:30 | That means we can rewrite this
of 2/x plus, minus, dx/x^2 plus dx/(x+2).
Ok? Those are all very straightforward integrals. |
| 19:02 | The middle one is minus.
Cause it's minus 1. I know I'm skipping some steps. You guys have to fill in the intermediate steps. Ok? Alright, I'll do one completely. Oh god, what were they, 24 and 30? Are we gonna do 30? |
| 19:32 | Oh 31.
Alright, I'll do 31 completely. That's gonna take a few- that's gonna take a couple minutes. And then I wanna get to the new material. |
| 20:06 | Maybe you guys will get this.
x(x^2+4)^2 Alright, this one's gonna really not be fun. Are you ready? So notice, I have an A over x |
| 20:32 | then I have a term with x^2+4 so that needs a Bx+C.
But wait, x^2+4 is squared so I need Dx+E, yeah I'm not happy either, over (x^2+4)^2. Which by the way is (x^4)+(8x^2)+16 |
| 21:01 | Just for future reference.
So this is 1/(x(x^2+4)^2) is gonna be A is gonna need to be multiplied by (x^2+4)^2. So A is gonna get multiplied by this (x^4)+(8x^2)+16. Bx+C already has an x^2+4 so it just needs to get multiplied by x and x^2+4. |
| 21:36 | And Cx+D is already over (x^2+4)^2 so it only needs to get multiplied by x.
Not as bad as it looks cause remember, over here everything is 0. 0x^4, 0x^3, 0x^2, 0x. There's only 1. So maybe it won't turn out so bad when the time comes. |
| 22:01 | Alright, so I've got (Ax^4)+(8Ax^2)+16A+ Bx times x times x, oh wow. Let's see.
Who picked this problem? |
| 22:31 | And of course, the left side's really only 1.
So you get 1 is (Ax^4)+(8Ax^2)+16A +(Bx^4)+(4Bx^2)+(Cx^3)+(4Cx)+(Cx^2)+Dx, wait I lost something. |
| 23:04 | This should be Dx+E. Sorry.
Is that what you're gonna ask me now? |
| 23:33 | Ok, so far so good.
Who's lost? Everybody. So again, because you have x^2+4, you need a something x plus something. And then because it's squared, you need another something x plus something over the term squared. Then your common denominator is x(x^2+4)^2, so when you multiply through, A gets multiplied by (x^2+4)^2 |
| 24:04 | Bx+C gets multiplied by x and x^2+4, and Dx+E just gets multiplied by x. Yes Paige.
Well, the common denominator is x times (x^2+4)^2, right? So A already has an x. Right? Bx+C gets multiplied by x and x^2+4 |
| 24:32 | and Dx+E already has an (x^2+4)^2, so it only needs to get multiplied by x.
Why aren't these all together? No, never mind. Why is it Bx+C and not B? Yeah, why does it have an x on it? Why does B have an x on it? Why is it Bx+C rather than just B? Yeah. Well we did this last time. When you have a quadratic factor that can't be reduced, you need an x term and a constant term. Ok? |
| 25:09 | So if you had squared, you'd still need Bx+C.
Well, not really. You've got x squared plus something. Ok? You need a Bx+C. And if it's squared, you need a Dx+E term. So this mess turns into, let's see. The 4 power terms are A+B, they have to be 0. |
| 25:31 | The cubic term, C has to be 0.
That makes life easier. The quadratic terms are 8A+4B+D, that has to be 0. Everything is 0 except for the constant term. The linear terms are 4C+E, that has to equal 0. And then 16A has to equal 1. |
| 26:08 | So now we know A is 1/16.
And since A+B is 0, B is -1/16. We already know C is 0. And 4C+E is 0 so E is 0. |
| 26:30 | And D, let's see.
That would be, 8/16 is 1/2 minus 1/4, so that's 1/4+D is 0. So D is -1/4. Ok, I'll let you solve those again on your own. |
| 27:00 | Those are your equations.
That means that this integral, dx/x(x^2+4)^2 equals A/x, so 1/16 dx/x +Bx+C, well C is 0. So -1/16x +0 |
| 27:32 | over x^2+4
+Dx+E, so -1/4 x/(x^2+4)^2
Glad you came to class today, huh.
I was hoping for snow. I was. |
| 28:03 | So again, I just pulled 1/16 out and pulled 1/4 out.
Ok, that's why these are sitting out here and C and E both came out 0. So when you had your Bx+C and your Dx+E, those both go away. You did turn the camera on, right Michelle? Yes. And you turned the sound on? Yes. Ok good. Alright, so this is 1/16 ln|x| |
| 28:35 | minus 1/16. Alright, you should practice making sure you can do that middle integral.
Alright, that is a basic u substitution integral. Let u=x^2+4 du is 2x dx. So 1/2du is x dx. |
| 29:03 | Be sure you can do one of those integrals. Integral of x over x squared plus something.
This integral now becomes 1/2 du/u. And look, I can use the same substitution for this integral. This just becomes 1/2 du/u^2. |
| 29:37 | Some of you are nodding your heads, others of you are crying.
Ok, -1/16 times -1/2 is -1/32 ln|u| -1/8. What's the integral of du/u^2? Well that's just u^-2 du. So that's u^-1/-1. |
| 30:08 | Also known as -1/u.
So this becomes -1/(x^2+4) + a constant. And then you substitute back for u and you get (1/16) ln|x| - (1/32)ln|x^2+4| |
| 30:36 | +(1/8)(1/(x^2+4))+c.
Ok? You're allowed to hate these. If you practice the partial fractions you will get good at them. They're really just a lot of algebra. They're actually not that hard calculus wise. |
| 31:05 | It's just hard algebra wise. I can't imagine you would ask one like this on the test.
I don't know since I'm not writing the test. But it's a lot of work. I mean look at all those steps. And I skipped a couple of steps in the middle. Alright, time to learn something new. I know, I love learning new stuff. |
| 31:34 | Any other questions before we get going?
If I do have a review session it might be on a Saturday or a Sunday. I'll warn you because I might not be able to get the room during the week. Ok, if I have to choose between Saturday and Sunday, Saturday? Sunday? Oh, there you go that's easy. Alright. Who didn't vote? You didn't vote. What did you vote for? |
| 32:00 | You vote for both?
Can't vote for both. Gotta pick with Sanders or somebody. Oh wait, there's another guy running. The orange guy. Um, anyway. America's a great country. One of my ex high school students went to [inaudible] and he's emailing me and saying "haha I'm not American anymore." And I said don't come back for a long time. |
| 32:37 | He's a Canadian citizen. You know what college costs in Canada if you're a Canadian citizen?
Nothing. You even get free books. Alright. Now, new type of integral. Now this, the integration isn't actually very complicated. It's just you have to learn to think about another type of integral. So what if I wanted to integrate something very simple? |
| 33:09 | Just wanna integrate 1/x^2.
But I'm gonna integrate from 1 to infinity. You go "infinity?" What do I do if I have infinity? Ok, cause we can integrate 1/x^2, right? The integral of 1/x^2, we're gonna do it in a minute. It's very straightforward. |
| 33:31 | But what do I do about the fact that it's infinite? So this is something called an improper integral.
This is the last integration technique thing we do. After this we're gonna do applications. I know, you're very excited. You're gonna learn areas and then you're gonna spin those areas around an axis and make pretty volumes. |
| 34:02 | So how would you do that?
Well, you can't really plug in an infinity cause it's infinite. So what you do is, you make up a letter. What letter should I use? How about C for Chris? So I do- cause he was yawning just now. So what I'm gonna do is I'm gonna do the integral from 1 to c and then you're gonna take the limit as c goes to infinity. Ok, so when- I'm just gonna do the integration. Then I'm gonna do it from 1 to c and then, when I'm all done, I'm gonna take a limit. |
| 34:38 | So what I recommend is that you break this up into pieces.
So first, just do the integral of 1/x^2 dx. That's the integral of x^-2 dx. So what's the integral of x^-2? Well that's x^-1 |
| 35:03 | over -1.
I had it over there. Also known as -1/x. You all see that? You can differentiate it if you're not sure. So I'm ignoring the 1 to c and I'm ignoring the limit for a minute. So now I've done the integration. So the integral of 1/x^2, I make that x^-2 and the integral of x^-2 is x^-1/-1. x^-1 is 1/x. |
| 35:34 | Now, I'll do the limit as c goes to infinity of -1/x from 1 to c.
That is, you plug in the c you get -1/c + 1/1. |
| 36:01 | And now, what's the limit as c goes to infinity of 1/c?
Can you see that down here? Well, as c goes to infinity this limit is 0, 1/c. This just equals 1. Ok? So that integral is 1. Because it's minus... where is it? I plug in 1, it's 1/1 but it's minus minus so it becomes plus. |
| 36:30 | Remember it's the top one minus the bottom.
I'm plugging in c, I get -1/c Minus -1/1. Then I take the limit. We'll do a couple more of these. Don't worry, I won't abandon you on the first one. Ok. Yes. So, this convergence to 1. So the magic thing you're trying to figure out is whether the integral converges |
| 37:01 | or diverges.
And for those of you who are gonna take MAT 127 you're gonna see converging and diverging a lot. So one way to think about converging is if you imagine some horizontal asymptote somewhere. Ok, it means that when you calculate the area as the top gets bigger and bigger and bigger it approaches a horizontal asymptote but it doesn't get above that horizontal asymptote. That number, that asymptote is what it converges to. |
| 37:33 | Ok, that means it converges if it has a number that it gets closer and closer to without actually getting there.
Ok? So if you did from 1 to a trillion, you'd get very close to 1. Ok? If you did 1 to a google, you'd get really close to 1. But you wouldn't actually get to 1, you'd just get lots of 0.999's all the way up. Ok, if it diverges, that means the area just gets bigger and bigger. |
| 38:01 | Ok, so you really just need to figure out, does it converge?
Or does it diverge? Ok? Let's do a couple more. I'm having problems with this book. Cause it lacks for examples. |
| 38:33 | Alright.
|
| 39:04 | What if I do the integral of 1 to infinity
of dx/(x+4)?
So again, you rewrite this with a limit. |
| 39:30 | Let's see.
I'd give partial credit for writing the limits. But, you know depends on the professor. Let's make up a new letter. How about m? It doesn't matter what letter you pick. I advise against s. Ok? So the limit as m goes to infinity of 1 to m of dx/(x+4). |
| 40:09 | Alright. So again, ignore the limit and the limits of the integral for a minute and just do the basic integral.
So the integral of dx/(x+4) is ln|x+4|. Ok? So this is gonna become limit as m goes to infinity |
| 40:35 | of ln(x+4)
from 1 to m.
That's gonna be the limit as m goes to infinity of ln(m+4)-ln(5). |
| 41:05 | Don't need the absolute value bars cause the absolute value of 5 is 5.
You don't get partial credit if you know that the absolute value of 5 is 5. Sorry. Ok, now what happens as m goes to infinity to the log of m+4? You do the log of infinity, and that's infinity. Right? So this becomes infinite, so this diverges. So you're done. It doesn't matter what the other number is. |
| 41:33 | Ok? So once you get to the integral, and part of it becomes infinite, then the whole thing is infinite and forget it.
Ok? Yes. You should write diverges. Ok? Well it's tricky because divergence, you have to do infinite sums. Divergent series can have all sorts of different solutions when you sum them. |
| 42:02 | Ok? So write diverges.
Like the movie. Ok, convergent, divergent, allegiant. Insurgent. Detergent. Now, here's a rule of thumb. This isn't true for everything, but if you have If you have that kind of integral, ok, and by the way, x^4 could just as easily be x. |
| 42:35 | Ok?
If Ok? You will find that that is generally true. Ok? Oh, sorry. Now be careful, because there's lots of other types of integrals we can give you. |
| 43:03 | But if you have the integral from 1 to infinity of dx/x^p,
and p is less than or equal to 1, it'll diverge, and if greater than 1 it'll converge.
Alright, let's have you guys practice one. |
| 43:45 | There you go, let's see if you guys can do that one.
Be careful. |
| 47:43 | So again, you would wanna write this with a limit.
Let's see, how about I use b. Later, somebody's gonna ask why I use different letters. |
| 48:01 | Answer is, why not?
There's no reason to use different letters. You can use the same every time. As I said, I would stay away from letters like x, y, and z. Don't you hate it when cell phones ring in class when you're talking? Ok, so I would make this That's The Beatles. You guys are too young. You don't remember The Beatles. |
| 48:31 | Alright.
So now, let's just integrate dx/(x^2+1). That's inverse tan of x. Ok? There's no x up here. It's not a log, it's just x^2+1 on the bottom. So this is the limit as x goes to infinity. B goes to infinity. Of tan inverse of x |
| 49:05 | from 0 to b.
Ok, so that's the limit. As b goes to infinity, tan^-1(b) minus tan^-1(0). What's the tan inverse of 0? Tan inverse of 0 is 0, because the tangent of 0 is 0. Right? |
| 49:32 | Here's the fun part. What's the tan inverse of infinity?
Who got that? Tan inverse of infinity is π/2. Because the tangent of π/2 is infinity. And why would I do a problem like that? Because I hate you, and I wanted to make you suffer. Because you should expect something like that, ok. It's fair game. So, when you see infinity, the tan inverse of infinity is π/2 because tangent of π/2 is infinity. |
| 50:08 | Ok. That was entertaining.
Let's do some more. Yes. This converges. |
| 50:30 | It converges to π/2.
Ok, if you get an answer it converges. It's only when you get infinity that it diverges. Alright, what if I had |
| 51:11 | integral from -∞ to 0 of xe^x?
You guys wanna try this on your own first? No? Yes? We have some no's on the left, we got some yes's on the right. Some maybe's in the middle. |
| 51:31 | Well let's see. First I should set this up.
I will use "a" this time. Should we do -∞ or ∞? Let's see. We'll do -∞. a, 0 xe^x dx. So I'm gonna have to integrate |
| 52:02 | xe^x dx. How do I integrate that?
Integration by parts. So much fun. Hey, we taught it to you for a reason. I know, you really wish you were still on vacation. Working on the tan. |
| 52:30 | Well, ok why don't you try integrating that for a minute? Make sure you can integrate xe^x.
I would let u=x. But you know, you can do what you want. Paige, you got it? Hey. No? Let u=x and v=e^x dx. |
| 54:22 | You got it?
Who got it? |
| 55:00 | Alright, that's long enough.
So let's let u=x dv is then dx. du is, sorry, dx. dv is e^x dx. And v is e^x. So you can rewrite this as u times v minus the integral of vdu. |
| 55:31 | And that is
xe^x, and the integral of e^x is e^x.
Ok? So this becomes limit a goes to -∞ of xe^x - e^x from a to 0. |
| 56:01 | Plugging in 0 is always fun.
So this is limit a goes to ∞ of 0 - e^0 minus ae^a - e^a. |
| 56:32 | That's to -∞ not +∞.
Sorry. Ok. So -1 is easy. When I do -∞, what is e^-∞? That's 1/e^∞, which is 1/∞ which is 0. So this is gonna go away. That's just -1. So this is the only tricky part. So how do you integrate ae^a as a goes to -∞? |
| 57:10 | I'm sorry, not integrate. How do you take the limit?
|
| 57:32 | Got any ideas?
Well, what you can do, it's gonna come out 0, is you can make this the limit as a goes to -∞ or just a plain old ∞ of -a/e^a. |
| 58:13 | That make sense?
Because -∞, right, that's -1/∞. If you do it this way, then you do L'Hopital's rule You remember L'Hopital's rule, just take derivative of the top and bottom and that equals 0. |
| 58:35 | So this is 0, this is 0, the whole thing comes out to -1. Yes.
Well you can't do 0 times infinity, ok, that's the problem. So you have to make it, you have to be able to figure out the ratio. But as a general rule if you have a term like x or something in the numerator, and you have e^x in the denominator |
| 59:03 | as you go to infinity it's gonna go to 0. Cause the e term is gonna get there, is gonna get much bigger than the numerator. Yes.
Here can you factor out the e^a? Oh here? Yea, sure you could take it out and you'd end up in the same spot. You would. Try it and see. Alright. |
| 59:30 | We'll do one other type and then Wednesday we'll do some more of these cause these are way too much fun.
Well that was fun. Ok here's one for you guys to do. |
| 60:02 | There you go, that'll keep you busy for a minute.
|
| 60:33 | Now, think about this geometrically.
What does that mean if I'm doing from 2π to ∞ sin of theta? Right, got the area, and I'm gonna have a piece below. Piece above, then a piece below, piece above. Right? So aren't they gonna all cancel? Well what if I had one left over at the end? How would I know? How do I know where I stop? |
| 61:01 | Do I know where I stop? It's going to infinity, there is no stop.
Right? That should diverge. Ok? It should diverge cause it's just, if you imagine this well Every one of these areas cancels with every one of these areas. This is the problem with divergence. That's why you can't really write infinity or things like that. |
| 61:30 | So each one of those is gonna cancel, but the problem is they're going to infinity.
So there really isn't a stopping point. So if I stop there I would get a positive area. If I stopped, well, I can't ever really get it negative. Right, if I stopped here, I would get a different area. Well, do I have more negatives or do I have more positives? You don't really know and therefore it's gonna diverge. Now, you can integrate it and see. Ok? Where do I wanna do it? I'll do it over here. |
| 62:12 | So the limit as w goes to infinity
2π to w sin(θ) dθ
is the limit as w goes to infinity -cosθ from 2π to w.
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| 62:34 | That's the limit w goes to infinity -cos(w) + cos(2π).
So you need to know cosine of 2π is 1. The limit as w goes to infinity of cos(w) is infinity so you can stop. Ok? So this one is diverging. |
| 63:02 | Is the new movie any good?
Cause you know, the first one wasn't so great. The second one wasn't so great. The books were better. Somebody must've seen them. Ah, what is cosine of infinity? It's divergent, you don't know what cosine of infinity is. Right? It just oscillates so it goes on forever. Well it has some value, but you don't know what the value is. Since you can't pin it down you can't say what the integral's gonna be. |
| 63:32 | Ok? As I said, you can't, you can't pick when the area's gonna come out.
Ok, I think that's enough for today. We'll do some more of this stuff on Wednesday. |