 We are at lecture 7 today. The goal today is to wrap up improper integrals. In the textbook it's chapter 5, section 10. So we have kind of the type 1, not that type 1 is anything to be remembered, but where one of the limits of integration is some version of infinity. Either we're starting at a number and headed toward positive infinity, or we're starting at negative infinity and going up to and including some number. So the type today will be different. Similar in a sense that it's still counterintuitive, that we can let something run all the way to infinity and expect to find the area under the curve. It is possible some of the time kind of depends on the rate at which it's approaching the asymptote. So we have vertical asymptotes today. So here's what we had in our type 1. So we had some curve beginning at some value, let's say it starts here at a, and we've got a decreasing curve in its asymptotic to the x-axis. And sometimes we can find the area, finite answer, therefore it's a convergent integral. Sometimes we can't. So you can either say it does not converge or it diverges. That's kind of the visual that goes along with type 1. The visual that goes along with type 2 is that we've got a curve that is undefined at a certain value. So we've got a vertical asymptote and we're either going to, I'll just draw one version of it, come away from the vertical asymptote up to and including a certain value, a. So notice we're not headed toward infinity in terms of x, but we've got some, I'll call that b, b for bad. We've got some bad value for the integrand that causes the integrand to be undefined so that probably for most cases is in fact a vertical asymptote. So similar in the sense that we've got this asymptotic effect and we're letting it run all the way up against there, is it possible to find this area if so it converges if not it diverges. So let's start with a type 2 problem. And I think I've commented on what we call things that we do in math class. It's probably not a good idea to call them problems because it already has this connotation that it's going to be some negative encounter. So it's really an opportunity. So let's look at this opportunity to explore this type 2 type of improper integral. I'll probably call it opportunity today and then every other day I'll call it a problem because sometimes they are. Do you see why 3 is the bad value? So we don't see the positive infinity as a limit, we don't see negative infinity as a limit. Say how it just looks kind of mundane going from 3 to 8, 3 is a problem. So with this function 1 over x minus 3, x equals 3 is a vertical asymptote. So we're going away from 3 up to and including 8, 8 is not the issue. In fact any other number really here between 3 and 8 is not the issue. 3 is the issue. So we do what we did yesterday on improper integrals and that is we delay the bad value. The bad value here is 3. So let's give it some temporary name, capital A. So we're integrating from capital A to 8 and we want A to approach 3. We have to be a little more, it is 3 but we need some more, 3 from the right, 3 from the right. Because we're starting at 3 and going to 8 so we're going beyond 3. So we don't want to approach 3 from the left, that would cause us to cross over the bad value. We want to start right on the other side of 3, the 8 direction. So in the direction of 8 we better start with numbers that are a little larger than 3 and come in from that side. So the picture is very similar to, if you can kind of visualize the picture, here's 3, here's 8. So in this picture, when x is 8 what do we get? 1 5th. So it's coming way down here. So we want to come in from this side of 3, so from the right side of 3. That's not that big of an issue to be honest with you but since it is a bad value we probably ought to clarify which side of that bad value we're converging from. The integration part I don't think is a bad one. What's the integral of 1 over x minus 3 integrated with respect to x? Natural log. Absolute value, hopefully that won't come into play. So let's go ahead, we integrated, now let's go ahead and evaluate. That's just going to be something numerical. So now we've done everything we can do. We've integrated, we've evaluated, we've simplified as much as we can simplify. Now it's a matter of checking out what happens to the term that has our temporary letter in it, capital A. What happens to this term as A gets closer and closer to 3 from numbers that are slightly larger than 3? I guess it comes into play here. If we had numbers smaller than 3 we'd have a problem wouldn't we? I mean not like this is a problem. This is a problem, okay? Because we'd have a natural log of a negative number which is not a real number. So we want to have, we want to be talking about real numbers. So we've got to have a number a little bit larger than 3 to even discuss this. And as x, excuse me, as A approaches 3 from numbers that are slightly larger than 3, this is gradually approaching the natural log of 0. Think about a natural log graph. What is the natural log of some number that's getting very, very, very close to 0? Negative infinity. Here's a natural log graph, right? The mirror image of y equals e to the x on the other side of the line y equals x. There's a natural log graph. So as x gets closer and closer to 0, what's happening to the natural log? It's headed down toward negative infinity, right? So it may not look like it when you first look at it, but if you kind of think through it and say I'm letting A approach values that are very close to 3, but yet slightly larger than 3, this thing inside here is getting closer and closer to 0. The natural log of 0 itself is not defined, but as it gets close to 0, the natural log is a larger and larger number in magnitude, but it's negative. So this term approaches negative infinity, which means what is the decision on this problem? This opportunity. Diverges. Even the fact that we've got this nice number, natural log 5, it's not going to help the fact that we've got negative infinity here. So our integral diverges. We've got a little musical something going on. What was that? Questions on this one? Okay, see if you can guide me through this one. Very similar. I don't know that the answer is similar. That was divergent. I truthfully can't remember what this one is. My guess is that it converges, but that's a guess. So this is the type 2. Why is it the type 2? There's no infinity that's actually visualized here. There's no negative infinity, no positive infinity. So if it is improper, which it is, it's got to be the second type. And what's the so-called bad value? 2. Because it causes the function in the integrand to be 1 over the square root of 0, which is 1 over 0. So there's our bad value. So what's our first step now that we've identified the type and the bad value that goes along with that type? Okay, get rid of it. Delay that till the end of the problem. 2 being the bad value. And okay, that's the next issue. We want B to approach 2 from which side? From the right, because we're heading into the direction of 4. So in a sense, we're kind of coming back from 4 toward 2 in our analysis. Not a natural log. If it helps, you can think of this as x minus 2 to the negative 1 half, right? It's only going to be a natural log when it's something to the negative first. We've got something to the negative 1 half. What is the integral of... We need to do any corrections? This is the kind of stuff I've mentioned thus far in this class that if you're not doing it in your head, I think it's probably a pretty good goal. U is x minus 2. We need to multiply 1 over 2 square root of x minus 2. I like some of that. I don't know that I like all of that. 2 square root of x minus 2. I heard a 2, right? 2 square root of x minus 2. So the 2 was really division by 1 half. So we've got u to the negative 1 half. Do we also have du? Yes, because u is x minus 2. Derivative of x minus 2 is really 1 dx, right? So we don't have to do any corrections. So this is u to the negative 1 half, which is going to be u to the 1 half over 1 half, right? So division by 1 half is the 2 that was mentioned a minute ago. So we have integrated. We've got a 2 out in front. That's the division by 1 half. We've got an x minus 2 to the 1 half, which is just square root of x minus 2. Okay, I think we're all right. Let's plug in our values. Simplify what we can. 2 square root of 2. Here we don't know what this is until we start plugging in numbers that B is approaching. So if that's ugly, we've probably got a divergent integral. If that is something we can evaluate and in fact disappears, then we're good, right? It converges. What about that one? What happens to the square root of B minus 2 as B approaches numbers that are slightly larger than 2? Close to 0, right? So this is slightly larger than 2. We're going to subtract 2, so we don't have much. The square root of something practically 0 is practically 0. So this term is headed toward 0. So we do get an answer. 2 square root of 2, it doesn't really matter what that finite answer is. It matters that it's a finite answer. Therefore, the integral what? Converges. Question on that one. So two very similar looking problems in a sense that we had this asymptotic effect. This one apparently approaches the asymptote in such a manner that we can come up with an answer and the other one did not. Questions? Issues. All right, third example. I'll kind of jump start you on this one. And then see what you think needs to be done after that. Some days in here, sitting down kind of actually gets to me. I'd like to be standing. But part of the reason I don't use that a lot is for me to write large enough where everybody in here can see it, I don't have a lot of screen to work with. And part of you that were in here last year know that there's some more reasons probably that I'm not technologically that savvy, but we won't, you know, we'll just kind of keep that a secret. So most days I'm kind of antsy today. I'm so thankful to be seated because I played full court basketball last night and it was a college length court and I'm kind of old, you know? And so today when I was coming down the stairs I was holding both rails this morning in my home as I was coming down. So it's a little better now as I walked over here from Harrelson but I'm glad to be sitting down today. Here's my jumpstart on this problem. Let's make a picture over here too. Everybody recognizes Pi over 2 for a bad value for the tangent function, right? Tangent of Pi over 2 is undefined. So we've got this visual image of the tangent. Tangent of 0 is 0. Tangent of Pi over 4 is 1 in the tangent graph. So I hope that's this visual image that you have of the tangent. Is it possible to find the area under that curve? Again, we look at the picture and we think there's not but we really don't know until we evaluate this integral. So the bad value, let's save it till the end of the problem, is Pi over 2. Now some of you probably don't need my help on this problem. My first step of rewriting tangent as sine over cosine, you may say I know how to integrate the tangent. I've already done that. You may have spent a lot more time with that in your Calc 1 class. We didn't but we have a way I think of doing it is to rewrite tangent as sine over cosine. Does that help? What would be the method that we're going to integrate sine over cosine? U substitution, let U equal cosine. So we're going to do this off to the side here. Now if we let U equal the cosine, where are we headed with this? We're going to have a U in the bottom. What's the top going to be? Du kind of, right? With a slight correction. So if we're eventually going to have a du over u, what's that mean? That's going to be a natural log integrand. So with our choice for u, cosine of x, what's derivative of u? Negative sine x. So is the numerator du? Yeah. Almost. Negative. We need a negative there and there. So we're stuck with a negative one out in front. Now that we do that little correction and compensation, we've got du over u, which is kind of where we thought we were headed and what's the integral of du over u? Right. Natural log of u, right? So for us our u was cosine, so we're integrating sine over cosine dx and we're getting negative one times the natural log of the absolute value of cosine. I think I know what your question is, Nicole. Why didn't we put pi over two from the left? Oh, we need to. Yeah. Right? We need to clean that up. Thank you very much. That's not what I was thinking. What's going to be your question? Because we're starting at zero and headed toward pi over two. So we're coming in toward pi over two from the left. Thank you. Did anybody else have an issue with how we left this answer? That's where I thought Nicole was headed. Negative one, natural log of cosine of x. This integrand actually has another answer. I'm already too cluttered on this page. Close. Can't we take a coefficient and put it back in this position in terms of the exponent position? I mean, we do that all the time. If you have the natural log of two cubed, isn't that the same as three natural log of two? Right? If it works that direction, it certainly ought to work the other direction. We could take a three that's here and basically replace it in the exponent position. So this ought to be equal to natural log of cosine of x to the negative first. Well, what is cosine of x to the negative first? Seek it. So that's where I thought Nicole was headed with her question. If you don't, that doesn't look familiar. Negative one times the natural log of the absolute value of the cosine of x, maybe it's because you've used that other version of the answer. Do we need to do that? No. No. All right. So we integrated it. We now have it integrated. We need to evaluate. So the upper limit is q, so we don't really know what's going to happen there. And the lower limit is zero, so we plug it in there for x. We get cosine of zero. That, I kind of hesitated there because I thought we were going to have a problem with that as well, but we're not. Because the cosine of zero is what? One. And the natural log of one is zero. So that term just becomes zero. So that becomes minus, this is zero, negative one times zero is zero, so minus zero, I'm not going to write it down. So now we've simplified everything we can. Let's throw in the value of q coming in toward pi over two from the left. That happens to, that gets very, very close to zero, right? Cosine of pi over two coming in from the left is positive, right? That's good. But it's getting closer and closer to zero. So if this, in fact, is very, very close to zero, what's the natural log of some number that's very, very close to zero? Negative infinity. Negative infinity, again, right? So here's our natural log curve. So as we're getting closer to zero in terms of x, or in this case, q, then the natural log is getting headed down here toward negative infinity. So this piece right here is negative infinity, and then multiplied by negative one is not going to help the cause a whole lot. Now it's going to be positive infinity. So in either case, negative infinity or positive infinity for our answer, we don't get an answer. Basically, there is no limit. Therefore, the integral diverges. So back to our picture. Is it possible to find the area under the tangent curve from zero to pi over two? No. So the answer keeps getting larger and larger and larger, and in fact, it's infinitely large the further or the closer that you get to pi over two. Questions on that problem? Yes, sir. There's like three dots you're putting in triangles. That's the answer. Yeah, that's a kind of a math lingo thing for therefore, which I always say should be there three, because there's really only three dots. But that's a therefore, the integral diverges. Anything else? A lot of times you'll see, this is kind of an old fashioned math thing, you'll see a QED at the end when we're done with the problems. That tells you that I'm old again. You probably have never seen that at the end of a math problem. It's Latin for it has been demonstrated, okay? So we won't be putting our QEDs at the end of the problem or proof. Most of the time it's at the end of a proof. Questions or issues? Don't laugh at me because I'm old. That's not nice at all. I'm going to write that down. That'll hurt her grade. Okay, I think we're going to have time for this. This is slightly different because the limits, neither one of them is a bad value. Zero's okay, right, in this function. One's okay in this function. But unfortunately, as we travel from zero to one, we encounter a bad value. What's the bad value? One fourth. One fourth. So let's break this problem up into pieces. Zero to one fourth, as long as we pick back up at one fourth and go to one. That's legal, isn't it? So we like it when there's an infinity there. That's pretty easy. That's type one. Type two, normally, kind of the easier version is when one of the limits is itself a bad value. This is kind of the only other thing we're going to have to deal with is when we travel from starting point to ending point, we encounter a bad value. Let's look at this one. If we determine that this one diverges, can this one save it? No. So if you look at either one and it diverges, you're done. The integral itself diverges. So I think we can go kind of quickly. There is a little minor correction on one of these that we haven't had today, but we've had numerous times before. So we've discovered one fourth is the bad value. So we're going to go from zero to r. r is going to approach one fourth from which side? From the left on the zero side, which is from the left. So temporarily we're ignoring this one. All right, let's do the integral problem out here to the right. Let's see if we can do it on our heads. Okay, now you said one fourth. Where'd the one fourth come from? I put it out at the beginning where you could substitute it. Okay. So we're going to have a mix in here of kind of what you're used to. You're probably Jacob, right? Yes. Jacob is saying if my u is 4y minus 1, what is dy? Well, it's a fourth of. Am I correct in your... I pulled the one fourth out. Okay. This is a fourth of du. Because to have du what do we really need? Don't we need a 4 dy? So I'm used to doing this. So if you're kind of used to this process, I need a 4. So I can't just manufacture a 4 without also manufacturing a one fourth. So this is now du. And this is now 1 over u. So are we getting closer to the point where those are being done in your head? Yes, would be a good answer to that. Alex. So you're saying like, if we're to the point where we can take the integral in our head like on a test, we can just write straight... Yes, especially these that we've seen in calc one repeatedly and we've also seen here in the first seven days of class. I'm not going to actually show how. Just kind of make sure and be careful that you're not saying, I need a 4 out in front. No, you need a 4 in the integrand. Therefore you need a one fourth out in front. So we've got a one fourth. We had a 1 over u. This integrated to the natural log of u. Natural log of 4 y minus 1. What do you think? Converge or diverge? Any guesses? Converge. Converge? It'll be l into 0, so it's going to diverge. It's going to diverge. Because when we put in a fourth, 4 times a fourth, actually a number that's a little tiny bit smaller than a fourth. But practically a fourth. This is going to be practically 1. Right? A little larger than 1, so we're legal. But practically 0, natural log of 0 for the third time today is negative infinity. That's the first time this is coming to play. What happens in the other one? We end up with a 0 minus 1, but thankfully we have the absolute value of that. Right? So we can actually not scrap it and go to the next problem. Absolute value of negative 1 is 1. The natural log of 1 is 0. I don't know if it's you guys just being quiet, but when I ask for the natural log of something, tell me what you think, because I can't read your faces yet. There you go. If you're not thinking that, that's what you should be thinking. Because when you're asking for the natural log of something, you're asking for the power that you would raise E to get that thing. Is that right? The natural log of 8 is the power that you would raise E to get 8. That's what that means. These are base E logarithms. So you're looking for a natural log as an exponent. Any logarithm as an exponent. So you're looking for an exponent. The base is E. So you're looking for an exponent or a power that you would raise E to get that number. So the quick question you ask yourself is what power would I raise E to get 1? That's 0, right? E to the 0 is 1. Doesn't that seem odd that something to the 0 is 1 if you ever kind of thought, you know, math is kind of weird, but that's really weird. E to the 0 is 1? How is that? Maybe we need to change the rules. Somebody give me a valid reason why something other than 0 to the 0, so 1 to the 0, E to the 0, 5 to the 0, why is that? Couldn't you do, I mean, you could always add something to an equation multiplied by 0. So I mean, never mind, I wouldn't refuse. But basically it seems like if you didn't have it that way, then it would end up changing equations just by... Oh, it would change a lot of things. Absolutely. Well like, if you differentiated E to the 0, you'd just get 0. Okay, I can go along with that. If you differentiated it, if you differentiated E to the 0, you'd get 0 times E to the... Well, I don't know if we're going to get there with that. Let's go an easier route than that. What if you had E cubed over E cubed? What is anything non-zero over itself? Everybody agree that's one? But don't we have another way of handling division by two things that have like bases? What can we do with their exponents? Oh, there we go. But what is E to the 3 minus 3? It's E to the 0. So it kind of fits in with some other rules that we, you know, we don't ever question at all. We never question that. You have division of two things with like bases. The exponents, well, if they're the same exponent, when you subtract them, you get 0. And do you see why 0 to the 0 is a problem? 0 to the 0 cannot be validated like this because then you'd have 0 over 0, which is an indeterminate form, and we can't say what that is. But this one, we can back it up with this. There are other good legitimate reasons, but something to the 0 that's non-zero is going to be one all the time. But when you're asking yourself, what's the natural log of something, ask yourself in a way that you're trying to find the power that you would raise E to get that number. All right. So this becomes the natural log of 1. The natural log of 1 we decided is 0. So we've got a minus 0 here, so I'm not going to write it. Now we're ready to throw R in there, R getting closer and closer to 1 fourth, which means this thing gets closer and closer to 1. So we've got something that's larger than 1, but practically 1. We're subtracting 1 from it. So it's very, very close to 0. So again, we're back to what Jacob said. This thing is getting closer and closer to the natural log of 0, which we've seen several times today is negative infinity. So our integral diverges. Now back to the original problem. We split it up into two pieces. The first piece diverges. No reason to go on to this one. Okay? So the entire integral is diverging. Okay, we started some things yesterday with the comparison test with one of my inane examples. So if we have, let's say this is, I don't know what I called the top curve, F or G? F. And we've got another curve that's underneath of it. So we're going to actually come up with some... Now the bad thing about this test is we can decide convergence and divergence, but if it converges, we don't know the value to which it converges. So it's a decision only, no actual numerical value. So we decided that if F of X, the area under F of X from A to infinity converged, and we have this other curve that is less than or equal to in value over the entire range. Now that's, by the way, that's not absolutely necessary. I'll come back and qualify that in a minute. But for the part of the picture that we're looking at now, from the same starting point all the way out to positive infinity, if this is true, then we should be able to determine that from A to infinity, what is the decision about the area under the G of X curve from A all the way to infinity? It's also convergent, right? And the stupid example I used yesterday is small. If we decide what small is and someone comes in the door that's even smaller, don't we put them in that same bucket, the small bucket, right? They're in the small bucket, okay? So if we can determine that a function is smaller than an existing convergent integral, it must also convert, Katie. If athletics is greater than zero, does G of X also have to be greater than zero? There are some ways that this picture is not exactly this way, and we'll get to those after the kind of the basics, I hope, today. But hopefully I'll address our question in a little bit. Now, let's take the same picture. So we've got this upper curve, and we've got some curve underneath of it. Let's say that we know from A to infinity the one underneath, the smaller curve, the lesser curve, is divergent. We know that the G of X, same relationship, is underneath the F of X. So if G of X is divergent, divergent, you think of the word big or tall or something, right? It's too big to come up with an answer. Therefore, that's why it's divergent. So this one is already divergent. What about the one that's in worse shape than this? It's also going to diverge. So this one is big. I think I used Todd Fuller yesterday in my example. Seven foot tall. Here comes somebody else that's even taller. If we've already put Todd Fuller in this big bucket, aren't we also going to put the next bigger person into the big bucket, right? Bigger than big is what? Big, isn't it? That's big. I mean, you wouldn't say bigger than big is tiny. I mean, that's kind of absurd. So if this is already divergent, we've got one that's even bigger than that. It's going to also be divergent by this comparison test. Now, here's what we can't decide. We've got one that is convergent. We've got another one that is slightly, slightly larger than that. Does that mean it's also going to converge? Because if something, we've got this threshold. We say, here is small to continue with my poor example. Here's small. Everything under here, we also consider small, right? So if this converges, so does everything underneath of it. I don't know where large is, but let's say large is up here, okay? This diverges. Doesn't everything above that also diverge? Yes. What about in here? You don't know. So we can't make judgments about every problem we encounter just smaller than existing convergent integrals also converge. Larger than existing divergent integrals also diverge. These, we don't necessarily know, and the test isn't going to be that sensitive, so to speak, to make a decision on these. All right, what do these look like? So we don't want to know if it converges what value that it converges to. To which value that it converges. Do we have anything that we've made a decision on that is similar to this that we know for a fact converges and this function is in fact smaller than that? There we go. So isn't this smaller than 1 over x squared? Make sure that we're ready to go beyond that. Smaller means it has a larger denominator, right? So if we take this term and add something to the denominator, it makes it smaller over this entire interval from 1 to infinity. Because this is true on this interval, then we can go ahead and make our conclusion because I guess I should say this. That converges. We've actually done this problem. I think we did from 2 to infinity, but it's basically the same problem. Or we could say this is one of those integrals with 1 over x to the p, and p is larger than 1. We summarized that too. So we've got two kind of validations why this converges. Therefore, this one also converges. And if we need to put a justification, we could just say by this comparison test. Chandler. Now will the value of that be smaller than 1 over x squared? Probably. But that is part of where this test is not going to give us that value, but we could make that guess that that's true. Questions on that? Not a whole lot of justification. It's just a matter of... I mean, this is probably the key statement here. The one in question is smaller than an existing convergent integrand. Therefore, it has to converge. So we have a simpler existing divergent or convergent integrand that we could work with and compare this to it. The other one we added to the denominator, so it probably is headed toward convergence, right? Here we're adding to the numerator. So, right, it's probably going to be... There we go. 1 over x is the one that we already know something about. And this... No, that's not correct, right? It's the other way around. This one is even larger than 1 over x. Scrap that. If you add something to the numerator, you're going to make the value of that fraction larger. Because this is already divergent, again, we've got two reasons to justify it. One, we've already done this problem, kind of. We did 2 to infinity. Very validated that it was divergent. So this is the 1 over x to the p. P is 1. When p is 1, it diverges. And because the 1 in question is larger than 1 that's already been proven to be divergent, we know this one also diverges by the comparison test. Okay, remember that question. I drew a diagram that probably overly simplifies this at the beginning of class, where we've got one that's completely under the other one for the entire duration of the interval. And we'll address that along with Katie's question about being above the axis and below the axis. Just to clarify that it doesn't have to be exactly like that picture in order for it to work. So we don't meet tomorrow. Monday is a holiday, so I will see you on Tuesday.