 Welcome back. We're lecture six today. We left a problem kind of pending yesterday I think we set up all the arithmetic for it in a Simpsons rule. It was problem 10 421, does that page sound right? What kind of decimal approximations were we able to come up with? I've got one, but I'm not going to tell you tell I get one from you and see if I'm in the ballpark because I was Okay 0.898 014, okay, that's what I got to any problem with that decimal for the Proximate area under that curve or is it 1 over 1 plus t squared plus t to the fourth right from 0 to 3 Just to See that Simpsons rule kind of gives a better number the I Looked up the trapezoidal approximation is point eight nine five probably pretty good Different in the third decimal place and the midpoint Approximation was point eight nine five also so seems to be that This is giving is a different answer probably a better answer I Also took a few moments to try to generate the error bound for this But I didn't make it to the fourth derivative I started and got to Into the second derivative. It's kind of an ugly series of derivatives So let's just suffice it to say that it's a pretty good approximation. Okay. I didn't maximize the value of the fourth derivative This would be interesting I think interesting to me, maybe not to you Take this problem That we were given our old friend do it Okay, put that problem into maple and see what it gives you I Kind of like to see how accurate our point eight nine eight. I feel pretty comfortable with those three decimal places Maybe not quite so comfortable with the next set of decimal places, but take a few moments and Do that and I kind of wonder to what maple is actually using If you did this problem without the Limits of integration and you did an indefinite integral That might give us a clue as to how maple is actually doing this problem And then we throw in the limits of integration in it I think I'll run that through report on that in tomorrow's class Okay, yes To evaluate the integral because you point eight nine five three seven. Okay So it doesn't do it very accurately on calculators, I don't think because it uses just It doesn't use as many rectangles Okay, so that would lead me to believe I'd have to look at the Manual that goes along with the calculator, but sometimes you can it'll tell you how it's doing a problem Like a lot of manuals and calculators will tell you that it's obviously your calculator doesn't know what a sign is You know, it's not a very smart machine. It's been kind of Programmed to do problems certain ways. So it's doing Signs and we'll do this as part of this course Taylor series and Taylor polynomials It's doing a sign in that fashion So I don't know if it goes to like the seventh power the ninth power the 13th power or whatever but that leads me to believe that that calculator is probably doing either midpoint or Trapezoidal approximation, which is kind of interesting that you find this stuff out after you You know, you study this stuff and then you say, oh, well, that's what my Calculators been doing So I don't know I just mentioned something that we'll touch on later and some of the things we'll do today We'll come back to revisit them later in this course when we get to chapter 8 slightly different but similar Sign of X for example, I just mentioned a Taylor series or a Taylor polynomial You don't need to know this now, but since it was brought up Sign of X is that an even function or an odd function? That's an odd function, right even functions have symmetry with respect to the y-axis What type of symmetry does sign have? Not the x-axis origin symmetry with respect to the origin if it adds symmetry with respect to the x-axis It wouldn't be a function, right? So symmetry with respect to the origin. It's an odd function so we will discover with Taylor and Maclaurin series and generating Taylor and Maclaurin polynomials if any of you had those before in a math class You've had some Taylor series anybody else It does involve higher order derivatives. That's how you get to these But there's the pattern that your calculator is probably using to do the sign of let's say pi over four So whatever goes here pi over four Pi over four goes there pi over four goes there and there and there so I don't know how Deep into this process your calculator is working Whether it goes to the ninth Power over nine factorial or the 11th power over 11 factorial But this is a way of evaluating the sign of an angle that clearly That's not trigonometric. In fact, that's pretty polynomial looking in it x x cubed x to the fifth x to the seventh Kind of looks like a polynomial. And in fact, that's Where it gets its name the Taylor polynomial or the Maclaurin polynomial, so that's what I meant when I said that's how your calculator is doing the sign. We'll revisit that in Chapter eight actually near the end of this course All right new material today I guess any further issues with Trapezoidal approximation or Simpson's rule Approximation for area under a curve. Okay, so the next section deals with improper Integrals, there are two types of these. Well, I know get to the first type today hopefully get a fairly thorough treatment and then I don't know how the time is going to work out, but we may get to the type two Today, if not, we'll have plenty of time for that tomorrow So what is an improper integral an improper integral of The first type We're used to having a number as the lower limit of integration and another number as the upper limit of integration you wouldn't think a problem Was going to be an issue if you Encountered this we're integrating 1 over x dx from 2 to 7 it's starting it to it's stopping at 7 We're going to have a numerical answer. It's the area under this curve. In fact, we know what that is It's the natural log right of 7 minus the natural log of 2 or The natural log of 7 halves would be a way to put those together But what we're going to do with improper integrals is we're going to go from 2 And in fact, we're not going to stop it at all on the other end We're going to let it run all the way to infinity kind of seems improper, right? Doesn't seem like we're going to be able to find the answer to that Sometimes these problems have an answer and sometimes they don't and we'll try to look at some examples today Where they have an answer what that curve looks like Ones that don't have an answer what that curve looks like so we'll get some visuals not that visuals are going to tell us the answer But there's our curve 1 over x and we want to go from 2 All the way out to infinity. We're not going to close it off on the other end I know what I think I know what I thought when I saw these the first time I thought well, there's no way it's going to have an answer because we're not stopping it at a certain number None of these problems is ever going to have an answer This one in fact is not going to have an answer, but let's evaluate why it's not going to have an answer Let's slightly change it and The curve looks very very similar to this one and in fact it will have an answer So how do we approach? This problem so that we can evaluate things kind of in the right order and let things approach Infinity when they need to be approaching infinity the first step is to For lack of better terminology is to delay the bad value to the end of the problem the bad value is infinity So let's go from 2 to something a And then the very last step of the problem will let a approach infinitely large numbers So we integrate Saving this approaching of infinity till the end of the problem We integrate 1 over x dx That's the natural log of the absolute value of x We evaluate it from 2 to a So that is the natural log of the absolute value of a I Don't think we have to worry too much about absolute value here because we're starting it to and moving to the right So it's almost irrelevant when we put in 2 we get natural log of 2 So part of it we're going to know and part of it we're not going to know So now we're done integrating. We're done evaluating. Let's see what happens as a approaches infinitely large numbers so we want a to be 10 Not very large a to be a thousand a to be 100,000 and so on What happens to the natural log of a as a? Gets larger and larger and larger It gets larger right isn't the natural log of 10 pretty small number when you put a thousand it Don't we get a bigger number when we take the natural log of a thousand? How about the natural log of a hundred thousand is it even bigger than this? So as you keep putting in larger and larger numbers for a the natural log of a continues to get larger and larger so this thing Approaches infinity not necessarily at the same rate that a is approaching infinity But this thing gets larger and larger and larger. It's not bounded at all So if this is not bounded are we going to save it by subtracting the natural log of 2 from it? No So the limit does not exist So we will say that this is a convergent or let's just I'm sorry not convergent Divergent this integral diverges. We don't converge on a Finite number for our answer if we don't get a finite number for our answer Then it does not converge therefore it diverges so it doesn't have an answer So can we find the area under this curve f of x equals 1 over x from 2 to infinity? We cannot So that kind of seems the answer seems like what we thought it would be prior to doing the problem You better cap it off on the other end or you're not going to be able to find the answer Questions on this one Okay, I'm a slightly change the function. It does change the visual appearance, but not drastically But we're going to change the function now to Instead of 1 over x 1 over x squared This is why the visual appearance of the function is not going to really I mean you need to know that it's decreasing It doesn't have a chance of converging if it's not descending on The x-axis or asymptotic to the x-axis So what is the critical issue is how fast is it? Closing in on the x-axis, so let me actually Let's say this is one over x And let's plot a few points here. Let's say this is one What's the corresponding y value of? What's y when x is one? It's one Okay, is that going to be any different on this one? No, it's going to be one over one squared So we've got this point in common between the two and and we're not really starting till two So let's see where we are at two in this curve one over x When x is to where are we? One half that right one over two and in the other curve the new one one over x squared where are we? one fourth So it appears that we're kind of underneath this one Let's continue out here at let's say five Where are we on the one over x curve? One fifth right where are we on the one over x squared curve? 125th isn't it a very similar looking curve Kind of descends like this one does So We're starting closer to the axis and we get there a whole lot quicker now Does that guarantee the fact that it's going to converge it does not? But in fact, we're going to find that this one does converge and we can get an answer to this from to all the way to infinity So what do we do we delay the bad value to the end of the problem? To to something and we'll let that something approach infinity This is not a natural log Never has been a natural log is not currently a natural log and just for future reference never will be a natural log When is when do we have a natural log? integrand Okay, a number constant one over what? X to the first right it's got to be x to the first down there to be a natural log Well, we don't have an x to the first So that kind of follows the normal Exponent rule where you add one to the power right divide by the new power So if it helps to write this as x to the negative second so that we're not tempted to call this a natural log integrand So what's the integral of x to the negative second dx? That's right No Add one right x to the negative one x to the negative one over negative one right That's our rule add one to the power divide by the new power We were integrating x to the fifth we'd get x to the sixth over six So we're integrating x to the negative second x to the negative one over negative one Revaluating that from two to be Let's rewrite x to the negative one over negative one What's an easier way of kind of putting things where they belong negative one over x negative one over x? So we just rewrote it. That's where stuff belongs anyway Will you find the fact that we've got the limit as be approaches infinity? Will you find that somewhat annoying to have to write down several times? Yes, you will I also find that somewhat annoying to have to write that down But I am going to save that value to the end of the problem So I'll put in the upper limit negative one over B Put in the lower limit negative one over two And then anything that I need to kind of simplify I'll do that in my next step and I should be done except for the so-called bad value Minus negative a half that's a half And then we've got minus one over B You don't have to write the positive term first and the negative term second Keep them in order. They were in the original problem. It's fine. Now. We're done simplifying except for Allowing B to approach infinity What happens to this term? One over B as we allow B to approach infinity It disappears doesn't it approaches zero So this term in the previous problem as we kept putting in larger and larger numbers The natural log of that larger and larger number itself got larger here because of its location denominator One over tremendously large numbers that in essence disappears So we end up with an answer we get an answer to this so what's the area under this curve from to all the way to infinity? It's one half So it does converge Now that seems counter-intuitive That you could start this thing at two So kind of the red region right here started at two and let it run all the way to infinity And if you let it run all the way to infinity you would get an answer of one half, but it's true Suppose we did this instead suppose we took the same curve and let's start at two same value and let's end at 600,000 Let's get draw a diagram Lance not okay Tell me what the answer is Very very close to one half, but less than one half right So I'll just call it one half with a little Superscripted minus there. It's very close to a half, but it's less than a half because what do we have to do to get a half? You have to let it go all the way to infinity that's not infinity It's way on out there But so we're going to get some number that's very very close to a half The only way we could get exactly one half is to not stop this thing at 600,000, but let it go on to infinity So it is possible they can diverge in fact that seems pretty intuitive It is possible which seems counter-intuitive that they can convert This is kind of skipping a little bit actually let me put this up here first So these are our type one Integrals hopefully we can kind of get through this today and look at the type two the fact that it's type one Don't remember that they're just two categories that we're going to deal with So if the function exists for every number t greater than or equal to a what we want to do if we're going from a to infinity is farm that bad value out to the end of the problem and then Make that the last step of our problem provided this limit exists Or the other way We could have the bad value on the other end in fact We're going to look at a problem that has a bad value on both ends next If we're going from negative infinity to be then it's okay to start at t and Go to be as long as we last step of the problem allow t to approach negative infinity So we're going to have potentially a bad value up here or a bad value here Or in some cases we're going to have a bad value on both ends Now what we're going to do on the next example because we're going to have a bad value on both ends is Break it up into pieces and go from negative infinity to some Convenient value and that's good as long as we pick up at that same convenient value and go the other way out to positive infinity So these are kind of the three categories that Fall into this type So we actually see the bad value or in this case the bad values in the problem And then we try to delay that till the end of the problem So let's see one With a bad value on either end probably should Not do this example next but I think we'll be okay to do this and then we'll try to Summarize some of the things we've been finding along the way. Does that integrand look at all familiar? Inverse tangent right So we know that the derivative of inverse tangent is One over one plus x squared therefore the integral of that one over one plus x squared Is going to be inverse tangent so we know that both directions. It's certainly going to come into play on this problem This may or may not be helpful On all problems, but it is going to be helpful on this problem Does this problem have any if we were to picture this? Does it have any symmetry one of the tests for symmetry is The f of x equal to the f of negative x doesn't who remembers what symmetry we're testing for right there You get the same y value when you put in x as you do when you put in negative x That's symmetry to the y axis right is that true on this particular function is the f of x Do you get the same thing when you put in x that you do when you put in the negative of x it is true So this thing does have symmetry to the y axis If you were to either Physically plot enough points or put this in on your graphing calculator When x is zero, what do we get we get one? And if you plot some more points, I think you'll see this graph do something like this Therefore, it's symmetric image over here does the same thing right? So because we have symmetry we can take this problem, which has two bad values and Rewrite it in terms of a problem that has one bad value, which is kind of what they said on that summary We're allowed to do so if we find this area if it's possible to find it What do we have to do to get the whole area? Find it and add it to get a multiply by two right double it we've got the same bounded Number of square units here as we're going to have over here So let's change this problem to Let's go from zero to infinity And double it So technically what's behind this is we can go from negative infinity to zero and Then again from zero to positive infinity we've split it up into two pieces But the two pieces happen to be exactly alike, so we don't need the two pieces Let's just talk about one of them and double the answer So I'm going to rewrite that up here and let's do what we're supposed to do I'm just going to bring the two all the way out in front. Let's go from zero to something What is the integral of one over one plus x squared? Inverse tangent so I know sometimes in math class, you know, we get that we work with something We know the derivative rule. We throw it in reverse. We get this anti derivative rule We are going to be forced in this problem to think about what inverse tangent is and how is that function inverse tangent? Different from but yet somewhat like the tangent function, so we want inverse tangent r minus Inverse tangent zero. We don't know what r is technically yet. We know what's eventually going to happen to r How about inverse tangent of zero? What's the question you ask yourself when you have inverse tangent of zero? There we go what has zero for its tangent, right? What is the angle in radian measure? That has zero for its tangent or as he stated Jacob, right? Tangent of something equals zero. What is that? What angle can we take the tangent of and get zero? Zero? Right, isn't the tangent of zero zero? So this itself is zero Why? Because the tangent of zero degrees or zero radians is zero. So this disappears Now here comes a similar question, but it's a little more bizarre Because now what's going to happen to r? r is going to approach infinity What angle can we take the tangent of and get in this case positive infinity and it does have an answer Okay, 90 degrees, which would be pi over 2 isn't the tangent of pi over 2 radians positive infinity, right? So as we approach so this is where this comes into play. It's never actually infinity It's approaching positive infinity. So as r gets larger and larger and larger What angle can we take the tangent of that allows us to get this? Number that is more and more and more positive It's pi over 2 right just a tangent graph the tangent of zero is zero. We kind of visited that up here, but we know it's Asymptotic here, but what happens to the tangent as we get closer and closer to pi over 2? Don't the values get larger and larger and larger. So we're reversing that as these values get larger and larger and larger What angle are we getting closer and closer to pi over 2 right? That's kind of bizarre that you've got infinity in this problem inverse tangent of infinity Maybe that's not the correct way inverse tangent of something that's approaching infinity is approaching what it's approaching pi over 2 This might be a way to get a computer to Come up with a good 200 decimal place approximation for what? Try it try it in maple and see what it comes out of maple defaults to 15 or 10 Think 10 decimal places and you can get it to do more decimal places by what digits? colon equals 15 or digits colon equals 20 or whatever But this would be a pretty simple. That's not a real complicated supercomputer-looking problem, right? pretty simple-looking problem and The answer to it is pi so it'd be interesting to see what that would look like in maple So if we could back to our rough sketch We evaluated this region And in a sense, we got pi over 2 therefore this other region is also pi over 2 so the area of This entire region under this curve from negative infinity All the way to positive infinity is a nice clean value pi So converges right this integral converges if you don't then you're going to have to find a value Where you're going to kind of stop one region and start the other one right So that her question is if we don't have some natural symmetry What are we going to do it does help to sketch it because even if it's not symmetric like this This is an even function. You might come up with one. That's an odd function. Let me let me give an example But that third category so let's say it's not So I don't have the symmetric image on the other side of the y-axis But I've got the symmetric image on the other side of the origin we could still break this up at zero, right? and this in essence would be Positive right because it's bounded between the curve in the x-axis this region would be negative So the net is going to be zero if in fact it is The curve that looks just like this, but we do have to actually deal with that slightly differently Because if this piece diverged and this piece diverged We can't even though visually we think they cancel each other out if one of the pieces diverges It can't really be helped by the other one regardless if it converges or diverges So you could split this up into pieces and then ultimately the answer would be Possibly if this diverges so does this therefore the answer diverges even though we think they balance each other out But somehow find a place where you can break that up Okay, this is probably a little out of place, but We've taken a look at two of these situations. Let's see if we can kind of summarize these in general What that I don't want that either. I guess I just didn't write this up What if our integrand is of the form 1 over x to the p now? We've looked at two of these 1 over x to the 1 and we've also looked at 1 over x squared Those were our first two examples today. Let's see if we can kind of Categorize what we should expect for an answer Based on the numerical value of p is there kind of a cutoff where when p is smaller than that it? Diverges p is larger than that it converges. Let's see if we can come up with this So let's just say from a Where a is some number. I don't want to put a specific number in there All the way to infinity so we kind of have to handle this with cases There's one category that we already looked at When p is equal to 1 that's kind of often a category by itself because if it's just 1 over x to the 1 That's natural log right? This is our first example We decided that was natural log and we ended up with natural log of a Value that Just put this down here. Let's call it r Where r was approaching infinity? Minus the natural log of this number whatever the number is this is always going to be a Number the natural log of 2 that was a number the natural log of 8. That's a number. So that's never going to be a mystery What happens to the natural log of r as our approach to infinity we decided that got infinitely large Therefore the result was the fact that it was divergent. We did not get an answer the limit does not exist So when p is 1 we were kind of forced to integrate this as a natural log If p is something else other than 1 so larger than 1 smaller than 1 can't we integrate this? by adding 1 to the power Dividing by the new power right as long as it's not 1 We we use the power rule So I'm 1 over x to the p. I'm going to call that x to the negative p If the little a is bothering you just put a number in there. It doesn't matter I just don't want to kind of zero in on any one specific number As long as p is not 1 which we've already handled that What's the integral of x to the negative p? over negative p plus 1 Okay x to the negative p plus 1 Over I think this is what I heard negative p plus 1 right add 1 to the power divide by the new power And we've already ruled out There's no way we're going to divide by 0 here if p were equal to 1 it'd be negative 1 plus 1 Which we'd be dividing by 0 we've already handled that case Up here, so we don't have to worry about that So if we plug in our Let me bring this out in front I'll just bring in that bring out that coefficient out in front so we can look at what we want to look at here so don't we plug in That ours approaching and we plug in our for the x value right? Got too many letters here apologize for that again a's a number, so don't worry about this. This is a number Two to a power eight to a power 11 to a power it doesn't matter. This is what we're concerned about If that hangs around and gets larger we're in trouble in this diverges is that correct? If it somehow gets thrown to the denominator and gets larger That's a good thing because if it gets thrown to the denominator and gets larger What happens to the value of that fraction? It goes to zero and in fact disappears What value for p would send this our thing that's getting larger and larger and larger? down to the denominator If p is Greater than one is that right if p is greater than one just pick a value Let's start with two if p were to This would be our some number that's getting larger and larger To the what negative two plus one. What's negative two plus one? Negative one. Where does that send it? To the denominator and it can get as large as it wants to in the denominator. What happens to that term? It goes to zero So if p is larger than one well, I looked at two. I don't know how about one point two Does that still send the number that's getting larger and larger and larger down to the denominator? Well plug it in negative one point two Plus one. What's the result? Negative point two. Where is it if it has a negative exponent? It's in the denominator So as long as p is larger than one What's this thing going to do? It's going to converge and if p is one or less It's going to diverge so we've already got p equals one. It's divergent if p is greater than one. It's convergent and you could try a number that's Smaller than one just try zero if p is zero What do you get? You get one which is r to the one. Where does it stay? It stays in the numerator if it gets larger in the numerator It's divergent if it gets larger in the denominator It disappears and causes the integral to converge. So for this particular integral if p is Less than or equal to one. What's the decision? Diverge and if p is greater than one It's going to converge. We will revisit this problem in a slightly different manner later in this course when we get to chapter eight and we'll talk about series of Progressions of numbers that are called a p-series for this very same reason one over x to the p and We'll see similar results So the fact that we've done this problem kind of generically let's suppose that we had This example right here. We can't completely do this problem But we can make the decision is it convergent or divergent just based on what we have right here It's convergent right There's p one over x to the p So p is equal to three halves. We've got the same format set up seven to infinity So as long as p is greater than one that's going to eventually send the bad value the number that keeps getting larger and larger Down to the denominator when that happens. That's a good thing. It disappears therefore the integral converges now What we can't answer with this is okay. We know it converges What number does it converge to we'd actually have to actually do the problem? To do that, but we can determine that it's convergent Okay with this today. Let's say Let's say that we have some curve here and I'm not going to specify what f of x is but we know we can start at something and We know that if we let it run all the way out to infinity that this is convergent Okay, we have another one that starts at the same value and we know for a fact that it's Either equal to or less than the one that we began the problem with What do we also know about this problem it also converges I heard that from a couple people does that make sense So if the curve that I wrote in black is a convergent integral from a to infinity anything less than or equal to that is Also convergent so certainly We're going to be able to find the area under the black curve from a to infinity We will also be able to find the area under the red curve from a to infinity So this is going to seem overly simplified and you'll think less of me for saying this but if someone walks into the room and They're a diminutive human being they're small Okay, so I don't know what that conjures up in your mind. Let's say this person walks in and they're I don't know Four foot eight. It's a little it's a little person. It's a third grader. Okay, they walk in the room and they're four foot eight And you'd everyone in the room would say oh, well, that's kind of a that's a little person. Okay, they're small Okay, someone else walks in right after that little third grader and they're even smaller Wouldn't you say that person's small too? Right so if that first person was small and the next person's even smaller you're not going to say oh, that's a huge Person that's that's they're humongous. No, you're going to say they're smaller than the one that came in before Clearly they're small right Smaller than small is what? Small okay, that's what this is if this one converges and we've got one underneath of it See, I thought you'd think less of me This one's even smaller than this one. It has to converge Okay, now. Let's let's get the analogy on the other side I taught Todd Fuller here when he was Basketball player. He's also a math major Lives in Charlotte now really really neat person. He's seven foot tall So Todd Fuller walks in the room. Okay, and we all think oh man. He's tall Okay, and then we have someone else minute bowl walk in the room who's seven foot nine Okay, do you think oh minute bowl? He's just a tiny little person. No What do you think about him because he's taller than Todd Fuller? You say man. He's really tall. He's tall So what's the analogous situation to this problem? And we're right. We'll start class with it tomorrow If we've got an integral that is divergent We've got another one that's even higher up on the axis system than this one. What does it also do? It also diverges because if it's considered to be large What would you call the one that's larger than it? It's also large. Okay? You want us? Okay. We'll continue this inane thought tomorrow