Transcript for:
Improper Integrals Overview

so improper integrals my goodness what in the world does this even mean well typically what we need for integrals to work so normally for any sort of definite integral from Like A to B of some function what has to happen for this to exist for us before right now for this to exist and let not have any problems with it f ofx must be bounded and finite on the interval A to B that had to happen what that means is that if I give you some curve right I say remember what an integral means what's a different integral actually mean for us the area good area under a curve between here and here basically it says if I want to find the area under a curve of this curve from these two between these two numbers our curve can't do weird things like between here and here go up forever well that doesn't really make a lot of sense or it can't go up like this forever and have an ASM toote or the there can't be holes or weird things that happen uh basically it says it's got to be bounded doesn't go up or down forever it's got to be finite can't continue on in One Direction forever so basically it's got to stop it's got to start it's got to stop and it can't do crazy things in the middle that's really what this this sort of thing means that do you guys understand the idea of bounded it has upper limit has a lower limit uh it has right side limit has left side limit A and B so basically anything that's not like this is called an improper integral we're going to study what those scenarios are how in the world we're going to deal with them so now we're going to talk about two basic scenarios so here in section 7.6 we talk about these two cases case one case one happens well suppose if you had an integral you had some curve here's our curve our curve goes like this forever and I say I want you to find the area into the curve from a to and I don't give you a stopping point if I did this check it out if I said A to B from A to B this would be fine we'd have a curve that's bounded above and bounded below above by the curve below by the x-axis we have a starting point we have an ending point that's what this would say it's bounded and it's finite that would be cool does that make sense but if I say no no no now what I want you to do is take this function f ofx and I want you to find the area under the curve from a to infinity and beyond just joking uh forever and ever can we do it well the problem is the answer is yes we can I'm going to show you how to do that the problem is this is no this no longer falls into this it's boundage sure it is do you guys see what I'm talking about being bounded it doesn't go above a certain number it doesn't go below a certain number it's going to be asmed to the xaxis as X approaches positive Infinity but it's not finite it starts at a no problem but it doesn't end so it it fails these conditions of being a normal definite integral that's what makes it this idea of an impr an improper integral do you guys get the idea of an improper integral it's not finite it's going forever and likewise I could go to negative Infinity if I wanted to so I could go this way but this case is going to cover both them or case number two case number two so case one was the interval is infinite case number two happens like this suppose you have some function this is just for an example here and I see here's my function f ofx and I asked the question hey uh can you find me the area under the curve from A to B can you find the integral basically of this function from A to B can you do it no well see what happens here um is the interval finite yeah interval starts at a interval stops at B that's cool it's a finite interval does that make sense to you is the function bounded is the function between two actual numbers that no we have a top bound top bound would be whatever whatever that number is but the bottom bound it does not exist this function goes to negative Infinity on the y- axis that's that's an issue right here so case one is well if the interval is infinite we have a situation it's not a proper integral if the function's not bounded so here f ofx is not bounded that's our other case for an improper integral so again we got this idea of a definite integral it works when A and B are finite numbers so when we go from one spot to another we have a starting spot we got an ending spot it works when f ofx is bounded this doesn't go up forever it doesn't go down forever within this range okay so it it can be a polinomial for sure which does either go to positive infinity or negative Infinity it's not ASM totic at all but within the range between this interval A and B the function has to be bounded did you guys got that the interval has to be finite no problem that means that's what we're talking about for definite integral if we don't have one of those cases so in other in other words if the interval is not finite if it's infinite goes to positive infinity or negative infinity or both of them you can do both in both directions that's one case where we we don't satisfy these conditions hence it's called an improper integral it's not the way we want it or we have another case where f ofx is not bounded within that interval so so we say hey here's our interval but the function goes positive infinity or negative Infinity within the interval basically we've got an infinite discontinuity here does that make sense not a removable discontinuity that would be different it would be still be bounded but if we have an infinite a removable discount did you guys remember what those are those are just holes right those are fine but if we have an infinite discontinuity that's an issue where the functions going into positive infinity or negative Infinity hence you have an ASM toote somewhere on on the interval that's an improper integral so again two cases interval is infinite or the function is infinite it's unbaled soents be okay with our two cases now let's talk about how to deal with these with an example we're going to talk about the first case first we'll talk about for infinite intervals for infinite intervals let's do it with an example see if you understand what I'm talking about here let's take the uh let's take the example of the integral from one to Infinity oh that already looks new doesn't it that's something we haven't ever seen before probably unless you had this class before you've never seen this before of one/ X2 DX now of course this the first time we're going through this so I'm going to really teach it to you try to make you understand what's Happening Here the first thing we look for and really you haven't ever had to look for this before um in your in your integrals but now we're going to because we're trying to do different integral right well it has to be defined on this entire this entire interval so let's check it out what's a number that would not work or would be undefined in this problem zero zero would be an issue so so am I Crossing zero on this interval no okay so we don't have this situation but check this out if I wanted to go from -1 to Infinity do you see how I'd cross over a point where I'd be undefined basically I'd cross over an ASM toote they'd go like this at x equals z does that make sense that would be this case we'd have some sort of a infinite discontinuity at that point at at X would equal zero that'd be an issue that would be case number two does that make sense to you now we don't have that we're lucky we go okay no no no let's just do positive one for this for this to work well no problem it's got to be defined on this interval it is is defined everywhere from one to Infinity now' you have you understand that we're good we don't have this situation on this interval because well we missed the one point that's bad for us we missed the zero now let's check out the other thing let's check out the interval itself is the interval itself finite no finite means it starts and it stops is it finite no it starts at one where does it end it don't end well that's an issue for us so which case are we are we in right now are we in case one case twoe one case one so for an internate interval we're going to think about this a little differently say you know what let's consider this so let's consider not Infinity because that's really hard to think about okay we don't want to think about Infinity right now let's just call it B so 1 two I don't know some m b of 1 x^2 DX where B is just some number that's bigger than one so notice something I'm not saying B smaller than one because this interval starts at remember how uh intervals work for integrals starts at the small one goes to the big one so let's consider this to be the integral from 1 to B where B is simply some number bigger than one basically bigger than where we're starting at well if that's the case here's our picture our picture if you know what the graph of 1x2 looks like it looks like this from both sides now I don't have to worry about this to worry about this side because I'm going from one to Infinity our graph basically looks like this where we're starting at one now here's the little idea the idea is it's going to be really hard for us to legitimately go from one to Infinity because how can you actually plug in Infinity for a definite integral you guys didn't see the problem you can't really plug in Infinity you can't do it which we're we're going to cheat a little bit we're going to say no no no let's not go to Infinity let's just go to B now since we're starting at one B's got to be some number bigger than one does that make sense so let's call this B over here somewhere one to B then our what our integral represents is this area from one the area under the curve from 1 to B where this curve is 1/x^2 so if you you feel okay with that so far I really do want to make sure you get the idea not just be able to do it but understand it so the idea is if we have an interval interval from 1 to infinity or negative Infinity to some number let's break this up a little bit let's let's not go to Infinity let's call the the infinite bound here let's call it b or if it was negative Infinity we call it a down here let's call it something well maybe we go ahead and we do this integral the way it is now instead of plugging an actual number like four we're just going to plug in the B and then we'll deal with it later so we know that we can write this as not one to Infinity but one to b b is some number just to the right of one we're representing this area right here that's what we're doing quick head note if you're okay with with that so far let's go ahead let's see if we can do this interg I'm just going to rewrite it so one to be how would I rewrite the integral of 1x2 to actually do the integral what would I do yeah by the way don't start doing silly things things hard things when you have easy integrals don't be doing weird things like partial fractions I have one overx squ partial fractions no don't do that okay don't make things harder than they are which some of you have a tendency to actually do okay uh so make the integrals as easy as possible we know all the integration techniques right now just so you know we got them all now we're just applying them well all for us can you do the intequal sure what is it don't not talk at once I mean that'd be rude okay x to the I me over 1 now of course with definite integrals we evaluate these things well we're going to change this around a little bit because right now looks kind of nasty but we're going to change it around uh where's this integral start interval start where's it yeah and we're calling it B really it's going to be Infinity but we'll deal with that idea in just a second okay let's rewrite this so this is going to be --1x from 1 to B Isn't it nice to deal with some of these easy integrals right now after doing all that stuff that we just did in this chapter can you evaluate 1X from 1 to B yes what's it going to be good we plug in the top number first minus what okay let's clean it up -1/ B + 1 so f youc with that so far that's easy right I mean we've been doing this from calculus one this is nice now here's the issue like a 10-second recap 10 recap is uh this is improper because it's not a finite interval this is cool because it is bounded on that interval that's good news that's the good stuff so bounded yes finite no infinite call the infinite portion of that call it b or call it a uh call it some letter usually we use B for positive Infinity we use a for negative Infinity cuz intervals go from A to B that's why so call that B all right no problem what B is is some number bigger than one so since our interval is starting at one B's got to be to the right of that and what our integral represents here is the area under this curve from 1 to B no problem that's exactly what this idea is about do your integral plug in the B plug in the one simplify it but then let's think about what's happening now when we did this what did we substitute the B for so what do we want the B to go where do we want the B to go we want it to go to Infinity now the way that we deal with going to Infinity is this idea of a limit limit that's right with a limit so as we want to go X goes towards Infinity we can't do that by just plugging in Infinity it's we kind of can we kind of can't um we can because that's really what how we're going to evaluate this thing but the way we need to evaluate with is a limit so what we do now is go where where's the B go well since we want B to approach Infinity we do that with a limit so now we'll just let the uh limit do its work so we're going to do a limit as X sorry not I'm not I made a mistake how do I make a mistake it's not X approaching Infinity it's the B is what we substituted in for our Infinity so now we just got to substitute the back so B goes to posi infinity1 b + 1 well now it hopefully it makes some sense let's let B go to Infinity what happens if B goes to Infinity here so can we find the area under the Curve from one to Infinity of this curve 1x2 that's really cool how much is it one yeah that's the idea so um this idea is we're we're kind of cheating we're not really cheating we're really doing it perfectly mathematically um but the idea is if you're going to Infinity just call it B take a limit as be approaching Infinity of whatever the integral is that way you can actually evaluate you can't plug in Infinity really you you can plug in B then take a limit as B approaches infinity and that's where we we change the problem a little bit show up hands if that might made sense to you is it a really difficult concept kind of interesting right kind of cool um so let me write some notes out for you on what we're going to be doing every time questions I always forget do you have any questions on this okay good did I explain it well enough for you all to understand it yes okay we'll deal with the unbounded version um probably another time but so here's what we know some notes let's say we had an integral from a to infinity or we have an integral from in negative Infinity to B of some function of some function what we do we don't really do every time okay because this what I've done this is technically very inaccurate uh because what I'm calling this is B but I haven't said anything that b is going to Infinity does that make sense to you so what we do right off the bat with our with our improper integrals after we identify what's going on whether it's discontinuity or whether the interval is infinite uh we we do that first then we go ahead and we write out the limit right now you see if we're going from a to Infinity then what I'm going to do right now is I'm going to say which which is the a problem by the way is it the a or the infinity infinity so I'm going to say A to B but in order to do that right away what you should do is say yeah you know what I'm going to take the integral from A to B but I need to let B go to Infinity if it seems trivial to you it's not you're defining what's happening to the B you have to do that it's not that I just get to randomly plug in a b figure it out and then plug in Infinity you can't do that mathematically you do that with a limit so we say no we don't get to plug in Infinity really what we're thinking about is this B it's approaching Infinity it's getting really close to it it's plugged in bigger and bigger and bigger numbers until it makes no difference anymore as compared to Infinity does it make sense to you so this not trivial idea it's sure call it a b but then Define what's happening to b b is going to positive Infinity you do that with a limit now the integral itself this is where you guys get a break the integral itself is it's doable I mean it's everything that we've done so far so basically I'm teaching you today is this that's it that you can do this that instead of an Infinity you put a limit as B approaches Infinity of B this is still Infinity when you think about it we're just kind of Mak it a little bit more doable so we can actually evaluate the integral that was a lot of words did you understand them all all right now next up let's say that we're not going from a to Infinity we're going from negative Infinity to some B what's our problem here okay so the B is not a problem we call negative Infinity our a and then we say okay cool I'm going from a to to B but what's happening to the a where's the a going to go so a approaches negative infity that's right it should match up so a negative Infinity a negative infinity and then the integral no problem you'll know how to do this I'm not saying it's going to be easy but you'll know how to do it because it's something that we've covered before you guys okay so far now we're going to come up with two different words here two uh two words I'm sure you've heard them somewhere in your life before but for integrals we're going to Define them right now these limits are either going to be what's called convergent or Divergent now if the limit exists what that means is that you can add up all the little sums and a finite sum that would be a convergent integral that means that the the area actually exists if the integral adds up and it keeps getting bigger and bigger and bigger and bigger well then that integral is not going to exist that limit won't exist actually the integral probably will exist but the limit won't exist if that limit doesn't exist well that's Divergent either it doesn't go to a finite number it goes back and forth or it goes always up or always down it doesn't go to one single number if that limit doesn't happen then we call that Divergence so these things things are convergent if the limit exists Divergent if the limit does not exist does that make sense to you yes no so we call these convergent if limit exists we'd say it's Divergent if the limit does not exist convergent basically means this I'll actually write that down look up here so I know that you guys are done writing I can actually talk at you convergent basically means this it means that the area that you're adding keeps getting smaller and smaller and smaller and smaller as you're approaching infinity or negative Infinity smaller and smaller and smaller to the point where when you add it it's so small that it doesn't make a difference does that make sense to you so it keeps smaller and smaller and smaller and smaller Divergent does this Divergent says it keeps getting bigger and bigger and bigger as you keep adding it or it doesn't keep getting smaller and smaller and smaller and smaller and smaller it doesn't do that um or there's no pattern to it it's bigger than smaller than bigger than smaller than bigger than smaller something like that so then the limit will tell us that so don't worry too much about it basically the idea is if limit exists it's going to be convergent because that is happening it's getting smaller and smaller and smaller quick enough if it's Divergent if the limit doesn't exist that means it's not happening so we're getting bigger areas or we're getting areas that uh don't go back to one single number it keeps getting bigger and smaller bigger and smaller does that make sense to you okay would you like to do some examples of course you would so let's do something thankfully these examples won't take 30 minutes A U 30 minutes example they're fairly quick question question just kidding was it on this oh you were joking funny guy is there going to be like in case two where it be taking two like from negative Infinity to Infinity absolutely yeah I'm going show you how to do every case I can think of okay so this is this why this lesson's going to be about three days long is is I'm going to show you everything I can think of as far as what cases go okay we're starting very easy right now but we will have limits that go from negative Infinity to Infinity we have negative well limits that go from negative Infinity to Infinity where they cross over infinite discontinuity and that's about the worst case scenario you can get okay so if you're going from negative Infinity to infinity and you cross over if you cross over more than one that sucks even worse but if you cross over an infinite discontinuity then you have to break up that limit in sorry the integral in at least two of them sometimes more Okie okay so let's start off very basic firstly we want to identify what type of an improper integral if anything we have do we have an improper integral here absolutely well firstly let's see if there's an infinite discontinuity are we defined for an entire domain no are we defined for an entire domain I don't care about the zero if I'm not Crossing zero I don't care I care just where I'm I'm going to so on my interval and my F yes then then we're good there's no infinite discontinuity do you guys get what I'm talking about how about the interval is the interval finite or infinite infinite so what that's called is an infinite interval that's one of these improper forms so if we have an improper form and an infinite interval let's write out exactly what we're going to do with it so tell me how should I rewrite this problem so that I can do it okay do you need the limit I'm not going to check for that or anything on your tests am I of course I am uh limit hello a limit from where to where oh say that correctly from where to where one to and then B goes to where perfect do we change this no pull the three out yeah you can pull the three out absolutely in fact because it's a limit and because that's a constant you can actually pull a three way out here doesn't even matter pull it here pull it here don't care wherever you want to go does that make sense so we can pull the three out for sure can you do this yes integral they're all going to be doable yeah of course for you so we have a limit we have B approaches Infinity I got to warn you um in the section because we've covered all of our integration techniques I'm going to be doing a lot of them pretty quickly uh because I'm I'm going to assume that you have studied the previous four sections that we've done on integration techniques okay so I'm going to cut some steps out uh just for the sake of time because we're going to get some integrals that are kind of nasty you know and I'm not going to repeat all those techniques that we just just learned I'm going to use them here well this a pretty basic One to begin with so what's our integral going to be here 3 Ln absolute value X oh absolute value did you get that from where to where one one to B now be smart about the absolute value look at your interval your interval goes from one to Infinity tell me something about all those numbers they're all positive so your absolute value is kind of irrelevant here but you do need to show it so in our case we've got a limit do you remember something about limits how we always write them until the very last part where we actually get to evaluate them same thing happens here so we will have 3 lnb minus and you do this a couple ways you could pull the three out and have a bracket here or you could have 3 Ln B minus 3 Ln one it really doesn't matter as long as you have the three in the appropriate spot in both both places or being factored out make sense some algebra so I'll probably do it like this because I don't want to have to write the three again I'll do 3 * lnb minus Ln 1 so this is limit as B approaches Infinity of 3 Ln B how much is Ln of one do you know okay so check it out simple integral no problem we've evaluated no big deal we've simplified it as much as we can to the point where now we're going to attempt to find out what happens as B approaches Infinity you guys okay with this one awesome so what happens when B goes to Infinity what's Ln of b as B approaches Infinity oh dear did you know again read it what's Ln of b as B approaches Infinity no Ln of one is zero how can Ln of infinity also be zero doesn't make sense it's a it's it's a increasing function the entire time it's always increasing say what now infinity infinity it's Infinity that's right Ln goes like this remember Ln hope you remember Ln goes to negative Infinity as we approach Zero from the right goes to Infinity as we approach Infinity it's always increasing what that means is that this is infinity that's Infinity tell me something about this integral is it convergent or is it Divergent it's Divergent what that means is that look at this this limit keeps getting bigger and bigger and bigger doesn't it means the area keeps getting bigger and bigger and bigger it doesn't go to a finite number therefore our area under this curve we can't say what it is because it's an infinite area called Divergent you guys see the difference between convergent and Divergent the one first one we did we actually got an area out of it second one we did no no this right here says that this is Divergent and I wanted to give one to you right off the bat that way you see what I'm talking about when I say Divergent if the limit doesn't exist or if the limit goes to infinity or negative Infinity then we say that that we can't find the area under that curve it's always getting bigger than uh than some finite number so hands of that one made made sense to you okay can we keep on rolling not too bad right not yet at least let's keep going what we're going to do now is um we're going to figure out how to do this in general for any power of X so instead of doing example after example let's try this one let's go from 1 to infinity 1/ x to the n DX you know what I'm guessing that most of you can probably help me out on this one uh what's the first thing that I would do with this particular integral right right limit yeah because what tell me what's wrong with it do I have a discontinuity or an infinite interval here it's infinite interval do you notice how in this case the only number that I have a problem with is zero still I miss the zero on my so I'm good so yeah you know what we just had it on the board a little while ago this will be a limit as what approaches what everybody approaches infity and then we write from where to where please one to and then one okay yeah okay we'll do another step sure can you guys do that first step hello yes no no big deal right you just take care of the infinity make sure you have a limit going to that variable and then take your variable and make it look like it's finite right now make it look like it's going to actual numbers just something bigger than one now let's go ahead and do this in general I know that some people don't really like doing things in general but this is going to save us some a lot of headaches later so check it out first thing we could do always it's going to get annoying I know but you always got to write the limit you always have to do that until you actually Val so do not neglect that are you listening don't neglect that always write it now with 1X to the N I can write that as x to the N DX you see if I have X to any power besides one I can always move that up and I can always do this integral notice that we already took care if you didn't notice this from before we already took care of the case when we have one over x to the first don't we mhm I took care of it in fact the three is irrelevant if the three was a one you would still have something that's infinite do you see what I'm talking about hello yes no so even if this was a one no problem so basically what's 1 to Infinity of 1x DX it's also Divergent it's also a problem okay so we already kind of dealt with that case here let's look at every other case if we do this we'll get a limit as B approaches Infinity of when we do this integral what do you do with like x to the 3 or x to the 3r would be a better you add one so this would be x to the n + 1 n + 1 sure that's exactly what we do we add one to the remember doing that we add one to the power divide by the new power so we do all the time and then we evaluate it from one to B well if we evaluate it from 1 to B we have a limit as B approaches Infinity tell you what I'm going to change this around because I like the look of 1 - n better than n + 1 just me we're going to have remember where the B goes the B is going to go in terms of not N N's going to be a constant okay and it's B is going to go in terms of X so this would be B to the n + 1 over 1 - nus 1 the n + 1 over 1 - n are we okay with that one I want to make sure that you guys are are good with this before we go on any further are you you sure so hey add one divide by the new exponent no big deal plug in the B we got it here plug in the one we got it here it's got a common denom let's simplify this just a little bit so if we simplify we got a limit as B approaches Infinity well in here we have B to the n + 1- 1 notice that one to any power is still one over 1us you know what we can do also anytime you don't have the specific variable that this is approaching oh sorry uh the specific variable is right here like B you can pull out everything in front of your limit so I'm going to pull out the 1 1us you can rationalize this because well n's going to be a constant isn't it so this is always going to be just a fraction pull out the fraction now let's see how I want you guys to see this next I'm going to do a little bit of fancy Fant math because it's going to be easier for you to see what happens next are you ready for it I'm going to force this to factor out a negative so what I'll end up getting is I'll move down here to 1 over 1 - n time the limit B to the infinity I'm going to Force this to be B to the nus1 B to the N minus1 do you guys follow the math on that okay minus one probably need some parenthesis here just to show what that what's happening I'm also going to write this as a fraction limit as B approaches Infinity because this is now a negative exponent what I'm going to do is I'm going to write this as 1 over B to to the N -1 - one we okay with that one are you sure some of you guys are zoning out don't zone out now man you got to be with me here so you okay with uh factoring out the negative from the exponent if we have a negative exponent hey you can write this as 1/ B to the positive exponent that's all I've done here so 1 over 1 - n no problem limit limit this is this -1 - one so far so good yeah now let's think about this for a second when is this going to converge when is it going to diverge because we want to do this once and for all so let's think about this let's say that n let's say that n is well how about we get n is one if n is one think back it just stop with me for a second okay if n is one then this was one correct are you listening yes I need your attention here if n is one then this was one we've already dealt with that case we know it's Divergent does that make sense so we're going to have anything that where n is one is Divergent if you got that here you'd actually get infinity to the zero it turns out that when you do remember our our indeterminate forms when you did that you'd end up getting undefined out of that you get infinity so that would be an issue for us so any = 1 we know it's Divergent how about Nal 2 let's think about Nal 2 if Nal 2 what's 2 - 1 this would be like Infinity to the first Power that's Infinity what's one over infinity this would be convergent you'd actually get a number out of that how about three what if n's three same deal or four same deal or five same deal now we already did the one correct what if n was Z or 1 or -2 if n is zero we get 0 - 1 how much is that1 so basically we have B to the 1 wouldn't we that would be B to the first up here that would be an Infinity do you guys get what I'm talking about so we get this uh we get this statement I'm going to write it right up here so from all this stuff we know a couple things so let's think about the integral from 1 to Infinity of 1 over X2 I don't care what it is you can do it now without well you can you can't evaluate it but you can see whether you should evaluate or not so check it out let's say that uh X was to the first Power we already dealt with that that was diversion let's say x was to the second power so n was two we talked about that that would be convergent or three convergent or four convergent basically this is convergent if n is greater than what number please one one yeah not equal to one but greater than one if n is two or three or four or five or six then we'd have Infinity to the positive power Infinity to a positive power would be Infinity 1 over infinity is 0 0 - 1 is oh netive 1 no big deal that would be 1 time whatever this is does that make sense now that's kind of cool right it gives you almost a formula for doing that cool uh now let's say the N was equal to one we talked about that at n equals 1 I have an Ln we just did the Ln that diverged if n is less than one if n is less than one we'd have zero Orga 1 or -2 we'd get B to a negative power is that do you with me B to a negative power is the same thing over under one is the same thing as B to a positive power by itself Infinity to a positive power is infinity infinity minus one I don't care what you do it's still Infinity that's an infinite limit that that limit is not going to exist for something that's going to be convergent so basically this converges at n is bigger than one diverges at n is less than or equal to one that's the idea so here's the here's what I'm trying to tell you if I give you this should you bother to find out whether this thing has a limit or not yes absolutely because that's a positive exponent that's uh that n equals 3 no problem that's bigger than one but if I this one would you bother no heck no you know automatically that's going to be Divergent for sure don't even worry about it just say oh look at that uh n's less than one hello divergent or if it equals one if it's that you go don't even need to worry about it that's going to be a Divergent you with me okay so that's that's kind of a nice little thing that we can we can do here just look at that didn't make sense to you this whole explanation there's one more thing I was going to say I don't remember what it was mustn't that important oh yeah I know what it was what that basically says is that because all these negative powers are going to be Divergent change that into a positive and what you end up getting is X 4th DX don't you you get a polinomial do you realize that if I take an integral from one to Infinity of any polom pols don't have negative Powers okay any polom pols go like this they go like or they go like this that's their end behavior see the positive in or negative Infinity are any of those limits going to be convergent for us no they're going to exist but they're going to be positive infinity or negative Infinity they're not going to exist to a finite number therefore these integrals will all be Divergent does that make sense to you okay we're going to stop there we'll continue with this next time all right let's get started on our next example the next example is another case of where we have these infinite intervals where we're starting at a number or we're starting at an Infinity but we're going we have one one section that's defined like one part of the interval is defined and one section that's not so here we have yeah we're starting to one but we're not ever ending we're going to Infinity do you remember what we do right up the bat when we have a scenario like this what's the idea it's a limit that's right so what we're saying is that no we really can't go to Infinity the way we want to because we can't really plug in Infinity since we don't know what it is but what we can do is we can say all right well I know I could take a limit from what's our problem here is it the one or the Infinity infity of course it is the one's great the infinity we call that some letter we typically call the upper bound B and the lower bound a so if your Infinity is in the upper bound call it B and then what we say is all right well this is fine this now becomes something that we can use a definite inte interval for sorry definite integral for where B is a number the limit idea says okay now you're going from one to some number bigger than one this is the idea then the limit says okay now take that number that you're going to where we called it B and move it forever to the right so move it forever towards infinity and that's why we have to have this limit here and then the rest of the integral well that doesn't change after that the nice thing about these well we know integration techniques is we spent all this time doing which is fantastic question uh if we just avoid writing that to the end is that technically incorrect yep yep we got to you have to write the limit you have to have everything I'm showing you is what you need cuz if you just go from 1 to B that doesn't make sense oh I meant just avoiding the B Al together until you do the actual um you could do the integral first I I prefer you doing the limit uh this is what's what's going on here we're changing to a definite integral uh where B is just some number to the right of one or whatever number you've given then you do it and then you evaluate with b and then you take your limit uh really that's a proper form super big deal for some people I don't know I'd like you to do it this way though okay so continuing can we do an integral like that one hello can you do the integral and that's pretty easy what are you going to do with it to I I'm deal with a lot of people who come up with questions going man how do I do this and I'm look at man you guys are making this way too hard way too hard start with the easy stuff calculus is supposed to be well how can you make it easier all right so when we look at this when not we're not thinking integration by parts are off the bat or whatever we're thinking well do Basics is it in your integration table well no can you do it with a Uub yes yeah do a Uub then do something simple if you can so with us we go okay cool well we can do Uub so U = -2X du = -2 DX or du over -2 equals DX something really nice what that means is we get the limit for got the limit we get the limit as B approaches Infinity of the integral from 1 to B uh what's going to be on the inside of our integral u e to the u e to the U that's right e to the U and then DX no no no we got du over -2 this is actually kind of a a pleasant break for us as far as dealing with some simple integrals isn't it kind of nice uh where we we know how to do these they don't take four pages for one problem uh what are you going to do now what do you think okay so you know what you can actually do I don't know if you know this but you can bring that negative one2 to the front of your integral but you can also do it to the front of the limit so whatever you want if you want to bring it all the way out front and just do - one2 time the limit as B approaches Infinity U I'm also going to do the integral so we pulled out the the negative one2 what's the integral of remember this has be gone right this is our negative one2 what's the integral of e to the u e u I love that integral and then where are we evaluating to yeah one to B notice how if we pull out that negative - one2 it makes what we're evaluating just a little bit easier it's kind of nice we don't even have to worry about that it's just going to be neg 1/2 times the limit quick Head n if you're okay with with that so far now we get to evaluate so what we know is that we're going to have - 12 time the limit as B approaches Infinity of let's evaluate what goes first the one or the b e to the B minus E to the 1 now if you're going to do this correctly you do have to be very good at understanding what's happening to these functions as you approach Infinity some of you struggled with this when we dealt with limits the first time would I mess it up what oh my bad uh it's it's actually going to matter here too I got all wrapped up and talking about Infinity I forgot my substitution you mean I was just testing you you passed fantastic job yeah what do you do here then okay yeah so before we evaluate you should have caught me earlier thanks a lot guys limit of e to the -2X yeah good why is that important because you know what typically typically what I I do in my sub in my substitution problems I change balance so if you're used to doing it like I was don't make the mistake that I did uh where you forgot forget to substitute back in for that you that would have been a really big mistake I'm glad you guys caught me on that uh did you did everyone see the mistake that I made okay it's on video watch it again you can see the mistake that I made all right uh yeah make sure that you're actually doing that substitution back because if you haven't changed your bounds that's a must you have to do it so very good job nice catch so negative 1 have limit as B approaches Infinity what we end up getting is e to -2b minus E to-2 you still okay so far yes well maybe what we want to do is make this so it's a little bit more understandable on what's going on with our function so if you can do it from right here cool go ahead and evaluate your limit from right here if you're the type of person that says you know what I really don't like dealing with all these negative exponents that that's hard for me to do well then what I would do if I was you is I'd keep thega 1/2 out front I'd write the limit as B approaches infinity and then I'd rewrite these two things because maybe that's going to be a little easier for you how can you write rewrite e to the -2b what could you do [Music] minus yeah maybe that's a little bit easier to see what's happening now back to my point that I skipped over important point to make my point is this you're going to have to be really good at determining what's happening to your function as you approach infinity or negative Infinity it's really important to do that so when we look at this what happens to e to the 2B as that b goes to Infinity what's what's happening to that this little piece goes to Infinity remember what e looks like e looks like this so as X goes to Infinity e going to Infinity what that means is that if we have one over so Infinity 2 * infinity is infinity e to the infinity is infinity how much is one over infinity yeah this thing goes to zero 1 over infinity is zero for sure now what's this thing go to this is it it's a constant so what that means is that we're going to have this - one2 times notice how when I evaluate the limit when I evaluate as B goes to Infinity I stop writing the limit but I don't do it until I actually evaluate it you guys clear on that one we do have to have it every single time until we get down to oh yeah okay when B goes to Infinity this thing goes to zero then in here we have what okay 0 - 1 e^ 2 which is equal to let's do it 12 * 0 - 1 e^ 2 how much are you going to get negative positive positive 1 over 2 e^2 what I find just amazing about this is that what we really did here is we found the area under a curve what curve this curve we found an area under this curve from one to Infinity so start at the number one and going forever and we got a legitimate area underneath that thing isn't that interesting it means that we're getting so close to probably zero as we go to or some some numbers you go to the right that it doesn't make any difference when we add that that's kind of that's cool to me I don't know why that's cool but it's so cool is it cool to you that we can add something up from one to infinity and get an actual number by the way did the example make sense to us good you guys want to keep going we going make it a little bit more difficult as we go well not more difficult but uh more involved as I want to give you some different cases so let's continue so I guess long story May real short um the integrals are going to be very similar to what you've done before the only idea is you need to identify what type of improper integral we have whether we have an infinite interval or an infinite uh sorry emphasize the wrong should level an infinite discontinuity there so whether we have a discontinuity or infinite interval so in our case here when we get this one what do we got do we have an infinite discontinuity or do we have an infinite interval which one both both neither one or the other what do you think do you know I want you to think about sign are there any problems with sign ever do you ever have a discontinuity on the function sign never it goes like this forever and ever so from 0 to Infinity are we going to have an infinite discontinuity no we're good it's going to be defined for this entire re range of numbers because it's always defined for the entire interval of numbers does that make sense to you now we do have a situation here we have well we got the infinity this would be called an infinite interval so I want you to do the first step write this as instead of going from zero to Infinity we're going to go from zero to what INF and in order to do that we add this little thing in front of our integral what is it okay so you should have done that already or be doing that right now so we think about this and we go no no no let's just start at zero because that's good but let's go to some number just to the right of zero and then let's take that number and push it to Infinity after we've done the integral so we're not really cheating but we're we kind of manipulating our integral properly so we're saying hey let's go from 0 to B that way it's definite it's a defin integral this says now at the very end take your B and push it to Infinity that's the idea now I know that um we've been dealing with some really difficult integrals so don't make these harder than they actually are don't do anything fancy for the integral of sin x what's the integral of sin x awesome so we're going to have a limit as B approaches Infinity cosine X from0 to B let's plug them in we're going to have a limit as B approaches Infinity cosine B okay minus cosine of Z let's see what we get out of this thing so negative cosine B plus cosine 0 let's start with the easy part uh what's cosine 0 how much is that so we know this is going to be equal to one now think about negative cosine B think about just cosine B what is cosine b as B approaches Infinity what what is it back and for cosine cosine does I negative cosine negative cosine does does it ever go to a number no no no it's it's going to go up and down down and up forever and ever and ever correct so my question is as cosine sorry as B approaches Infinity does cosine B ever go to a single number goes to lots of numbers goes goes from a range of negative 1 to one so if it doesn't ever go to a number does the limit exist it's diverent it's divergent This limit doesn't exist because cosine does this forever it doesn't ever go to one or negative one or some number between there this is a DNA type of limit if you get a limit that does not exist what do we know about the area here what do we know about that is it going to be a convergent integral where we can find an area or a Divergent area integral where we can't find the area so this is what I meant about um those cases so where we had convergent or or Divergent if we have an integral that equals some function that when we evaluate we get a single number like we did here this would be convergent or Divergent so if the limit exists it's actually equal to a valid number like this one we have an integral that's convergent our improper interal integral would be convergent show F be okay with that idea convergent means the limit exists if you get an integral so that when you evaluate with our limit here we get something that doesn't exist like this one it doesn't actually get to a number or something where we have infinity or negative Infinity that's when we get this Divergence so there's basically three cases the cases are we get an actual number with our limit we're good that meant convergent the other two cases are Divergent where we have a limit that does not exist because it doesn't go to a single number like cosine doesn't or where we get an infinity or negative Infinity that limit it it exists kind of but it's not an actual number that we call Convergent it's just ever growing or ever decreasing show fans feel okay with the idea okay so here we'd say this integral is Divergent you know you might have guessed that just by looking at the function itself look at sign does sign ever go to a no solid place no really it's always going back and forth so if you thought about what integrals actually mean integral means area under a curve so if we're thinking about hey you know what I want you find the area under the curve from zero to Infinity of sign you're going all right well there mean area then very little area then more area then very what's going to happen there you know it's going to it's going to go forever like that you're never never going to be able to tell me how much the area actually is so far so good okay cool have I explained these two well enough for youall to understand them guys have any I'm going to ask you now before I erase them any questions on these before we get going okay so doesn't exist it's Divergent if the limit does exist it's convergent we're good to go we can actually find the area that's very cool let's continue can we ramp them up a little bit these ones can get kind of boring if we do simple integrals all all the time so let's see how this works with something a little bit funner okay first Duty first Duty we got is to Define what type of are you laughing because I said Duty uh you guys watch too much scrubs um anyway what was I say Duty first Duty the first Duty should be the second Duty anyway um first D is to find what type of improper integral we actually have whether it's an infinite interval whether it's an infinite discontinuity here this is all defined for for the entire domain X and time e x we're good to go we have no issues here so it's not an infinite discontinuity those should be kind of obvious to us when we get them you'll see them what we do have is an infinite interval we actually start though at negative Infinity so we're going to do something very similar to this idea only I'm not going to be going from a number to B I'm going to be going from how would I write this okay so because our lower bound is an Infinity I'm going to call that a zero I'm good with the zero we're just going to go from a to zero does that make sense to you now of course we got to Define what we're actually doing here so I can't just go from a to zero and not Define what a is doing what's a doing approaching infity good let's make sure we get that negative Infinity in there and that means the only way we can do that to let a approach negative in Infinity is with its idea of a limit quick head now if you're okay with that one so typically we call our upper bounds B's we call our lower bounds A's if they are positive infinity or negative Infinity respectively after that what's really cool again is that we know how to do all these integrals so let's go through this integral right now is the time we'd go okay well cool we defined What's Happening Here we now have a definite integral and then we're going to let a go to negative Infinity to to take this idea of negative Infinity to zero but we just got to worry about this right now so the integral of x edx DX let's go through what we know does it fit the integration table no can you do a substitution with it could you do a substitution if I did this yeah that would be nice so that's what we would do in this case if I don't have that then no we can't do a substitution can you do integration by parts yes you can it's a product we can do integration by parts with it so let's do integration by parts what would you pick what would you pick for do you remember how to do integration my parts it's been a while so this is one of those things that you should kind of uh do from time to time on on your own uh whether I give you homework or not because we haven't really covered that in a long long time all right but it is part of this chapter so what would you pick yeah U is X remember how to pick U you want to pick these u and v this way you want to pick so that you can do the integral of what's left over very easily and that the derivative or that the derivative of U becomes easier so here I'd say hey can I do the integral of x I can but it gets worse it gets X2 I don't want that can you do theal of e x uhuh andal e x is okay that's nice can you do the derivative of e thex that's e thex doesn't get nicer can you do the derivative of x that's one that's nicer so our appropriate Choice here is = X and then V equals way I always teach this I say hey cover up your u v is the integral of whatever's left over so dual DX V = evx because that's really nice and then do you remember the formula for for doing integration by parts honestly I heard okay very VD very good so so we have you know the annoying part about this is we still have to write the limit because that that's really still what we're doing okay so we've got a limit we've got a approaches infity negative Infinity that's right now in this limit we don't have the integral anymore we have well our integration but part says we do a u * V so what are we going to write first very good x * e x you guys with me still so far minus the integral d good so U * V minus the integral of V du but du is DX so we have integral of v and then du is our DX show P feel okay with that one by the way for those of you who are thinking well what about our balance here don't worry about it you're going to evaluate later don't evaluate this integral right now wait till you do it then evaluate the entire thing from a to zero does that make sense to you if you try to do it here just makes it too hard okay okay so wait wait on it for a second till the very end till your integral is actually done you got that otherwise you're you're really working overboard so I am going to put this here I'm going to put the a to zero so I don't forget about it but I know I have one more step in here to do before I actually evaluate that so I'm going to have a limit as a approaches negative Infinity I have X I've got e to X this is why integration by parts works so nice for this we reduce the power of our X by 1° if we've done that then that means the rest of this integral is really nice for us what is the rest of that integral for us from what are we going to do now what do you think we could simplify we could Factor if you want to sure you could Factor this you could do it right now if you want to it doesn't really matter to me how you do this U if we do decide to Factor this since you guys said that we'll have a limit as a goes to Infinity uh what perfect okay well now let's go ahead and evaluate that so what we've done so far is we' first thing we do we identify that we actually have an improper integral we we find out the type whether it's the infinite discontinuity we haven't even talked about yet or an infinite interval if it's an infinite interval no problem we know how to do that let's just take a limit and instead of the infinity we'll make it definite for like the right now it's a definite integral right now that means that we can actually do it okay by fundamental theor of calculus this actually works so we do this so that this is actually possible then we realize oh but you know what B wasn't actually definite B was this number that we want to now move to infinity or a was this number that now we want to move to negative Infinity so we're cheating a little bit we're calling this a definite integral when we know it's not but it uses the fundamental the calculus to actually do the definite integral and then says now use a limit idea now move that number to infinity or negative Infinity hey just evaluate the limit so when we do we're going to evaluate the integral but then evaluate the limit so a is going to eventually go to negative Infinity we got to plug it in first to see what happens which number do you plug in first make sure you do that okay make sure you you follow the the rules for evaluating your integrals here so we'd have e to 0 * 0 - 1 minus E to the a * Aus one okay let's simplify some stuff here right now because there's I have no idea what's going on with this I just got to I got to simplify this cuz now it's a little bit too confusing for me do you guys see what I'm talking about a little bit too much going on so uh what can we do with this thing e to0 is how much good 0 - 1 is so this whole piece it's 1 * 1 this thing goes to netive 1 very good also I'll tell you right now what's going to have to happen here because I have infinity and infinity I'm probably going to want to distribute this real quick so might have been nice if we hadn't have factored that but you know what not a big deal we take care of right now so in our limit we've got -1 we're going to have minus a * e to the A and then plus e to the a can you check my work I want to make sure that my algebra is correct too can you can you check that to see that I'm I'm okay on that one are we okay with that so we I've distributed this this is a e a minus E to the a but then my negative changes sign so minus a e to the and plus e so feel okay with with that so far now let's think about what's going on here do not neglect your knowledge of limits from your chapter six stuff that we've done okay you got to really follow that so if we look at this part of this is really easy so check out the uh the negative one what's that going to go to that's easy check out the e to the a oh let's think about this e to the a when a approaches infity so basically what's e to the negative Infinity it's zero that's right so this is one negative 1 this is zero that's fantastic now please listen carefully this is a * e to a correct we already know that this part is zero e to the Nea Infinity is zero but this Isa Infinity so we have an Infinity Time Zero that's an indeterminate remember that ohd it's coming back this is exciting it's an indeterminate look at how much stuff is involved in this had you not done indeterminate forms you couldn't do this problem had you not done integration by parts you couldn't do this problem and now we have improper integrals so we've got lots going on in this stuff this is why this calculus builds so nicely for us but you got to remember it all right so it's it's not like other classes where you can forget a section and ah you're probably okay you need everything to do these problems sometimes so for us we go all right well cool this is easy 1 this is easy zero but this one well that's an Infinity time0 type of indeterminant form what that means when we have Infinity time 0 we've got to rewrite one of those in order to get an Infinity over infinity or a 0 over Z in either case we can use ly remember ly talls our favorite word in limits ly tall so I'm going to do something here I'm going to separate these what I know is that this is this is cool this is NE 1 this is also zero so I'm going to do this I'm going to separate these limits I know I'm going to have1 I'm going to have minus the limit a * E A as a approaches negative infinity plus Z so basically I just want to focus on this guy I want to see if you guys are okay with the math on that are you okay on where the negative one came from yeah that's this limit uh limits are seable by addition and subtraction this is zero no problem I'm just focused here my minus is right here so I have a limit as a approaches Infinity of this guy just that guy quick show hands if you okay with with that one now if you don't remember how to do the 0 time Infinities or Infinity time zeros rewrite this as a fraction so take one part of this and move it to a move to a denominator now what I know is that it's going to be easiest if I move this e to the a to the denominator so I'm going to take a little little break here we'll call this asteris the limit as a approaches negative Infinity of a * E A can be Rewritten as a limit as a approaches negative Infinity of I'm going to keep the a I'm going to keep the a right where it is but the e to the a I want to rewrite that can you tell me how can I write a * e to the A as a over something to thetive yes in your heads is this the same as this stuff M now check it out remember whenever we change something in our limits I don't know if you do remember it was a long time ago but if you don't remember rewatch uh the end of chapter 6 okay on on the video and it will show you all this stuff this was the indeterminate forms whenever you do something algebraically you re-evaluate it and you see what actually happened so so if we re-evaluate this how much is a as a approaches negative Infinity now how much is e to the a as a approaches negative Infinity remember a negative * netive is a positive how much is this it's Infinity so we have a form of infinity over infinity does that make sense this is a prime candidate for yay l a fun word even so L doll can you explain to me please how you do a lals rule what do you do take ative of the top over of the bottom cool is it a quo rule no no derivative of numerator is this is why we left the a where it was okay if we hadn't if we made it 1 over a our a power would grow and our E power would not decrease so that's why we left a where it is because the derivative of a is that's beautiful the derivative of e to the a can you tell me what the derivative of e to the a is some said e the a some you said negative e to a which one's right what gives you that chain oh my gosh we even have chain R here so derivative means you leave it alone you leave e to the a alone but you multiply by the derivative of whatever functions inside there this a function of a der of a is1 now can you evaluate this limit now yeah sure how much is e to the Nega so what was what was that again e to thega a as a approaches negative Infinity oh my gosh so check it out if a goes to negative Infinity you got negative Infinity there right negative negative Infinity is e to the infinity is infity negative Infinity I don't care what it is if it's infin if it's a number over infinity you get how much Z this is zero now this was our asteris so we move right back up here so what our limit is is what do we have left on this thing this whole thing went to zero What's really in I I don't want you to forget what we're actually doing okay a lot of people in calculus they end up learning how to do this stuff but not really what what you are doing uh what do we what do we just do here what do we find area under this curve from negative Infinity all the way up to zero and how much is it equal to interesting that means it's below the x-axis that's what that area means the majority of it below the x-axis should Advance be loated with this example so we do hey infinite interval no problem make it definite just Define what's happening this means that we can actually do the integral by integration by parts or whatever we have here integration by parts saying hey u x equals integral the rest do integation by parts U * V minus integral of V to evaluate that thing once you've done the complete integral not right here but after you've done the whole thing evaluate it then simplify it and then work with your limit you've have you covered haven't you covered all this stuff before yeah oh nothing's really new nothing it's just putting it all together now sure you okay with it okay can we move on a little bit do it finally in a roll these problems don't take 45 minutes a problem like before which is nice too you're going to love that on your homework I'm sure well except for this [Applause] one what okay okay okay why F why do we do this why you want to murder us oh my gosh um let's think it through let's think this through first thing are we defined because we always check this first okay look at your function inside of your intergral is this function defined for your entire interval if you say no then you better give me a number that doesn't work here and you better rewrite so if you just guessed at it and said mm cuz it looks funny uh probably not the best thing to do all right so why don't you think about it is there anything that gives us a problem in this function is e to the X defined everywhere yes yes so this is fine is e to the 2x defined everywhere yes and if I add it to one do I have a problem in my domain no no this is defined everywhere but that's what I'm asking you to check you got to be good with find a domain this is why your teachers taught you how to find domain so you go oh you know what uh my domain here is all real numbers therefore on my interval of all real numbers I'm good to go I don't have to worry about any infinite discontinuities because there are none on this function I know I just talked fast but did that make sense to you are you sure there's nothing that makes this zero or negative therefore if I if this is always positive it's being squared so no problem it's positive if this is always positive and I add one to it it's still always positive it can't ever be zero can't ever give me any problems this is always defined they're all both they're both always defined so my entire rational function is always defined for entire interval now the problem is what's the problem here what's our problem yeah it's this idea of oh my gosh we've got this double infinite interval where I'm not going from negative Infinity to a number like zero I'm not going from zero to a number like Infinity I'm going from negative Infinity to Infinity so what in the world are we going to do yeah you know what we're going to do is instead of thinking of this as one big integral let's think of it from negative Infinity to I don't care what number you pick it's going to be the same whatever you want to pick um now I'm probably going to pick the easiest number I can think of to plug in I'm going to pick zero because zero for me would be the easiest number to plug in you guys know hopefully from your Calculus one class whenever you took it that you can actually separate integrals like that start from wherever your a is introduce some intermediate number c and then go from that number c to where your ending value is B so instead of from A to B we go from hey a to c and then again from C to B and we add those intergal together that's legal to do as long as you have no problems in between there that's legal to do quick head there if you're okay with that one hey have we seen stuff like this yeah have we seen stuff like this can you guys do the next step for me write out what you would do change these into our limits for us okay remember that lower bounds typically change to A's upper bounds typically change to B so go for it what is nice about this though is that we don't have to do the integral twice since we're just rewriting the same thing our integrand right there hasn't really changed uh we'll just do the same idea for both of these these integrals at one time so here's what I'm asking you to do I'm asking you to firstly understand that when I have this type of integral goes from negative Infinity to Infinity you do have to do this with two integrals here you have to separate it so we have to have at least some solid Bound in between there call it zero call it one whatever you want zero is going to be easier to plug in remember that we do have to evaluate these things right so we're going to have to plug in this number twice zero is going to be easier to plug in than7 or something like that you will oddly get the same answer out of that but it's going to be easier to plug in if you have a zero so we'll have instead of going from negative Infinity to Z what's my bounds of integration going to change to now I'm going to Define it limit as perfect the inside of my integral my integrant is not changed Plus what's the next thing I'm going to write as good well if I've called B positive Infinity Zer doesn't change with infinity I'm going to call that b e to X over 1 1 + 2 that's not right e 2x okay quick show hands be okay with this idea just a combination of two ideas that we've had before and here's a nice part once you've done this I don't want you to do each of these integrals independent of each other understand that they're exactly the same you guys get it so off to the side I'm not even going to worry about the eight to zero I don't care I'm not going to worry about the zero to B I don't care about that I'm going to just do the integral of e x over 1 + e 2x let's just do that take this idea and run with it afterwards and then we'll evaluate whatever we get twice from a to zero and then from0 to B do you understand that idea subtit integral off to the side now hopefully you can do this integral let's walk through it is it a substitution yeah if you say no okay if you say yes explain that to me = e oh my gosh someone's good at substitution true yes hey let's be smart about this don't jump to a conclusion if I don't know how to do it rewrite it think about these things think of can you find a derivative somewhere in your problem if you can then you have a substitution here so e to X not a problem e the o e 2x no let's think about that as e e X2 if that's the case a substitution of u = e to x man it works nice du = e x DX the derivative of e x is e x check this out here's e x DX here's du it's right there you guys follow me on this one okay so integral that looks really hard right now not so hard tell me what's going to be in the integral tell one okay so this goes away this is basically you can write du even if you want to I'll show you that you can write du because e to X DX is Du e to X DX is Du uh over what 1 now if you don't like the DU being there okay call it a one put the D over here that's no big deal I just want to make sure that you are okay doing our substitution e 2x is e x quantity s and then doing = e x hey no problem this is 1 + u^2 = e x DX we got a 1 du show okay with that one now you've got a couple options here you can do this the hard way or you do this integration table way the hard way would be a trig sub the hard way would be and this would be just fine if you want to do this remember doing this even though you don't have a square root you can put one in there you square root of this whole thing squared you would have square root of 1 + U ^2 you'd have U and you'd have 1 you would have tan th = U does that make sense you do it that way or you could recognize that this right here is actually in your integration table now when we're doing trigonom sorry when we're doing integrals the first thing we should be looking for is our integration table right we shouldn't make things harder than they have to be we should try substitutions and integration table first well if this is in our integration table that's brilliant how much is it 10 verse terse terse up a new word terse tangent inverse of U yeah absolutely that's exactly what that is now let's not make a mistake I did before by the way would you guys know how to finish it off if you did the triangle let's say you didn't see this on your test could you do that sure you know trig sub tan Thal U dual see s d Theta you'd have a little bit of work to do here but what you'd end up getting is a see squ Theta over see Square Theta You' get one the integral of 1 is Theta Thal tan inverse of U bam it's right there it's exactly the same thing that we would get so tan inverse U what's the next thing I'm going to do substitute very good so if I substitute I get tan inverse uh oh tan inverse of what perfect tan inverse of U no no no tan inverse of e x so your fans feel okay with that so far now I'm going to rewrite this because I want to be a a little bit more clear than what I'm being here so do you have any questions on on this side no all right I'm going to show you exactly what's going on with this thing here so what we ended up getting was these two integrals and then we Chang them to okay some definite integrals so we can actually evaluate them but we made our limit so that these bounds go to negative infinity and positive Infinity respectively so I'm going to rewrite this piece then we're going to look at what happens in conjunction with that so we had a limit as a approach negative Infinity the integral from a to 0 e to x 1 + e 2x DX big plus limit as B approaches positive Infinity 0 to B of the same integrand you guys okay with that so far now we've already done the integral of e x 1 + e 2x how much did this integral give us so let's just change those parts so if I know that the integral of e x 1 + e 2x is actually tan inverse of e to X which it is can you tell me where I'll be evaluating this thing from sure is that pretty straightforward to you guys to me it makes perfect sense I want to make sure it makes perfect sense to you so we say hey this integral is this thing therefore if I want to evaluate this integral from a to zero I just evaluate this thing from a to zero now I can't forget what limit the plus yes but I also can't forget this this limit plus now I get the same integral which is why we only do it once it's silly if you do itce this the same thing this integral is also tan inverse of e x I'm going to evaluate that from where to where please and I'm not going to forget the there's only one right answer here limit as B approaches inin okay what I want to know is can you make the jump from doing our integral here to evaluate our integrals here show things if you can make the job okay what do we do now so are we still going to have limits right now [Music] yes have you evaluated the limit yet no no right now what we're doing is we're look it look if I didn't have this limit I'd still have to evaluate wouldn't I me you haven't done limit yet so I'd still be plugging numbers in so I'd have tan inverse of e to what comes first the a or the zero zero the zero comes first minus tan inverse of that's very common for people to make that mistake it really is they go oh well zero always comes no no no the top top bound your B always gets evaluated first and then you do your lower bound so if hands feel okay with with that one are you sure okay so here's one big plus limit as B approaches Infinity tan inverse let's try this again tan inverse of E2 that would it be zero first or B first minus tan inverse of e 0 now are you guys okay with that one notice why we still have our limits have you let a go to negative Infinity yet that's why you still do limit now we get to let a go to negative Infinity we let B go to positive Infinity so we're going to stop writing limits right now let's think about this how much is e to the 0o hello so this is one how much is e to the a as a approaches negative Infinity z z very good this is zero as a goes to negative Infinity e to the a this is e to the negative Infinity e to the X goes like this so as negative Infinity we're going to zero how about e to the B as B goes to Infinity how much is that going to be Infinity very very good e to the infinity is infinity how about e to the Z again so now that I let these go to their appropriate places I stop writing the limit this this was a value in limit saying hey look at that uh e to the a as a goes NE infity that's zero and this one's going to be Infinity so we have tan inverse of 1 minus tan inverse of 0 plus tan inverse of infinity that looks a little weird to write but we'll write itus tan tan inverse of 1 now we got to be good at tan inverse let's see what happens we're almost done with this thing how much is Tan inverse of 1 very good so this piece Pi 4 it says tan of Pi 4 gives you 1 therefore tan inverse of 1 is pi 4 how about tan inverse of 0 Z tangent of 0 gives you zero therefore tan inverse of 0 gives you zero plus how much oh this is an important one how much is Tan inverse of infinity so basically tangent of what value will give you infinity pi/ 2 will remember tangent looks like this at pi/ 2 that make sense so if tangent of Pi / 2 is ASM totic uh then we have a Infinity so therefore tan inverse of Infinity is/ 2 it's a little weird because we can't actually plug in Infinity but we think of it like that would think of how can I get really close to Infinity with a tan inverse it's p / 2 minus we already did this one it's Pi 4 what's our final answer what Flo me what's really cool is that we have this function right with e to X in there e x 1 plus e 2X and somehow we get a pi over two out of that thing as a solid area isn't that kind of cool yeah so if that made sense to you good deal all right so the last little bit that we've got to cover in this section is instead of an infinite interval we got to find out what happens with an infinite discontinuity what that means is that instead of going from like a negative Infinity to a number or a number to infinity or even remember the last one we had negative Infinity to positive Infinity somewhere within our interval whatever it is our function is not bounded what that means that we're going to to infinity or negative Infinity along the Y AIS so basically we just have an ASM toote somewhere along the interval on which we're trying to integrate do you guys see the difference in the idea yeah infinite intervals go this way for Infinity infinite discontinuities go this way to Infinity the function is actually going to Infinity let me give you an example and it's going to clear everything up so first example let's say I want to go from 0 to 9 find the area of the following function from 0 to 9 now right there is that an infinite interval or a finite interval sure that's that's great so we don't have to worry about limit of B going to infinity or a going to negative Infinity we're not worried about that however remember the first thing I told you to check when you're talking about your your definite integrals is well uh look at your balance of integration and see if your function is defined for that entire interval is it defined for our entire interval No No in fact What's Happening Here with this function is this thing looks a sketch here looks kind of like this where as we go towards Infinity yeah cool we're going to get you know smaller and smaller values because we have a fraction over a number that's growing even though it's a square root it's still a growing denominator does that make sense to you however as we get towards zero What's Happening as we get towards Z it's going up it's going towards Infinity because when we're dividing by smaller and smaller numbers we're getting this oh what well we got an undefined Zone here and that's an issue so as we get closer to zero we have this issue we have what's called an infinite discontinuity it's not continuous it's going towards Infinity the function is not bounded show feel okay with with this one okay now on the left side of zero well we have nothing because we can't even plug in negatives because we got square root so this is basically what our graph looks like you got it yeah so here's our idea instead of going from 0 to 9 what's the problem here is it the zero or is it the nine or it somewhere between there Z it's the zero the zero is the problem so tell you what we're going to use a very similar idea to what we did last time instead of going from 0 to 9 maybe will go from I'm sorry what was the problem again zero is the problem or nine is a problem we'll go from C to 9 and we'll make C some intermediate value tell me something hey you ready tell me something if this was a number like one could you integrate this thing very easily piece of cake if it was like 01 you could still do it 0.1 you could still do it zero not so much we're going have to have a different idea so it's really similar to the idea of well we can't actually plug in Infinity but we can plug in a number right and then let that number go to Infinity with a limit so the idea here is all right well let's let's call it not zero but C now the question is what do we want to happen after we're done with this thing what do we want to happen to our C where is our C supposed to go okay our C so we want our c C to approach zero yeah that's right we want C to approach zero now there's a problem with this and this is why we talked about where the functions to find begin with is there any function on the left hand side of our zero so what this says if you remember anything about limits what approaches zero means is that the function does this from both sides remember that we have to make this not a both-sided limit but a one-sided limit so we're going to have to approach Zero from a specific side right side from the right side so we do use a plus or minus good what we want is let me recap this real quick so we understand everything about it then we're I'm going to give you the definition for it good to go the idea is well you know what the interval is finite that's fantastic but our function is not defined for the whole thing we have an infinite discontinuity wherever that happens listening wherever that happens change it to a c what we're saying is that okay I can't define the zero here it's undefined so let's call it C that way I can even do my integral we're basically changing this to a definite integral right now from a number to n no problem can't you do a definite integral from a number to n we just talked about how we could after we're all done then we'll say well what do we want to happen to our c what happens there approaches zero we don't want the C to be here we want our C to be there we're going to let that c Goes To Zero from the right hand side this is the idea should hands feel okay with the idea all right so basically this is where we get our our limit again so we'll go right fantastic I can do an integ from C to 9 I just have to let the C approach Zero from the right hand side of the same function so we're saying 0 to 9 no no no we'll go C to 9 that's a definite integral fundamental the calculus says we can evaluate definite integrals that's why we're doing this right now so we go Okay C to9 cool deal do the integral plug in nine plug in C subtract the two values and then at the very end go wait a minute let's trick it again let's say no no I don't want just c a finite number I want to say that thing is going to go to Zero from the right hand side it's going to get closer and closer and closer and closer and closer quick head KN not if you're okay with the idea now it's just a matter of evaluating it's going to be a limit again can you guys evaluate this integral what would you do with this it yeah so C Goes To Zero from the right remember we still have to write that C to 9 by the way if you didn't if you didn't catch where the C is coming from a is typically called a lower limit B is typically called a upper sorry lower bound B is your upper bound so if we had negative Infinity we' call that our a our positive Infinity would be our B anywhere in between these two numbers it's typically a c so that's where the C's coming from okay so y all said that this should be x to the what am I going to do here x to the what oh wait I'm sorry 12 orative 122 good i' move this up that'd be X to the2 DX cool deal let's keep going so this will be our limit as X as C approaches zero from the right of let's go ahead let's do our integral what's the integral of x to the2 tell me that one okay X to the2 I like over2 so basically two so you okay with that integral um I'm going kind of have to warn you now I'm going to go fast through a lot of these integrals because the idea is no longer hey can we do integrals it's let's look at how to do these these improper integrals does that make sense to you so actual integration we're going to go a little bit quicker now do I still have an integration symbol here you've already done the integral so all we're doing now is evaluating where are we evaluating from9 okay which way 9 to C or C to 9 which one do we plug in first so this would be the limit as C approaches zero from the right hand side 2 < TK 9 - 2 < TK C still okay so far let's talk about these things let's talk about what happens as C approaches Z what's going to happen to this part how much is that going to go to sure this is three that's two so we're going to get six notice how we're stopping writing the limit once we evaluate the limit but before that we have to have it every single time now what happens to the square root of C as C approaches zero notice we don't have a two- side limit here which is why we had to go from the right hand side only it doesn't exist from both sides so just think about what happens what happens to a square root of a number as that number gets really really close close to Z it gets really close to zero does that make sense to you if you pluged in zero You' get zero so we'd have minus 0 therefore what's really so cool I don't know why it's so cool to me but it's so cool to me it says that what's an integral even mean so there's a definite integral right between 0 and 9 of something that's not even defined is zero yet what happens is that if I have the area because of this because of our limit idea the area under this curve from 0 to 9 is equal to very cool idea very cool CH all right with with this so there's a few scenarios I want to discuss with you let's say that we have so for a discontinuity at B for discontinuity at a [Applause] or for a discontinuity at some intermediate value C so let's say we have some function and there's going to be a discontinuity here an infinite discontinuity at B so our function is defined at a no problem it's going to do something I don't know what it's going to do but somewhere around B there's going to be an ASM toote do you guys see what I'm talking about the infinite discontinuity at B hello yes no I don't care what happens past it my interval is going to be from A to B I don't care what happens before my interval is going to be from A to B does that make sense to you so if I tried to integrate from A to B of some function then the idea is well let's look for the problem area let's look at what's what's wrong with our integral what's making it improper now clearly hopefully it's clear to you my integral is finite it's going from A to B it's not an infinite idea so I don't have to go to positive infinity or negative Infinity it's not a problem but what is a problem is that while a is defined B is not so my discontinuity would be where A or B so wherever that discontinuity is that's what we're going to call C or whatever whatever you want to do so uh you can call it any letter you want we typically use C because it's going to be an intermediate value between A and B does that make sense to you so for us we all right well well awesome so we'll have I'm sorry which one would we would we change here would we change A or B that would become a c so we'd say no no no no not B let's make some point C some intermediate value and then where are we going to let C go to we're going to oh not Infinity because I'm not going to Infinity anymore I'm going to go to B so the limit would be C approaches B now think carefully here so here's here's the idea one more time if I didn't cover this well enough for you the idea is you're going to have a problem with an infinite discontinuity it will either be at a or at b or somewhere between there I'll show you that one last if it's at one of your your end points of your interval A or B well then Define that endpoint as some intermediate value C so we go no we can't go all the way to there so let's go to here let's go to C not B does that make sense do our integral because now it's a definite integral that's defined from a to c do you got it after that we go all right well where do the C really want to go C wants to go this way a little bit C wants to go this way a little bit till it hits B now from which way from the left or from the right from the left from the left that would be the idea idea I want to show hands to see if you're okay with with this one now let's reverse it let's say that we have some function that has a discontinuity at a but is defined at B can we do it the way it is well no we have an infinite discontinuity the interval is great it's from A to B not a big deal so if we had an an integral that says I want you to integrate from A to B of whatever this function is where is our problem here is our problem at a or is our problem at b b a i mean a yeah it's at a B is fine right B is defined we're good to go so what we go is well um tell you what I don't I don't really care about going to be that's good good to know but I can't start at a if I can't start at a because it's not defined where am I going to start C some intermediate value C so my B would stay the same but my a would become a c what that lets us do again hopefully the last time that gives us a definite integral that we can actually evaluate it says now you're going from a finite number to a finite number that you can actually plug into your your integral when you're after you've already evaluated that's great we go okay plug in the B plug in the C subtract them good to go uh calculus tells us we can do that we can subtract that evaluation when we plug in B and then subtract off when we plug in a or sorry C in this case now the problem is well I'm not really starting at C I want to start where a I want to start at a so yeah this integral says from C to B but now this limit can take the C and move it which way to the right to the okay so let's just say two somewhere to where to a to a from the right side from the right side that's right that's actually the first example that we had that's right here they said okay I can't start at zero but I can start at C now at the very end we go okay look at that from C to 9 actually lets you evaluate it plug in the nine plug in the C that's great that's what different integrals are all about then it says well I want that c to not start here I want to start really close to here and that's what this limit idea says I want the C to start really close to here so towards Zero from the right hand side that's what this is show pants feel okay with that one now there's one situation we haven't covered what's the situation we haven't covered there a hole right in the middle holes I don't care about because that's why that's why they're called removable discontinuities they're not going to affect this um it's still B bounded correct so if we have a hole we can plug it with a single point does a point have any width to it that's is going to blow your mind if a point has no width to it then if I'm missing a point on my line so a hole basically since a point has no width it can't possibly have an area under that point does that make sense it's area is width times height if it has no width I don't care how high it is0 time anything is still zero so for that reason if I have a removable discontinuity that does not affect the area under the curve those points have no width crazy right what'll blow your mind is you can have an infinite number of those on your line and still get an area what anyway different class different class uh discontinuity at C so what happens if it's not hit did I just completely blow your mind like a mind grenade do you see what I'm talking about how that one point doesn't affect the area yeah that's so Neato because there's no WID so therefore we don't care we don't care about the holes we care when it's not bounded right because that's a problem we need to Define that a little bit better so it's not here anymore it's not here anymore it's somewhere in the middle so here's our a where we'd have something toed here's our B where a function would be defined but somewhere in here we've got a problem oh did I say infinite number holes I think I might have meant finite I'm not sure I don't remember anymore could be a finite number of oopsie that would be just a little bit of a problem uh there I don't know that exact thing well what's the issue here are we defined at our end points of our interval of at our end points A and B are we defined yes so then our infinite discontinuity doesn't happen here and it doesn't happen here it actually happens right in the middle whenever you get this case this is the worst case scenario this one's kind of nice cuz you just go okay it's one limit does that make sense to you one limit because we have one discontinuity from One Direction here no problem one limit one discontinuity from One Direction here one discontinuity but it's from two directions so whenever you have your infinite discontinuity between your bounds of integration the A's fine the B's fine somewhere in the middle we have a problem so somewhere in the middle we'd say okay where would we go where would we start from if I want to do this i' start at a where's my problem at c c yeah I have i at C I'd have to split that up then I have another integral starting from where and going to after you do this so basically here's the idea if you have it at one of your end points of your interval not a problem just call it from a to c or from C to B respectively and let's that c go to the appropriate number if it's in the middle you have excuse me you have to break that up so the starting point to where C is then C to where the ending point is where your ending interval is and then you do basically the above so basically do you guys see that this idea is a combination of these two ideas we'd get one of these and we' get one of these specifically this integral is this one this integral is this one and you do exactly the same thing makes sense are you sure you guys want to try a couple of them yeah okay probably let's do so any questions question what if it doesn't approach Infinity on both sides only one you can still try it has a point and then it comes down from what do you oh like a like a pie wise defined function like this one yeah that'd be great because this integral is defined so this would be like easy okay you say hey a to c no problem but then this side you still have to split it up because it's not continuous you have an infinite discontinuity hence the not continuity part okay so this would be good this part you say well I got to do this one here that's a good question so you just would skip the limit on one you wouldn't have to do it for the left side that'd be easy because it's actually defined that's a good question and any other questions what would the limits look like for uh the a to C and C to B it's easier to show you with um actual numbers but what you do is you call this an intermediate Point D and you say a limit as D approaches C from the left you'd say a limit as maybe e as e approaches C from the right so you'd have to have this is why I don't want to show you that many letters cuz like what in the world are you talking about here but youd go from a to some number D okay okay you do your definite integral and then you you let D go to C your undefined point from the left hand side here you to find some intermediate Point e you've already used C and D so e then you do your definite integral from E to B after that you let e go to C from the right hand side did you guys follow that same ideas it works easier with numbers because they don't have all these letters f around okay so let's actually get to the nuts and bolts of this [Applause] thing you know one of the most important things we can do as integrators of this these improper integrals is determine what we're talking about whether we have a an infinite interval or whether we have an infinite discontinuity which in this case do we have an infinite interval or an infinite discontinuity interval is great we're going from 1 to four that's that's a definite interval it's finite but within that 1 to four we've got a problem so you have to Define all the problems you have and there can be several of them you could have a problem at an end point you could have a problem at both end points you can have a problem at both end points and somewhere in the middle or several places in the middle now of course right now we're going to keep it kind of easy right there going be like one problem but you could have many many different problems does that make sense to you so everywhere your denominator is zero that's an infinite discontinuity provided you can't simplify out the uh it's not a hole basically um that you would get from calculus one the difference between holes and ASM tootes if you had me you should know that exactly what the difference is between holes removal discon ASM tootes so in our case uh where is our problem at four it's not a one one's great I can plug it in stop zero two three cool four h that's a problem so what we're going to do is we're going to say well you know what if it's not defined at four let's make it be defined somewhere at some intermediate Point C are you guys with me on this one that's uh which case is this is this the case for one two or three which case do we have one we have one we have this one it's it's uh it's fine it's defined at a it's not to find at B it's defin at one it's not defin at four so we're going to call this some point point right before the B right before the four we're going to call it C maybe it's 3.99 99999 forever I don't know what it is we're going to call it C and we're going to have exactly the same function we just have to say what's happening on my interval so we have to say where C is going so what how do we say what where C is going what are we going to use limit yeah limit is the way we say where a number's going a limit as C approaches where it's always going to be this number or this number or some number in between there it's always going to be a number so it's not going to be a random letter anymore okay we're defining what's happening to the C now you got to be good at this is a c from four from the right or four from the left left from the left so let's see if it matches up C is going to the top number from the left C's going to the top number from the left hey that looks exactly right can you do that integral it should be pretty easy what would you use in order to solve this integral yeah what would the U du = DX orative du = DX again I'm going to go kind of quick on this but don't do not uh neglect your substitutions okay your signs are important you guys see where the negative is coming from on that one no that's important okay so we have a limit as C approaches four from the left of integral from 1 to c u to the -23 * tell me what I can do here okay you can actually pull it all the way out front if you want to you can pull it in front of the integral it doesn't really matter U -23 du uh can you all integrate U to -2/3 how do you do that okay so we have our negative we add one we get 1/3 divided by what3 so divided by 1/3 so this would be 1/3 / 1/3 U to the 1/3 divid 1/3 no more integral we've already done the integral what's the 1/3 change to now we can evaluate right now from 1 to C or you could pull your -3 all the way out front I might choose to do that just to show you that it's possible so from here I can pull that -3 out in front of my limit and do U to the 1/3 from 1 to C now there's a big issue right now with plugging in the values what's the problem changing down we didn't change bounds so if you didn't change bounds instead of plugging into you what do you have to what do first we got to take this subtitution put it back so we're going to have to do -3 * the limit as C approaches -4 sorry four from the four from the left from the left hand side of what's it going to look like 4 - X3 okay so 10 20 second recap here we go we first identify what type of problem we have whether it's an infinite interval whether it's an infinite discontinuity maybe both of them look for all your problems here we're defined we're good to go here we're not defined at this number at that end bound end point of your interval we call that some intermediate value C we say Hey where's the C going well it's going to go to four from the left hand side we don't care about the right hand side that function is not even there we don't care we're going to evaluate our integral any way that we can we have lots of techniques to do that get all the way down to here and then oh my gosh well if you use the U substitution before you evaluate definite integral you definitely better plug back in your U so we have a 4 - x^ 1/3 now we get to evaluate so we've got -3 let's not forget that a limit as C approaches four from the left what goes in first the C or the one the C 4 - c to the 1/3 minus 4 -1 to the 1/3 this is the point where you can start to evaluate your limit notice we've had to write limit every single time till we get right down to here because we haven't thought about what's happening as C is approaching four yet so this right here look it this is your definite integral this is evaluated on your definite integral does that make sense this is where you say okay now what was happening C really wasn't just a number I wanted to stop at C was something I wanted to go to four from the left hand side does that make sense to you so let's see let's see about no pun intended let's see about the C uh what's happening to the C it's getting really close to four really close to four what's four minus something really I I mean really close I mean it get to four practically speaking how much is that so this thing is going to go to Z since C is going to four from the left hand side this thing is getting really close to zero 0 to the 1/3 is still zero so we're going to have don't forget about this3 it's still there so we have -3 * this is zero minus how much is this going to bebe Al together we have three cuot this part should be pretty easy show if F feel okay with with this stuff can we move on just a little bit okay so moving right along questions all gone already have I taught you what the integral of um lnx is have we done that in this question oh no it's the integral of 1/x not the derivative of L and X the integral of LX have we done that think we've done it once sorry U * V minus the integral of B du what am I doing up here magic magic I mean integration by parts U V minus integral of V duu this is one you know I I never missed those C's some did the wrong thing but I never missed those C's there we go we get x l and X - x do you guys see where the x is coming from it's just basic integration by parts I know we've done it at least one time before uh but what that means is that if I wanted to evaluate this limit from oh where do I want to go 0 to one of lnx DX now I can use this formula basically I wouldn't have a c but now I can use that so whenever you see the integral of Ln X if you want to memorize this that's fine integral of Ln x is by integration Parts X Ln x - x be okay with the integral I know I did it fast but we've done it one time before so let's look at the integral from this is our real example okay let's look at the integral from 0 to one of Ln X firstly is this an infinite interval or is an infinite discontinuity infin discontinuity very good where is the discontinuity zero one or somewhere between one one what's Ln of one Z then it's defined oops what's Ln of zero error sytax error syntax my calculator won't do it I don't know uh do you know what lnx looks like you better yes it's the inverse of e to X so if e to X looks like this l x looks like that now you all think that it's undefined at one because when you plug in one you get zero that's not something that's undefined it's actually defined the definition is L one is zero that's not a problem it's at zero what our problem is is at Zero from the right hand side that's our problem Ln is isn't even defined on the side not even defined at all okay so our problem here is oh your problem here I don't got a problem here I'm good all right but it's not at one one's good we're good at one we're not good at zero so help me out with it what do I call the zero what am I going to make it let make it c yeah I'm not making it a because I'm not letting this this uh this number go anywhere I'm not letting it go to infinity or anything I'm calling it C because I have a number between 0 and one right now does that make sense it's between Z and one now where do I want that c to go from the right I mean correct and a limit lets us do that I made a really big integral didn't I make sure you know that's a handed R quick show hand be okay with that one so far so one's good man it'd be kind of worst case scenario uh to pick the wrong thing right go oh this is completely undefined at one no it's not this is greater one it's not greater zero because it goes to negative Infinity that's a problem so we say hey 0 to one no no no call it a number c to one somewhere just before zero as we're going from the right just before zero do your integral because we have a definite integral now after that you go okay cool after you've done your definite integral where do we want the C to go now we want the C to go to Zero from the right hand side this is the idea do you get the idea are you sure you get the idea yes all right so now that I've shown you how to do this this integral no problem you got a limit as C approaches zero from the right the the integral of Ln X is X Ln x - x we do not need the plus C because it's a definite integral uh where is my bounds of integration from where to where which one do I plug in first very good so this will be a limit as C approaches zero from the right of C * Ln C minus C first I just check and see if you were paying attention I didn't even write the C's the X for some reason I don't know why one Ln 1 minus one very I even asked you that question didn't I silly old Leonard minus okay now we'll have C Ln cus C can you see why this is correct anyway so you all right with that so far now let's start simplifying this thing what happens with ln1 Z that's zero what happens with oh goodness ration you know what we're going to do one more thing with this so we know that's zero I'm going to rewrite this just a little bit I'm going to write this as limit as C approaches zero from the right of let's do this let's do1 minus C Ln c plus c are we okay with that so far hello yes no okay now you can't make up your own math on these things some are going to be easy little Parts some are not going to be so easy you have to do some things with it so let's look at this what's the limit of 1 easy what's the limit of C as C approaches zero from the right that's zero this one oh this is 0 * negative Infinity that's an indeterminate form of 0 time Infinity we'll have to rewrite that one you guys follow me on that one you can't do zero time you should know that Ln of C as C approaches zero from the right goes like this to negative Infinity you can't do 0 * infinity and know what that is without doing this okay you've got to actually go through are am I getting through to you a little bit okay you got to do the indeterminate forms like we've been prac that's why we did that just to get here so we're going to have this is still1 this goes to zero so I'll have 1 minus the limit as C approaches zero from the right of C Ln C and then well plus zero that's gone so really I just need to focus on this thing I'm going to take it off to the side here real quick use a different color here the limit as C approaches zero from the right if you remember anything about indeterminate forms like 0 * Infinities or zeros to the whatever we're trying to make a fraction out of it of 0 over 0 or Infinity over infinity explain to me why we're trying to make a fraction out of it to do very good so never want to move Ln to the bottom we found that out when we remember doing e to the Ln of the bunch of junk we never move the Ln we always move the other thing so this you leave this as Ln C over one over C which was the idea that you can but you're going to be doing a derivative right now now let's let's identify this uh what is um what is this as we go from this is infinity this is negative Infinity as we get really close to zero this thing's also Infinity if this is infinity over infinity we can use so this was Infinity over infinity a weird Infinity this was Infinity over infinity so with a lals rule I'm going to do lals really quick we get 1/ C over -1/ c s derivative of Ln C is 1 C derivative of 1 C Isa 1 c^ s are you guys okay with that one can you simplify that of course you can let's see it's going to be negative 1 C over 1 C sliate multiply C we get if you simplifly this you're going to get Negative C got it c c c c now what's happening are you are we really okay with oh can you see three different C's I see what you did there thanks so uh what's happening to our c as C approaches zero from the right this thing is going to negative Z well I don't care about negative Z zero so this whole thing is settled and we geta one crazy quick show hands feel okay with with that one are you sure so the idea is man you you can do all these integrals they're pretty easy uh be careful with how you set it up identify where the infinite discount nity is that's important if you can do that you get your correct limits okay after that math you can't cheat at your limits don't make up your own math do them the way I showed you with indeterminate forms okay uh really make sure that you know how to how to do those limits maybe refresh your memory on the section of indeterminate forms so let's do uh let's do two more we'll talk about one we'll actually do another see if you're really really with it here we're not going to finish this but I want to see if you understand what to do integral from1 to 1 of 1x^2 DX integral of 1 to1 of 1x^2 DX uh infinite interval infinite discontinuity neither both neither infinite interval infinite discontinuity is there a problem in our function yes where zero is a problem you see what what I'm talking about the zero zero is a problem this is what I'm talking about you can't just blindly integrate with definite integrals in calculus one you can because you don't cover this idea IDE you go okay I'm just going to assume I can do it but here you can't do that because now well can you plug in zero but goodness gracious folks we're going over an interval for where this function is not even defined what that means is that we have this third situation where we've got a -1 to 0 plus a 0 to 1 CH pant you feel okay with with that idea I'm not going to finish the problem for you I think you can do check it out ignore this side could you do this yes yeah this would be as C approaches zero from the left as C approaches zero from the right and then you do both of them add them together if they both exist thanks for the emphasis on that one okay last one I have 84% last one integral from zero to Infinity of that bunch of junk oh my gosh one of the most important things you can do is actually identify what is going on what is wrong with your interval or your integral or both of them so what's wrong with this picture infinity infinity right off the bat smacks you right in the face so boom got an Infinity that wake you up a little bit so that's an infinite interval we're going from zero that's zero is great all right well zero is great as far as the interval being defined to Infinity that's going to be a problem we're going to have an issue there does that make sense now there's another problem what's the other problem we're def find everywhere except at zero 1 2 3 4 boom boom all the Infinity is perfect except with a zero so here's how we and you can take these ideas and run with it okay I'm giving you kind of the basics here so if you had started like you couldn't start anywhere negative but like on example like this one if you had other scenarios you got to break them up to where your interval is defined and do limits for where you are not defined is that clear so in our case we're going to say not a problem no we can do this not a problem but here's how to split it up if you have an infinite interval we're going to have to have somewhere where we can start at a finite number and end at an Infinity for here for our our infinite discontinuity we're going to have to have somewhere where we start at zero for sure but we end at a finite number so what we're going to do here is we're going to split up our integral with some intermediate value that's relatively easy to plug in so you guys with me on the idea we know that we can separate integrals as long as they're well continuous or as long as we separate them make a discontinuity with our limits now so basically we can separate them anywhere we want to we typically will separate at the discontinuity if one exists I'm sorry if one exists between our our ending bounds of our interval however if we have no infinite discontinues between the bounds of our interval just pick a random number or pick one that's easy to plug in what number is easy to plug in between zero and infinity one one does that make sense add them together now tell me something now that you've broken it up can you do this integral this is proper it's definite interval here it's just undefined at that point so we could write a limit as uh what number is going to change real quick what number is going to change to uh one C third times a chart okay now where's that c going to go from the from the right from the right listen if it's a left hand bound you're going to have to get there from the right if it's a right hand bound you're going to have to get there from the left because there is no other side of that show fans feel okay with with that one do you guys see how this idea is the stuff that we literally just covered it is this all you got to do is break this thing up so now next one we're going to have a limit okay let's see if you remember this from last time the one's not going to change what's the infinity going to be [Music] b as B approaches e x over of X now be smart about this don't do the same integral twice two different times just do it once and use it for both cases okay so for us uh I'm going to really rip through this thing very quickly uh because I have about a couple minutes left and I'm I'm going to assume that you know how to do this you I guess you could have made that -2 let's do that so U is X get -2 x to2 move over the -2 of X du that is our DX bam square of x's are gone pull out a -2 our integral becomes -2 e to U so far so good now of course before I start evaluating this thing I'm going to want to make sure I have my function in terms of X and not U should be D not DX oh yeah thanks yeah thank you for that so -2 e to U -2 e to thex your P be located with that that one basic basic UB okay if you need to watch that again Watch it again now we've got both of these pieces so what we end up doing is once we've done this integral now take that idea and put it here so we have from uh from this side we've got a limit as C approaches zero from the right this integral is this -2 e x it's just that we're evaluating from C to one okay I really want to make sure that you guys get this idea are you okay that this integral is this piece and I'm just going from C to one yes or no yeah are you sure is everybody sure on that one it's a big deal so we integrate once and we use it twice because we're just splitting up the same integral so F feel okay with it okay fantastic now we're going to add on to it a limit as B goes to Infinity of well this integral is this piece only on this particular integral I'm going from 1 to B so we evaluate limit as C goes to Z from the right which one do we plug in first one -2 e to theun 1 - -2 e to theun see this minus this so I'm I'm doing the plus here do you guys see where the plus is coming from I don't want to BL out of the water but I got to go quickly here on this one so this minus a negative that gives us a plus you okay with that S I would have messed up little little piece negative I'll do it over here that bother you 2 e to < TK b - -2 e to < TK 1 here's what I'm talking about this is what we did here so when I subtract a negative what does this become ladies and gentlem okay now we're almost done tell me what is what's the of one 1 so this is -2 over e does that make sense to you -2 over e now what's happening to the C it's going to zero it's going to zero what's the square root of zero so look at this is getting really close to zero you with meun of 0 is0 is e to the 0 is so this goes to well e 1 e 1 is over e so -2 over e this is zero e to the 0 is one this is this gives us two guys with me on that one plus let's go to Infinity what's my problem this is right right yeah B comes first plug in the B then plug in the one B comes first uh what's square root of infinity infity still Infinity this is negative in oh think careful do you guys see what I'm talking about negative Infinity e to the negative Infinity is thanks for the The Sting y'all zer okay now how about this one uh sare one one negative what's e to the negative 1 what happens what's our what's our area so this was this this is two this is zero because e to the ne Infinity is z 0 * anything is zero this gives us this again Bam Bam we get oh my goodness this is crazy okay crazy so we we do all this work from Z where it's not even defined to Infinity where we're going forever of this function and because of our limit idea we figure out this area under that curve is Bam our last little bit of section 7.6 of this section is dealing with something that's a little bit of a kind of a nice application that we can use improper integrals use to Advantage with improper integrals here's the idea let's suppose that we know something about two functions specifically we know that if we compare two functions one is always greater than or equal to another graph on this interval or one is always less than or equal to another graph or function on this interval if we know that then sometimes we can say something about the convergence or Divergence of these integrals without even doing them did you hear that without even doing them and the reason why that might be nice is is because there's some integrals that we can't do there are some nonintegrable functions so if we compare them to some that are possible we can find out some nice things about it does that make sense to you let me give you a picture of what I'm talking about okay so let's say that I've got some function f ofx and I don't care what this function looks like it's just some function and it goes forever and I've got some function G of X and it goes now here's the thought process I'm going to try it's a very simple and what I have to write here is very simple um it's going to be like one little statement one little statement but I want to make sure that you guys understand the concept that's what this class about right understanding the concepts here so think about this for a second firstly do you see that f ofx is always greater than G ofx for any number that I give you from a and let's assume this goes like this forever okay from a to Forever are you guys with me always greater than or possibly equal to it let's look at it uh the black line is f ofx the purple line is G of X is the black line above the purple line is a black line above purple line yes how about right here oh it's equal right there does that matter no it can be equal to it that's fine as long as it doesn't jump above it that would be a problem but there always equal to it or less than at so would you agree that f ofx is always greater than or equal to G of X for any number from a to Infinity so fan if you understand the concept okay now here's a plan think about what would happen for these scenarios let's say that we take an integral of G ofx how you start with FX let's say we take an integral of FX from a to Infinity remember the improper integrals that we've been doing watch the video If you from the beginning if you don't remember that if we go on this indefinite interval from a to positive Infinity we go okay let's take the integral of f ofx let's say that we can find it so the integral of f ofx from a to Infinity is actually convergent remember convergent remember the word convergent means the area exists the limit exists so the area exists so we do this and we say oh you know what I can find here's the key thing so Focus right now I can find the area of f ofx from a to Infinity you get that therefore the integral of f ofx from a to Infinity would be called convergent now here's a thought if the integral of the black line is convergent what would that tell you about the integral of the purple line of G of X it has to be convergent if the big one converges this one's bounded it's bounded by this that's what this says it says that f ofx is the upper Bound for G of X it's bounded so if this converges this one has to it's like it's squeezing it almost like it's squeezing it down there does that make sense to you so it's it's bound by the x axxis and by F ofx well this is always above this if this one converges a smaller function would also have to it's like it's getting squeezed down even more show as you understand that one so that's what we're going to say here if this integral is convergent then so is this one if integral from a to Infinity of f ofx DF DX is convergent then the integral from a to Infinity of GX G ofx DX is also convergent I really do want to make sure that makes sense to you do you understand that concept so if the F ofx converges that area exists then G of X has to it's smaller than that question so G of X just can't go down forever or is do it have to stay above some point yeah we're going to assume that it's bounded by uh this but even if it if even if it isn't even it's not x-axis if this went down below it as long as f ofx is on top of it you can still do it does that that makes sense there the only condition we have to have this has to be above that what can't happen it can touch that's fine this the area the difference in area is between there for the interval where they touch is zero it just if it goes above that that's a problem does that make sense to you okay so now next one think about the backwards way let's say we know something about G of X let's say that we calculate the area of G of X and it's Divergent so that means that we go okay the areaa under here oh it keeps growing without bound or we can't find it what would that tell you about the area of f ofx or under F ofx we don't know we don't is it possible for the integral of G ofx to diverge and the integral of f ofx to converge no no if this one grows without bound uh remember we're saying this is bigger than that does that make sense if this one grows without bound and F ofx is always is bigger than that how can f ofx converge if G of X grows without bound it's not convergent so if G ofx is Divergent and we know f ofx is bigger than that so G of X goes mm can't do it f ofx is larger than that can f ofx also converge no so if G ofx diverges and F ofx is bigger than G ofx then integral of f ofx also diverges I used F ofx instead of integral of f ofx but hopefully you get what I'm saying if the area of G ofx diverges the area under f ofx is bigger than that it must also diverge so if the integral from a to Infinity of G of X DX is Divergent then the integral from a to Infinity of f ofx DX is also Divergent quick show hands if you okay with with that basically says hey if the big one goes somewhere to a specific number G ofx also has to do that it's kind of squeeze between them it's bounded if G of X goes to Infinity then F ofx also goes to Infinity that's the idea here sure you okay with it okay would you like to see an example you get one example here uh because they it's it's honestly just this idea but here's why we we would use this so let's say that I give you an integral from 1 to Infinity 1 /x + sin^2 x DX and again here's why you might want to do this integral from 1 to Infinity of 1x + sin^2 x DX um is there anything that you know that will take care of that integral for you uh substitution it's not a sub for sure because there's no cosine anywhere to be found up there it's not a sub it's not a bip Parts because you have a plus there it's not separable by product it is not partial fractions because those are not pols it's not a trick sub you can't complete the square with that and there's an X not an x squ no no there's nothing you know how to do for this integral some things you can't directly do some integrals you can't do so and I know that's a shock to you wait a minute this is math math sorry uh right here there's nothing that we know how to do for this integral maybe somewhere else numerically or approximations for sure okay there's other things that we can do but here no uh we can't do anything with this in our class right now so what in the world are we going to do well the most we could say about this is whether it is convergent or whether is Divergent that's something we could probably do here so we're going to make a comparison to another function that we do know how to do something with now we might not be able to find the actual area of this but we can tell whether it actually goes to a number or whether it doesn't does that make sense to you whether we should spend time trying to find it or whether it's a waste of time so here's what we're going to do we're going to compare this to another function something that is going to be easy to do now keep in mind that if we're going to do this we do have to keep the bounds of integration the same from here to here just so we we match this up so notice that our a and our a they match up perhaps we think of this we're going to try to compare it to that now here's how we're going to compare it now where did I come up with that well I'll show you in a second uh how I came up with that U but a lot of times when you're just learning this stuff it's going to be given to you it's going to say hey compare this function to this one just to get you guys in the habit of of seeing uh how to make that comparison does that make sense so it'll it'll typically be given to you all I would say like on one of my tests and say hey you know what um take this thing compare it to this so what am I doing here well I'm getting rid of at least the sign I'm trying to trying to limit that um also I'll show you why the sign gets limited to this and the X I don't have to change that at all I'll show you that in just a second okay but the idea is let's try try to find something that is going to contain all the time so be bigger than this function or all the time be smaller than this function uh let me show you so the first thing we got to do show one function is bigger than the other [Applause] we start with something very basic just the idea of what in the world these things look like are you guys are you guys with me so far on the idea at least hello yes no the idea of course is we've got to show this first that way we can make some determination as whether this is happening or whether this is happening does that make sense to you so we start with something very basic like like this um I know and my I know just from my trigonometry that sinx is is always less than or equal to one I know that don't you know that okay now here's what's interesting U if I Square both sides of this inequality should not change that inequality so what this implies is that if sin x is always less than or equal to 1 then sin squar x is also less than or equal to one when I square a number less than one or equal to one it gives you a number less than or equal to one does that make sense to you cool okay you sure you're okay with that one right cool well uh let's just do something really easy you know we can do anything we want to as long as we do both sides of an equation and same thing with inequalities unless you divide or multiply by a negative then it flips inequality around so what I can do basically say hey if this is true then what I can do is let's just add X to both sides so if I add X here and I add X here I'm going to get x + sin 2 x is less than or equal to 1 + X quick show hands if you can follow the algebra on that one now do you see you see the way I came up with this function now I thought I thought about like this okay um you know what sign sign's always less than or equal to one s squ is always less than or equal to one so I'm trying to build something from this one so I'm saying hey how can I make this function from here well uh hey that's that's bounded by one therefore this thing is bounded by one plus X that's the idea now of course what we're going to want to do is we got to reciprocate these two things put them over one if you do if you reciprocate your inequality if I do 1 over x + sin^2 x and I do 1 over 1 + x tell me what happens to an inequality when I reciprocate both sides do you know changing 2 is less than three I reciprocate 12 is greater than 1/3 the inequality the inequality puts [Applause] around we reverse that inequality does that give us the relationship that we want to see it tells us that yeah cool okay we know that something's happening here that one function is always greater than or possibly equal to another show fans feel okay with this one that's the idea you need it to be all the time so we always know that yeah these could be equal at some points but other points this is going to be greater than this function does that make sense to you you sure any questions or comments before we we go any further I need to make sure this you guys get with this okay so the first thing we show is that I don't know right off the bat whether one's bigger or smaller I do because I'm doing the problem okay but you we not going to know that right off the bat we got to look at and go okay how am I going to get this thing from this thing well figure it out figure out some relationships start small grow your function grow it to what this side is then compare it to what this side is well this one turned out to be always bigger than that so what are we going to try to show let's think about it if this thing please listen carefully and think carefully on this if this thing converges if this well actually this if this thing converges does it tell you anything about this thing yeah if this converges does it tell you about this here's a scenario it says Hey suppose that of X converges G of X is the small one if this converges could F ofx diverge possibly absolutely you you don't know so if this small function converges we've said nothing about the big function does that make sense Contra positively if the large function diverges does it tell you anything about the small function no no it doesn't this could easily diverge and the small function could converge what happens is if the big function converges the small one has to if the small function diverges the big one has to maybe watch that about 14 times till it actually sinks in there okay because that's that has to happen so in our case if this one converges nothing nothing got it if this one diverged nothing now of course we can't do this integral so we're going to do this one and then make comparison if this one diverges now think here if this diverges what's it tell you about this one that's where we can make the jump so we can only show one thing here if if at all if this converges it's smaller than this one doesn't say anything it says yeah your small function converges the bigger one is obviously bigger than that I don't know where the converges or diverges but if this diverges if the small one diverges the big one also has to does that make sense at least a little bit M okay so let's go ahead let's see if we can do this integral from 1 to Infinity of 1 1 + x DX now that's something we know how to do what would I do first you got to change your Li I would say this is an infinite interval as B approaches Infinity they can change it to 1 to b 1 over 1 + x DX yeah you can do the UB we're going to keep this pretty simple because we're pasted all this stuff um what this changes to is a limit as B approaches Infinity of Ln absolute value 1. X from one to B yeah of course you can do a UB on that one you say u = 1 1 plus X it be 1/ U absolute Val U plug in the 1 plus X before we start our evaluation show should be okay with that one let's evaluate so this would be limit as B approaches Infinity Ln 1 + B absolute value absolute value here really is not that relevant can you explain to me why the absolute value here is not relevant for us say what now be negative never our interval is 1 to Infinity we can't get a negative we're adding one plus a number that's positive okay so really here kind of irrelevant doesn't really matter minus Ln 2 quick head on if you're okay with where those numbers are coming from just some evaluation now let's think about it where's B going Infinity okay what's infinity plus oneing you're a little kid I hate you to Infinity I hate you to infinity plus one it's still Infinity stupid kids anyway so infinity plus one still Infinity what's uh what's Ln of I always argue with stupid I just I really don't I ignore the little kids just like they should be ignored that was a joke by the way totally love kids um that's right back to math Ln of infinity is so this goes to Infinity what's infinity minus something really small so this whole limit goes to so you tell me something tell me something now now that we've done this this integral is this integral convergent or Divergent diver now we think back to why we did this whole thing right here if this function is always bigger than this function let me go through this one more time if this had been convergent convergent just not but if it had been conver doesn't say a thing about that don't know inconclusive okay but if this is diverent this says that this area goes to infinity or we can't find it or something like that this area is infinity area of this what's tell you about that one it has to be so should we even bother doing this integral heck no I know that if I did it if I was able to somehow magically do this right now our limit would also go to Infinity since this function is always bigger than this function the area under this curve is always bigger than the area under that curve so this is diverent this also has to be Divergent so that's infinity plus bigger infinity plus I don't know it doesn't matter it's still going to be Divergent what I want to know if is that stuff made sense to to you guys are you sure are there any questions about that at all question so this is just telling you if it's convergent or diver that's right it's not telling you what the area is so had we gone the other way right had this one been this pretend we were able to do this okay had this been convergent it would tell you that this would be convergent but it wouldn't tell you what that is yeah okay it say it's probably smaller than this area so if this comes out to a number you go cool it's probably smaller than that but uh I don't know what it is make sense question uh will you specifically ask us for that on an exam if you will use the convergent test to see if these are divergent or convergent sure I would say U use the comparison test or compare this to another function I probably would even give you the other function okay okay