Transcript for:
AP Calculus BC Review 1

welcome everybody to ap daily live review for 2022. uh we are so excited to be back this year doing some ap calculus bc review with all of you we're gonna have eight amazing days of review to make sure that every single one of you is fully ready to get a five on the ap exam uh just here in a few weeks i am joined by a long time friend of mine and a good uh great calculus teacher uh tony record tony how are you doing i am doing great and brian i tell you you and i we just haven't really been able to see each other much lately i know we've been busy working hard in our classrooms just like you guys have been busy studying your bc calculus and now it's all coming to fruition and we're going to get you guys ready to take that ap exam well we are super excited like tony said uh we'll officially introduce ourselves in case you're not familiar with us but uh my name is brian passwater i am a teacher at the world famous speedway high school in speedway indiana shout out to all my students who are i'm sure watching this and cheering me on uh and my friend tony is a teacher not too far away from me i'm not we're just uh oh just a few miles down the road i guess i'm at avon high school western suburb of indianapolis my name is tony record and again shout out to my students hopefully you guys are watching too along with the rest of the world to get better at your bc calculus so what are we going to be doing today well our topic today if you paid pretty close attention is all about the integration techniques that you would see on the calculus bc exam now of course that will encompass some of the ones that you learned in the beginning parts or the a b portion of the course and a few of the extension types of integration topics brian and i have put together a variety of material for you that is all brand new never been seen before in fact it is debuting right now on this particular broadcast and you can access all of those materials at this tiny url that you see on the screen and or by using the qr code as well and like i said we'll always have those documents out there eventually there'll be some solution keys available along down the road but we certainly want to make sure that you stay abreast with the material so that you can follow along with each of our videos now we talk about integration techniques the anti-derivatives that one would see on the ap calculus exam and we've got a fairly robust listing here that includes the main ones that you're going to be responsible for i know the list looks a little intimidating but more likely than not a few of these you've probably got pretty well under your belt any of these that look just a little bit unusual to you you might want to spend a little bit more time look at them whether you use electronic flash cards or physical class flash cards it will certainly make a difference in your confidence level when you're integrating of course if you want to follow along in ap classroom with a lot of the wonderful documents that you see there you can easily pull up this particular part of the video and you can see exactly which topic number corresponds to the various techniques that we're going to talk about today and of course the bold items are going to be strictly dc only type of integration techniques so with that brian i think we ought to start doing some practice what do you think tony i think that's a great idea and hopefully students you guys were able to download the handout like tony said there's so many great problems on there that tony and i worked hard putting together for everybody so we're gonna do a couple of short response practice here in the beginning and essentially we're going to highlight are the three unique uh bc type integration questions that would not be on an a b only exam and so we're gonna look at three of those things and we're not gonna do the entire problem right now but we're going to set them up and just talk about the skills that are needed sometimes students you know miss a question but there's so many different steps that you take on some of these problems right essentially integration where you have to do many many steps and you can do nine steps correct and make one mistake on the last step or the third step or any step and the entire problem is going to look wrong and so there's a lot of underlying skills that tony and i have identified that will really not make or break but there are these things that are going to prevent students from getting questions right if they don't have these skills so we want to make sure we target those up front and then we'll jump into some multiple choice and a free response problem here at the end so one of the bc techniques is integration by parts and we wouldn't see this part on the right hand side uh with the partial fractions but you know the question is going to ask us something like we see here on the left side of your screen uh the integral of 3x plus 1 over x squared minus x minus 6. and so as you're going through integration problems it's really easy to to follow along if we tell you what technique it is that you're doing right uh i think for students the and even for teachers like tony and i um the hardest part is just recognizing what type it is if we told you up front it was partial fractions you would have sort of a a lead in right a first step if we told you it was going to be integration by parts uh you kind of know where to start as long as you kind of knew the the rules and the formulas but uh when they're all on the same paper a lot of the problems look the same and you really have to start training yourself if you haven't done it already to think about what are you going to recognize what are you going to spot uh what are you going to look at in each question to help you decide first uh what kind of problem is it and then that will help hopefully decide what our best method will be so if i was looking at a problem like number one my two thoughts would be it could be a u substitution problem right if i see something on the bottom with x squared in it and i see something on top that has an x maybe a udu type situation might work and you could try it here but it's not going to work really but another big giveaway is if you see an integration problem and the denominator can be factored that's usually a very strong giveaway that it's going to be a partial fraction problem right so i'm always looking at these denominators thinking kind of factor this and in this case we can so if we factor this it's going to be x minus 3 i hope and then x plus 2 in our denominator and so we get this here on the fraction and we know it's partial fractions and so these problems integration is actually really easy but the first and most important step is going to be we're going to have to split this up into two separate fractions and then we can integrate them separately and so what we're going to do here is just that scale of how do we determine uh with partial fractions what these numerators are going to be a and b and there's a couple different techniques you can do you can do it algebraically and maybe in a problem later on tony or i might show you that method as well but there's also a method called the heavy side method not because it's heavy but because they're not for a mathematician named heavy side and partial fractions tells us we can rewrite this fraction as two fractions with uh those linear factors in each denominator and then a constant a and a constant b in the numerators and if you can figure out what those constants a and b are the problems become much much simpler and so we're trying to figure out those things before we can even do any integration so to do that i'm going to show us the heavy side method so we're going to cover up one of those two denominators right so if i cover up the x minus 3 denominator of course i won't cover it up here on the left-hand side because it's not going to overcover my ink probably but if i were to cover this up right i'm going to factor this here and put this x minus 3 over this a how about that and bring this back right here um if i cover this up and i'm going to plug whatever value into this fraction here on the left that would have been that solution so if i cover up x minus 3 i want to plug in 3 for x so we have this fraction 3x plus 1. i'll rewrite it over here right before this video tony and i were talking about how this method works really well when you have a board and students in front of you and you can use your hand to cover things up and actually talk about it and digitally i thought i had a solution with this little box to cover things up and i realized just now maybe it's not going to work as great as i thought but we covered up the x minus 3. so imagine this factor wasn't here and then we're going to plug in x equals 3 into the remaining part of that fraction and then that will be our coefficient for a so if i cover up x minus 3 under the a i'm going to plug 3 in in the numerator we're going to get what 3 times 3 plus 1 is 10 and then 3 plus 2 is 5 and that means that our value of a is 2. and then similarly we can do the same thing if we cover up the other fraction the x plus 2 then imagine that uh here we have the same fraction 3x plus 1 over x minus 3 times x plus 2 and if we had covered up the x plus 2 uh term in the denominator and this time we're going to plug in negative 2 whatever that solution would have been into the remaining fraction and 3 times negative 2 plus 1 is negative 5 and negative 2 minus 3 is negative 5 and we get 1. so we're going to get in this case b is 1. and so once you have those things uh you can rewrite the problem just find a and b which we did but i just want to show you what it would look like we would have uh the integral of 2 over x plus 3 and then plus 1 over x plus 2 dx and once we can break it up like this the integration becomes very simple and usually uh you're gonna get a lot of points from there now another type of problem is gonna be these improper integrals uh there's a couple different types of improper integrals and i think we always imagine improper as being the ones that have infinity somewhere in the limits which is very common and again we're not going to do this whole problem but we are going to set it up and so by setup imagine you're doing a free response problem the ap exam one of the four mathematical practices is communication and notation and so they want to see if students can use the right symbols and notation and proper mathematics right along the way and sometimes you know my own students we've kind of trained them through school before this year the answers are the most important thing of course answers are important uh but we also value your communication and you're using uh the symbols and notation properly and so a lot of students now as we grade exams they they leave easy points on the table you know i can read their papers and know they know what they're doing but because they just didn't quite show the right communication or symbols they weren't earning all those points and a lot of times those just those few little points that you leave on the table on the exam might be the difference between getting a 3 and a 4 or a 4 and a 5 on the exam so we want to make sure that you guys are writing things properly as well so we can't really take an integral with an infinity side in the limit and the reason why is because our fundamental theorem requires us to have real numbers in our limits and so infinity is not a a real number and so it's a really simple fix but we have to write it as a limit and so we're going to replace that limit with a just a dummy variable uh you know it's on the top of the integral i use the letter b often it was on the bottom i'll use a so i'm going to write the integral from 1 to b in this case and so then we're going to take that b value and we can approach infinity because infinity is you know an idea right a destination we don't quite reach and we can rewrite it like this and then all we would do from here is we would take the antiderivative we would apply the fundamental theorem with our limits 1 and b and then once we have that then we can evaluate the limit in order to let b approach infinity to solve from there so this little notation of writing it as a limit that's one point of these frq problems if you see one on the exam where if a student doesn't use the proper limit notation uh they're going to be at least one point short of what they could have gotten otherwise so later on again we will see both of these problems or similar problems to these uh worked all the way through but those two entry points are are just little skills that we want to make sure you have uh to make sure that you're getting all the points that you can um the third bc topic uh tony is going to walk us through and is going to be talking about integration by part so tony you want to show us the idea of here absolutely well done brian on our integration by parts one of the things that uh it always strikes me about this particular uh integration technique is that it's very formula driven it's one of the few advanced integration techniques that's very driven by much uh so by a formula and that formula you see right here the integration of a function u with respect to v is equal to u times v minus the integration of v d u and i know that looks oh very complicated it looks like kind of fell out of the sky but really all it is is an application of the product rule it's sort of the integration version of taking the product rule in reverse now what you would have to know on the ap exam is how to break apart the various components of that integrand that's provided into its you and its dv counterpart and then move forward now how would you decide which part was the u and which part was the dv and it's likely that your teacher at some point gave you some kind of a mnemonic some of you might use the mnemonic liais or maybe you use lipit l-i-p-e-t we're not asking you to change from whatever you're used to using but just find one that you can really follow really well and if we use liaid in this case that means that our u selection this drives our selection for you is going to first of all be the logarithm part if you have one and the reason is because it's very difficult to let dv be the logarithm part and integrate that that requires integration by parts all in itself if you don't have a logarithm maybe there's an inverse trig there that would be your selection of you keep going down the line something algebraic lipid calls that powers i believe and then you get to trichonomic trigonometric uh exponential and there's exponential twice we like it so much that it's on here twice but you really only have one exponential and by the time you get to the end to be honest the t and the e are quite interchangeable that's why lipid works very well as uh in addition and you can see that the e is placed before the t all right but our suggestion brian i think you should really lock into one of those mnemonics and just make sure that you're very well versed with that so our last little short answer problem is just a setup of a very standard integration by parts problem that has a difficulty level that's very appropriate for the ap exam and one other little trick that i share with my students is that when we do select our you and our dv i have my students write them a little bit sort of caddy corner now there's a reason for that that i'll discuss here in just a moment now the selection for you will follow the guidelines of the eight so we look through this and think oh do we have a logarithmic expression well of course we don't and we go to i inverse trig we find out we don't have one of those either we get to a and we see algebraic and we could certainly allow that 2x to be that algebraic piece and as we continue sine of x dx is the only thing that's remaining that's going to serve as our dv now we're not going to finish this problem you're going to see an example of an integration by parts a little bit later in the broadcast but the reason why i like to write these a little caddy corner is so that my students tend to think if they work downward they're taking a derivative and if they work upward they'll be taking an integral and so that's how they'll arrive at that du and that v that are all important in terms of using that integration by parts formula so we'll stick around and you're going to see another one of those here in just a little bit but i think i'm going to kick off our very first multiple choice question you're going to notice on the documents there's going to be a level brian and i a lot of experience with with working with students grading ap exams writing questions in brian's case for advanced placement tests and we have a pretty good handle on what we think the difficulty level for many of these problems will be they are a little subjective but we classify this one as an ap level three thinking that if a student's moving through problems like this pretty comfortably that they should be scoring at least a three on the exam mileage of course may always vary so in this particular definite our indefinite integral we have one over x squared plus 4x minus 21. now if you're paying really close attention earlier brian mentioned something that is very important about these kinds of rational functions the first thing that you want to do is take a look at that denominator and see if it will factor because that will determine what kind of path you're going to take whether it's a partial fraction decomposition or maybe it could be a complete the square problem well in this case when you take a look at that denominator you're going to come to the realization very quickly hopefully that it does factor thank your algebra teacher for that right and so it's perfectly acceptable if you want to do that work off to the side notice how i'm doing it independent of the integral i don't really care if i have the integral wrapped around it even if this was a free response question that would still be okay as long as we introduce that integral back in later because that's going to be really important try not to link things together with equal signs when you can avoid it so we're taking a look at this particular factoring right x squared plus 4x minus 21 factors into x plus 7 times x minus 3. now we're going to set this up into our partial fraction decomposition we notice that there are two distinct linear factors here which is going to produce two separate fractions the first denominator i'll write as x plus seven it will not make a difference which one goes where and because each of those are linear then we're going to have constants in the numerator and while you possibly may have been exposed to other kinds of partial fraction decomposition problems on the ap exam you will only see linear factors in the denominator and none of them will ever repeat so you're going to be able to use constants like a and b and well if you had a third fraction you may go as far as c but that's pretty rare and so at this point here what i like to do is say okay let's clear away these fractions because you know i don't i don't like fractions sometimes right they kind of be a little bit of a pain so if we multiply through by the common denominator what's going to happen is that the left side is gonna be completely free of the denominator just containing a one and then on the right side we're gonna be left with the a along with the x minus three and then the b is going to be residing with the x plus 7. now if you think about that again you know we could write the x plus 7 x minus 3 if we wanted to and apply it to each of these two terms but hopefully you can see that that x plus 7 pair will cancel leaving you with the x plus three and then say hey you guys pop out over here now and you can see the same situation happening with the x minus threes that's what brian was alluding to when he talked about this this potential for this cover-up method to really come into play here it's possible that you may have been taught how to do this a slightly different way where you would distribute the a and distribute the b in if that's your cup of t certainly go ahead and try that you're going to get the same constants as a and b if you've never tried the heavyside method we're going to give you another good look at it if you're not quite sure what the coverup method looks like another way to think about it is think of finding opportunistic x's what do i mean by opportunistic x's well x's that'll make the sentence a little bit easier like what if i let x equal 3 we certainly see how that's going to wipe out the a term and it would leave you with 1 is equal to b times the quantity 3 plus 7 or of course 10. and very quickly you're going to be able to find your b value without much trouble all right it's kind of the same as covering up this a quantity x minus 3 and inserting 3 for the x we'll do the same thing over here pick another opportunistic value for x i don't know about you but negative 7 is looking awfully good to me and we can see how that's going to wipe out this whole entire b term leaving us with 1 is equivalent to negative 10 times a and that is going to force my a to be negative 1 10. now that's just the algebra part we're going to put this all back together and we're going to finally be able to do a little calculus together right so we're going to take our original integral and look at it from a standpoint of this particular version of that fraction let's say and just because the a is in front you don't have to leave that there and this is a completely decision you know up that that's completely up to you but i'm a little bit more tempted to write the positive version if i have a negative here i want to write that one first and then i'll swap the order and just make that sign in the middle a minus now and then we can make sure that the one tenth that's negative is over the x plus seven and of course we're integrating this with respect to x and this is where things really kind of come to fruition because you use a a very simple integration technique formula if you will that dictates how you would integrate uh a one over u form if you will so i noticed that i brought this one tenth out in front i could actually bring it out in front of both terms to make things a little bit easier and then if i integrate 1 over x minus 3 i'm just going to get the natural log of the absolute value of x minus 3. now you can say that is there a little bit of a u substitution going on here yeah the denominator is a little bit more than just an x but because the derivative of that denominator is going to produce a one it's not going to cause for any need to offset with some constant out in front and we do the same thing here for the other piece after we drop that minus sign we get the natural log absolute value of x plus seven ah don't forget plus c so important with an indefinite integral and then at this point you start looking a little bit over here at the choices uh your heart sinks just a little bit right when you notice man i've done everything correctly i don't see any match here well sometimes on multiple choice problems you're sort of kept in check on being able to do a little bit of simplifying and so it's really important that you kind of dust off some of those skills from your logarithm days in your algebra class and remember that if you have the subtraction symbol between two separate logarithms that will allow you to consolidate these logarithms into a single quotient and so you would have an x minus 3 on top of the x plus 7 and we still would need absolute values around everything because we want to make sure that we uphold that particular domain restriction for this natural log and then our plus c pops right in and boy now you've worked even extra hard we sure hope that we matched and sure enough choice a is what we're going to go with all right uh brian i think we're going to turn it over to you or you're going to take a look at our next multiple choice problems thanks tony uh we're looking at another one here and we classified it as ap level three but as i was re-looking at it um i'm going to upgrade it to like a three plus because there is some nuance here that i think if you can do a problem like this you're probably more in that four range at least and a lot of the problems when tony and i label these as ap level three fours and fives we're doing that for you just to have an idea of how you're how you're pacing you kind of know what your goals are for the exam hopefully it's a five but uh you know as we've getting through the year we've had the last couple years of some rough you know schooling right um since code first began and we've had some missed classes and virtual classes and maybe we're not as strong on the algebra skills as we as students typically would be um in our calculus classes this year but as we put those in there to kind of give an idea of a benchmark of where you are and a lot of times students will misinterpret these these levels and they'll think i understand that problem so i must be at that level and what really is tricky on the ap exam is not so much the problems that students don't know how to do what really hurts scores on the exam is the problems that you do know how to do but yet you make a small mistake and the ap exam those wrong answers are not just there as random answers it's they're targeting those mistakes so all those mistakes that are calculus mistakes or or common mistakes those would be options on the exam and so you might do a problem halfway or kind of you know quickly and and see one of your answers there kind of the opposite of tony's dilemma on the last problem where you didn't have your answer where you might see an answer there that matches your answer and so you assume you're correct and you choose it and so uh the best advice tony i'm gonna give you is um it's important to understand it but it's even more important to be able to do it right you have to prove that you can do it from beginning to end correctly and it's one test to give you the credits that you deserve from an entire year of hard work and so uh don't take shortcuts on the exam don't make assumptions about the answers or move too quickly because as soon as you start doing that you'll get trapped on a few problems and so this is another good example of that here so it is known that the integral of f of x times secant squared of x dx equals f of x times tangent minus the integral of 6 x squared tangent of x dx which of the following could be f of x and so if i was reading this problem for the first time i might just think i don't know what does he's been talking about like what do they want me to do how would i know what f of x is and so you know always look at the original problem the integral of f of x secant squared and reminder like we don't have a product rule for integration you need a nice product rule and a quotient rule for derivatives we don't really have a product rule um per se we can just do it uh you know straight forward without having to you know do anything else to it but when you see a product and you sub won't work you know it's a really good indication that it's going to be integration by parts um and so i wrote the reminder here i like to write the formula down and i use u prime and v prime instead of d u and dv and your teacher might use d u and dv like tony writes and they might use u prime uh and v prime like i write so tony and i like to show different methods and different styles that way if you hear something different uh you can see it kind of both ways on on the paper so so this is really a integration by parts question but they're not asking you to just do the integration uh they're asking you um just to figure out what this f of x function could be and so we're kind of working you know inside out on this problem and and writing the formula down on your paper can help because i know if it's integration by parts one of those two things in the problem have to be u and one of them is gonna have to be v prime and they don't have to go in order so just so you know don't always assume that the first one is going to be u and the second one is going to be v prime there's no particular order of how they're written so we had to look at the right side of the formula and we know that whatever u is is still going to be there you know when we do integration by parts and whatever v prime is we'll see its anti-derivative afterwards and so we're starting with f of x secant squared and the first thing we see on the right hand side is f of x tangent of x and so f of x did not change right it stayed the same which means f of x is going to have to be the u of our integration by parts and we can confirm that uh by double checking with our secant squared meaning the secant squared turned into tangent on the right hand side and if you know your derivative rules uh that tangent's derivative is secant squared then hopefully you'll recognize uh the antiderivative of secant squared then would have to be tangent and so secant squared is going to be our v prime in this case so v prime is going to be secant squared and once we know those two things we're sort of able to continue using our formula because notice we have uv on the right hand side and then it's minus the integral of v times d u and so this right here tangent number was going to be the v in the problem and then we have here 6 x squared and 6 s squared then must be u prime because that's what our formula would dictate right and so u prime must have been six x squared and so you got a little bit lucky in the sense of that wasn't a choice here on a multiple choice that would be kind of you know one of those things oh i identify as 6x squared but it's not a choice because we're trying to figure out what f of x is and f of x is u and if u prime is 6x squared that we have here written down that means that the u value must be the antiderivative of 6x squared right we're going kind of up that ladder now notice how quickly if you just took 6x squared's derivative you would get 12x and luckily we don't have it as a choice but that would often be an option on the exam but we're taking the anti-derivative of 6x squared so we're going to add 1 to our exponent and then we're going to divide by 3. we'll put the plus c in here but you know we're not trying to get a plus c necessarily because we're looking at just a function in general so we're going to have what is that 2x cubed and that is going to be option d so notice that the idea is when i see a formula and i recognize the integration of parts it's really helpful to write the formula down on your paper so you can match things up and make sure you're not making an easy mistake that could be fixed by some simple clarification on the paper let's look at a couple more problems here now we're going to jump up to an ap level 4 question and it's a four i think because again the idea isn't that difficult but when you're in that testing situation you have to kind of run through this process of how do i solve these problems and so uh this one is a definite integral so we have now the integral from one to four of 1 over x minus 3 dx and so student's first reaction is just to take the antiderivative and apply the fundamental theorem tony just walked us through an example just like this where it's a natural log type solution and if you do that you're going to get an answer you're going to get one of the wrong choices and the trick is realizing this is an improper integral and it's i call it improper integral and disguise it really isn't in disguise it's just a real improper integral but what happens is with improper integration we see so many examples where they involve an infinity sign in one of your upper or lower bounds right that's what we kind of think is improper but there's more than one type of improper integration and on the handout hopefully you got that qr code at the beginning of the video where you could download the handout and follow along and also at the end of the video we'll put it back up as well where you can download it and print it out um but we have a nice graphic on there kind of walking through the techniques of improper integration and showing the different versions that you might see on an exam for calc bc but because it's improper um and why is it improper because notice uh the denominator uh we have a vertical asymptote when x is 3 right it's a vertical asymptote when x is 3 and you can't use the fundamental theorem through an in infinite discontinuity it does not work right and so according to improper integration rules we're gonna have to break this up into two integrals you know one kind of leading up to the vertical asymptote and one going from that asymptote until the end and so we would have to rewrite this first as the integral from one uh to b something like b um and b is gonna be like three in this case uh of one over x minus three and we're gonna have that limit like we did before uh with infinity but in this case b is gonna get closer and closer and closer to that vertical asymptote and so we're going to have b approach 3 from the left side because it's a little bit less than 3 because we're starting at 1 and then we're going to then have to write it another integral from that same value b to four in this case so we're going to split it up into two separate definite integrals and again we're gonna take the limit here as b approaches three from the right hand side kind of on each side of that vertical asymptote and so each integral by itself is not that difficult to to do but we're going to apply the limits and so in this case it'll kind of show you what a limit might look like on a free response question and what kind of notation we should use and we really can't evaluate the limit until after we've applied the fundamental theorem so the anti-derivative of 1 over x minus 3 is just the natural log of x minus 3. and we're going to apply this uh from 1 to b and so again we're not ready to plug the limit in yet i think students want to just really jump to plugging in that limit value whether it's infinity or 3 but really we have to just do the entire fundamental theorem with the b or our dummy variable first and then we're going to do the limit at the very end so we have the natural log of b minus 3 and then minus the natural log of 1 minus 3 the absolute value of this and then once we have this then we can start applying our limit now we have our value in there and we're going to plug in 3 and and watch what happens if you try to plug in 3 to this you're basically going to get the natural log of what 0 and then the natural log of you know two we're doing the absolute value here on the right hand side but the main issue is you cannot take the natural log of zero it is undefined value right it's like negative infinity the natural log graph as b approaches zero is going to be going down toward negative infinity and so it's it doesn't exist as a limit and so we don't even have to evaluate the other integral because with improper integrals if any part of the problem uh is you know doesn't exist the limit then the entire integral has no value and so this case that's a little bit trickier because it turns out because this doesn't exist it's an improper integral that has negative infinity for the limiting factor the integral must diverge in this case and so in general when you're seeing a problem you always want to double check especially on multiple choice you know there's no vertical asymptotes in the middle of this range right so otherwise if you do the problem and don't think about this concept and just ignore the vertical asymptote as if it wasn't there at all you would get a value because the you know the fundamental theorem doesn't realize there's a problem in the middle unless you recognize it so just double check is there a vertical asymptote halfway through or not and then make sure you do it and then general thumb uh that i've noticed and there might be some you know i don't want to speak you know in definitives at all times but um if i've seen a problem a definite integral and there's a vertical asymptote in the middle of the the limits you know one to four in this case there's a vertical asymptote somewhere in the middle and you're trying to pass all the way through it i've never really seen a question that would converge it just they've always tended to diverge and you know it's certainly possible to have a vertical asymptote um at the beginning of your area or your integral but if you see a problem that's kind of passing right through one uh you know on each side of it that should definitely raise some red flags it's probably going to end up diverging uh one more quick example for tony takes us through a couple more and looks at the frq is another ap level four and again i see this problem and i am running through my checklist of what to do and i see a product and i'm like what's the product what are we gonna do with this product um there's no product rule and i think it's very way to do use up with the x in front and it really isn't uh this is definitely another integration by parts and remember integration by parts rule is going to be uh the integral of u times v prime is going to be equal to u v minus the integral of v times u prime and we can identify u and v prime and essentially you want the u to go away right that's the whole point is we get an integral question and we still have an integral and our final you know piece right we have to integrate something again even after we apply integration by parts so the u hopefully should go away and in this case u is going to be x if i differentiate that it's going to become one in my life becomes easier and i like what tony did by kind of staggering these and so i'm going to put cosine of 2x as my as my v prime and then once i have this i can find u prime is going to be 1 dx and then v is going to be the antiderivative of cosine of 2x this is an a b topic right it's a integration by u substitution here but cosine becomes sine and you know the 2x in there if we do some u substitution uh we would because we're doing the anti-derivative instead of multiplying by 2 we're going to do what we're going to divide by 2 in front right so our v is going to be a one-half sine of of of 2x and so once we have all our components we have the right-hand side so we can sort of write this out it's going to be uv so x times one half sine of 2x and then it's going to be minus the integral of v it's another one half sine of 2x and then it's times u prime and our u prime was just dx and so we were able to to rewrite this and what i love about integration by parts is once you do it uh you think you're so satisfied and you're happy with yourself and then you realize it's just the gift that keeps on giving right it's like those infomercials we can order now and get a second one free because it really includes its own a second integral problem that you have to solve and so in front we have this one-half x sine of 2x and so we know for sure it's certainly not going to be a or b because we don't have that component of the one half x sine of 2x in front but we still have to do the antiderivative of this other component right we have to do the anti-derivative of one-half sine of 2x and i'm going to bring the one-half out in front that way it doesn't sort of cause any confusion for me and we're taking sine of 2x is anti-derivative and remember sine's anti-derivative is negative cosine of 2x and then because we're doing the antiderivative we're going to again multiply or divide by 2 right because we're not taking the derivative of the antiderivative and we get something like this and a little bit of simplification and we're going to get with simplification just a positive one-half because it's negative times one-half times a negative one-half and then we have times another one-half that's the positive one-fourth i believe uh cosine of 2x and notice that was going to lead us to choice d so i see as a student who did the first part and then kind of forgot about that integral sign uh in front of the second part and then i felt like they're done with the problems for what they had there so uh tony is going to uh take us home on a multiple choice problem or two possibly and then uh maybe begin us on a a neat sort of open-ended response type question yeah you know i i was kind of thinking here let's jump right to this free response uh we've got a lot of really neat things planned for you here in this particular problem now uh the rest of the packet the multiple choice that you see you're gonna find that that we've developed problems that sort of merge more than one idea together and so you want to be on the lookout for situations where you might see integration by parts coupled with an improper integral nothing to worry about just use the techniques that brian i have talked about in tandem with one another and you should be finding yourselves having pretty much success with those particular problems um i am i am probably one of the biggest fanboys of some of the free response questions that this gentleman that's working with me on this video writes this is one of brian's recent free response questions that's really rich it develops a lot of neat calculus topics from the beginning of your course and uh it certainly has a nice robust uh a variety of integration techniques so you can see that we've got a function uh defined as an integral you have 1 over x squared plus k from lower boundary 3 to x and we're given that k is a constant what i'm going to do is i'm going to work through the first two parts i believe brian will then take over on parts c and d and we'll give you a really nice free response experience so let's get started our part a says that k is going to equal 9. we're asked to find the function evaluated at the square root of 3. well one of the very first things that you're going to be able to do is lose these variable looking entities because they are not going to remain that way and so we can start this problem out very simply by replacing that upper boundary with a not with a square root of three and you know what that's really important right it's pretty easy to get these two things confused we want this to be our x of course and it's pretty clear that that's okay and so when we rewrite this guy it's going to look like this all right and we're going to integrate that with respect to x now we've talked about recognition we want to take a look at this guy and you first of all want to come to the realization that will not factor so we're not going to be delving into this integration by partial fraction d comp but what we are going to be diving into is another integration formula we haven't quite given much attention in the video yet but we are now going to talk about that inverse tangent form and so it's really important that you're very very much on top of things and you can see that an x squared plus a constant in that denominator is always going to result in that arc tan form maybe you've set it up before where you identify this as your a squared and this guy right here the x squared would be your u squared and then the principal square roots of each of those are just x and 3 for your una respectively and at this point it's just a matter of assembling that particular inverse tangent form you may have seen it on one of the earlier slides that we showed but you would simply get a one over a in front you always have to think which one of those two popular inverse trig forms has the one over a in front our tan or arc sine it's definitely going to be the arctan and then of course we have the inverse tangent you could write that with a tan and a negative one superscript instead and then you have your u over a in this case or x over three let's not forget this is a definite integral with boundaries from square root 3 down to 3. and so we can now insert those i'm going to factor out that 1 3 and toss in that square root of three here subtract for using our fundamental theorem something that you tend to use quite a bit of when you're working with integration techniques and three over three of course is one and then you just kind of deal with the trigonometry here as best you can now or it's worth mentioning could you stop right here could i put the little stop sign put on the brakes right there yes you certainly could nothing wrong with you wanting to go ahead and evaluate the tangent of 1 over the square root of 3 i believe is the 30 degree angle pi over 6 arc tan of 1 is pi over 4. if you were so motivated you could go ahead and simplify this all the way down to the negative pi over 36 but brian and i are going to be real honest with you why work any harder than you have to and it's perfectly fine if you can leave something in a form that you could easily find the answer to on a basic scientific calculator then you're going to receive full credit all right let's take a look at part b which reads somewhat similarly you've got a k value of negative 1 an x value of force but when we assemble the integral it's going to look something like this now where the denominator is one or x squared minus one now below that one now you don't wanna take the bait and look at this as being maybe one of those arc tangent arc sine inverse trigonometric forms because it just doesn't follow those forms you look at that denominator and you know whoa that guy's going to factor and so we're going to really quickly run through that factoring this decomposition and of course we we see that we would get 1 over x minus 1. i'm sorry let's let's do it like this yeah i'll go with that x minus one why not we'll go x minus one x plus one to get things started and then we can break this apart into its separate partial fraction components again it doesn't matter which one you put where i multiply through by that common denominator one is going to equal a quantity x plus one plus b quantity x minus one and then again the cover up method this heavy side method oliver heavyside developed this back in the 1800s if you let x equal negative one you're going to get rid of this term so 1 is going to equal b times negative 2 of course and if we let x equal positive 1 that's going to get rid of our b value setting the stage for us to find the a and we get b to be negative half and a to be positive half let's put that negative in there and so now this integration problem takes on a totally different look and we focus our attention on the a which is one half over x minus 1 plus well that plus probably isn't going to stick around very long because the b value is negative one half certainly nothing wrong if you want to go ahead and factor that negative out in front turn this into a subtraction problem and now this is going to integrate very similarly to the problem that you saw previously it's going to use the natural log forms again but the one big exception is that this is a definite interval and so we're not done until we evaluate at these boundaries four and three and so when we do that uh we might put a half out in front there we would just plug our 4 in we have the natural log of 3. don't really need the absolute values at this point it's nothing wrong if you put them in there but 3 is pretty confidently going to be positive all the time and then of course when we run our 3 in here when i when we run our 4 in here we get the natural log of 7 and then we subtract the quantity it's very important that you have the parentheses there because when we run our 3 in we initially you're going to get what looks like to be a couple of terms natural log of 3 minus 1 is 2 minus the natural log of 3 plus 1 which is 4. and yep it is going to be two separate terms and so we probably want to make sure that they're in wrapped in parentheses and tell you what i'm going to stop it right there we don't obviously don't want a plus c in this and we don't really see a need for us to to do any simplifying with this so i think i did that right brian does that look good to you i think we're we're good on part b yeah i think the only thing i would mention tony and this is what happens when you're live tony is trying to uh model for you uh what we do sometimes as teachers and even students uh when he evaluated that he plugged into four uh to the natural log of x plus one and he wrote down uh seven but i think the seven is not as confident as a seven it's probably more confident as a five let me think here one two three four five and i was just making sure that you were paying attention i thought i saw you dozing off there yeah you're right that is a natural log of five i got one more actually that one's great there so uh this is what happens when you're on live tv right you have to make sure that you uh at least see people are paying attention right so hopefully uh students caught that as well and then they will be able to uh let us know so uh tony mentioned um we're gonna look at just one more part of this free response problem but there is a part d so i hope that you will uh go to the link or the the qr code that we're going to give to you again at the at the end of this video and grab those files it's the same link the same url that we will use for all eight of these days and uh we put a lot of time into making these really uh long worksheets of activities that are really good ap style problems to make sure that you're ready so if you need more practice on these ideas or you feel like you just need you're starting to get it but you feel like you get more help definitely go back to that and use it and also don't fret because as we go through the other videos we're going to continue to spiral these things through all the videos and so you're going to see more integration techniques when we're doing series if we're doing parametrics if we're doing polars and we're going to keep bringing these things back so hopefully by the end you've covered all the bc things many times and also all the a b topics pretty much uh you will see throughout these videos even though that the names of the videos are sort of bc topic names so the part c here i put a little oops in there because uh somehow in all these problems i forgot the dx in the problem uh and at the end of it right so there should be a dx there otherwise the integral doesn't make any sense but uh part c says let k equals zero and then find the limit of f of x as x approaches infinity and so if i just you know rewrite this question briefly as f of x equals the integral from 3 to x of 1 over x squared dx right because k is 0 we're not going to have it and we're trying to find the limit of this and and notice um if x goes to infinity what's that going to do right this is really a kind of clever way right of writing an improper integral because we're taking the limit of this definite integral as x approaches infinity in this case and so we got to kind of a different roundabout way uh didn't just say put an infinity in the limit but it's the same idea and what we're going to do is the same thing we did before right we're going to go ahead and uh you know we're going to keep we can keep x on here now i don't have to use a dummy variable because i don't have the infinity symbol in my integration but we should probably integrate this uh 1 over x squared and i'm going to do a little bit of side work here because this is where i think we start going too quickly because we're getting pretty confident integration we make a little silly mistake with a simple integral problem it's always the easy ones that trip us up not the challenging ones but uh you know don't leave that x squared in the denominator right we need to bring it to the numerator before we apply the anti-derivative so we don't make any easy mistakes we're going to add 1 to the exponent and divide and that's going to end up being negative 1 over x as our anti-derivative so just making sure we don't go too quickly when the problems have a lot of steps and a lot of other things happening like limits and infinity sometimes we go quickly on the basic stuff and then we have no chance of getting it right so in this case we're going to have negative 1 over x and we're going from 3 to x essentially right and so we're taking the limit of this and we have as x goes to infinity and we're left with negative 1 3 or negative 1 over x right if i plug in x to x i should get negative or just negative 1 over x and then minus and we're doing negative 1 over 3 so it's going to be a plus 1 over 3. and then now when i plug in infinity for x and take this limit we're going to get i'm going to put this in quotes a little bit because i don't want any teacher sending me hate mail about some notation but we're kind of mentally thinking about 1 over infinity plus 1 over 3 when i apply my limit and a constant over infinity is just zero isn't it so the limit of one over x as x goes to infinity is going to be zero uh plus one third and we end up getting just one third for our answer so hopefully that made some some sense and you can kind of see how the improper integration uh and limits as x goes to infinity can apply into this problem and uh be sure to check out part d uh in the uh in the online handout so what are our key takeaways after our first session well number one uh integration by substitution u substitution is a big part of the exam almost all the anti-derivative problems you're going to see will have some kind of use up in them for bc partial fractions there's a lot of techniques that you might have learned but on the ap exam they're only going to be non-repeating linear terms if you're not 100 what that means look at the examples that tony and i had in this video and on the extra handout and they're all gonna be that kind of problem so they're gonna be much more basic uh partial fraction problems and then uh integration by parts is really one iteration so you don't have to loop them around and do it more than twice uh carl's board typically stays away from that as well knowing your basic formulas is going to be a huge a huge uh part of your success so make sure you know your stuff and no matter what we tell our students they always think they can cram them all in like the night before the exam but then you get that pressure situation you end up you know having a blank you know stare on your face you forget things so make sure we don't uh forget those things so reminder again where can you find these materials there's a qr code you can scan takes you right to the link there's also the the tiny url you can go there that way it's the same exact link for all eight sessions lots of extra problems on there and there'll be detailed solutions for each of them uh each one of those things uh we haven't decided yet tony but we'll have to figure out whether it'd be a couple days after the videos uh or maybe in two days from now or three days from now but the solutions will appear in there as well so uh what a great first session tony what do we have coming up tomorrow and what are our final thoughts here as we leave well i think we've got uh some differential equations on tap we're going to spend a lot of time talking about the logistic differential equations which is what kind of distinguishes the bc part from the a b but i know we're looking forward to it uh this was a great first session with you um i i just i can't thank brian enough for for working with me on this and we've had a blast working with you as well awesome thank you guys so much we'll see you next time you