 Welcome to the exam review for exam 2 for math 1220 calculus 2 for students at Southern Utah University. As usual I'll be your professor today, Dr. Andrew Missildine. So in this video I want to quickly talk about the contents of exam 2 for a class to give you the students an idea of what you should be studying for and how to better prepare for this exam of course. Now there are some specific information you definitely need to know about the exam like the time place manner which it's gonna take place but as these change from semester to semester that will not be included within this video. So things like what materials am I allowed to use during the test, when will the test take place, dates, those type of things. Please consult the exam syllabus that can probably be found on Canvas for the semester specifics. This video is just gonna focus on the structure and contents of the exam. Just so you're aware exam 2 is gonna cover materials from lecture 10 all the way up to lecture 21. So a quick summary of what's gonna be on this exam is there'll be techniques of integration. So this will include many of the new techniques that we've learned about like we've learned about trig integrations. So how we can use trig identities to help us compute anti-derivatives of trigonometric functions. But we also have trig subs that we've now learned about how the method of substitution can be modified to turn algebraic functions into trigonometric functions and use the techniques about trigonometric integrals to help you in that setting. We of course also have our partial fraction decompositions that help us with rational functions. We have things like a rationalizing substitution. But we also have some of the more classic materials techniques we've learned about previously like u-subs are still perfectly valid here. Integration by parts of course is also a valid technique that we need to be aware of. And so and then of course you can put all these together in a street fighter manner. And any of these techniques could be used oftentimes in combination with others. And so the vast majority of the test is gonna be about these techniques of integration. Now of course some other things that should be mentioned topics of numerical integration will also be covered here. Midpoint rule, trapezoid rule, Simpson's rule, improper integrals will also be discussed on this exam. And then also we have two applications of integration that will also appear on this exam as well. This includes arc length and surface area. And so let's say a little bit more about these topics in just a second. I do want to mention just like exam one there are going to be three sections on this exam. There's the multiple choice section, a short response section, and the free response section. The structure and policy related to these types of questions are identical to the previous exam. There'll be four multiple choice questions worth five points each. There'll be four short response questions that'll be worth seven points each. And then there will be four free response questions. Most of them which are worth 12 points but there is a 16 pointer that'll be coming up. And let's take a look at these topics. All right, so let's start with the multiple choice section. Four questions there, like I said, five points each. The first question that you're going to see on this exam is going to be a question about trigonometric substitution. But all that the question is going to ask you, it's not going to ask you to evaluate an anti-derivative, it's just going to ask you which trigonometric substitution should you use when considering an algebraic integral like this one. Now clearly this comes down to what is the square root doing? We're interested in basically three basic forms. So you have the square root of a squared minus u squared, where a is here a constant number, u is some function. This is like what we use for a sine substitution. We could also look at the square root of a squared plus u squared. That's useful for you would want to use a tangent substitution in that regard. And you also have the square root of u squared minus a squared. A secant substitution would be relevant there. But also keeping track of where do all the coefficients go and things like that. Question one, just ask you what is the correct trigonometric substitution you should use to start the start the anti-derivative process. So just the very initial step of a trig sub. There will be another question in the free response that asks you specifically to compute an integral using trigonometric substitution. This is just asking what's the very first step. What is the substitution in play? It's a fun question. Question two is very similar in nature that using the techniques we've developed we just ask what's the very first step. So imagine your task to find the anti-derivative of a rational function like the following. This is definitely more complicated than we usually do but this question only ask what is the correct template of the partial fraction decomposition. So we care about the factorization of the denominator. So do we have linear factors and how often do they show up? Do we have irreducible quadratic factors? How often do they show up? In which case we need to have a partial fraction for each of the powers there. So like with an x squared on bottom you have a partial fraction for x and for x squared. Similar thing can be said for x plus 2 squared and then for irreducible quadratics you have to deal with those as well. For a linear denominator you should have just a constant on top. For a for a irreducible quadratic you should have some type of linear factor on top. dx plus c or something like that. And so question two is again just a starter question. You don't have to do the complete partial fraction decomposition. All that you have to do is determine what is the right template to get started. Now if you need a review of any of these topics here remember that the topics of partial fraction decomposition were covered in lecture 14 and 15. And then likewise topics of trigonometric substitution were discussed in lessons 12 and 13. Please go back to your notes on those topics if you need some more assistance. For example the codex was introduced in lesson 12. That's exactly what you need to use for question number one. Question number two we saw we saw linear factors show up in lesson 14. Irreducible quadratics showed up in lesson 15 so you'd want to consult both of those there. So then coming to question number three. Question number three is going to ask you to compute a numerical approximation of an integral using one of these numerical integration techniques like the trapezoidal rule is a possibility the midpoint rule, Simpson's rule you should know all three of those rules here. Now how this question is going to be formatted is that your function is going to be given actually as a table. So we see the x coordinates are given here and then they're corresponding f of x values are given there as well. So the way you would read this is that f of 10 is equal to negative 12 and f of 22 is equal to 1. Using this table you'll be asked to approximate an integral in this case you're integrating from 10 to 30 and you're going to use five subdivisions in that regard. You want to use the and for this one you're going to use the trapezoidal rule t5 to then estimate what is the area under the curve. You're going to estimate this integral. Be aware that the way that this question is formatted you won't be given an algebraic function. You cannot use the fundamental theorem of calculus to evaluate this integral. You actually only have enough information to approximate it using the method we're introduced here trapezoidal rule midpoint rule or Simpson's rule are all possibilities here. Now if you need some more practice or need a review what's how these calculations are actually done as a reminder lesson 17 reminded us about the trapezoidal and midpoint rules and lesson 18 was exclusively about Simpson's rule. You'll probably want to go to both of those for some further practice there. And then the last question in the multiple choice section will be you'll be asked to evaluate an improper integral. So these are the cases we have integrals where one bound goes off to infinity or negative infinity maybe both or maybe the integral approaches a vertical asymptote and so how do you how do you deal with improper integrals? This is exactly what we learned about in lesson 19 and so question number four will ask you to compute to evaluate an improper integral. Be aware that the answer could be that it's divergent. If the integral is divergent that is the area under the curve goes off towards infinity you would select choice f that it's divergent but if the integral is convergent then you have to actually provide what was the value it converges towards and so then you would say whatever you think it is pi over two or whatever. Moving on to the short response section remember there's four questions here and each of them are worth seven points each. Question number five will be a second question about numerical integration and this will be the last question on the test coming from numerical integration. What it's going to ask you to do is to work with the error bounds associated to these numerical estimates. So the trapezoid rule has an error bound the midpoint rule has an error bound simpson rule has an error bound. I recommend you turn to the study materials or to your notes to find the exact formulas for those error bounds but let's say that we were working with an integral let's say the integral from zero to one of cosine of x squared dx. Be aware that that would be a very difficult function to find an anti-derivative for but we could estimate the area under the curve using something like the trapezoid rule but how do we know we have a good enough accuracy? Well this question asks what choice of n what's the smallest choice of n to guarantee that the approximation is accurate to four decimal places? Well there's an error bound associated to the trapezoid rule you would use that to solve for n remember you're always had to round up in that situation for simpson's rule you always have to round up to the next even integer because for simpson's rule it always has to be an even number there. These error bounds will be dependent upon how large second or fourth derivatives are equal to so for convenience you might be given information about the second or fourth derivative so that the exercise focuses on the error bound and not on the the implicit derivative calculations that are necessary as well. So like this one's format in such a way that the k value you can actually argue a six and that will help you when you try to determine what is the best choice of n in that situation. So these type of questions about error bounds showed up in these numerical approximation sections so again like we saw in the previous slide lesson 17 will give you the error bounds for the midpoint rule and trapezoid rule lesson 18 will give you the error bound for simpson's rule you should be prepared to do any of those three. Question number six you'll be given a trigonometric integral and you'll be asked to find the antiderivative of that trigonometric integral and it could be anything under the sun it could involve signs and cosines the periods could be the same the periods could be different it could involve secants and tangents I mean maybe even cosecants and cotangents right we've learned all about these techniques we should be looking for things like is there an odd number of cosines an even number of cosines odd number of signs an even number of signs tangents and secants similar questions can be asked there a whole litany of trig identities come into come into play there question six will be about any one of those what any one of those type of calculations this is the only question on the test that directly comes from lessons 10 and 11 just as a reminder lesson 10 was actually broken up in two pieces the first half ended our lesson about integration by parts but then the second half then introduced our topic of trigonometric integrals there's some important examples and identities found there lesson 11 then continued and then finished this discussion about trigonometric integrals so like I said question six is to be the only question on the test that comes directly from there but like I mentioned earlier there will be a question in the free response section about trigonometric substitution the whole method of trigonometric substitution is to turn an algebraic integral into a trigonometric integral and so all the techniques that are applicable here will be applicable there as well and that's why I only put one direct question coming from lessons 10 and 11 because the trig sub questions will ask you to do similar information as well now this is an indefinite integral do remember that whatever you calculate you do need a plus c there for full credit so don't forget that c the other two questions on the show response section are going to be questions that ask you to set up simplify but do not evaluate an integral and these are going to do with our two application topics question number seven is going to ask you to set up an integral to compute the arc length of some curve question number eight will similarly ask you to set up an integral that computes the surface area of some surface of revolution like we did of course in sections let's see what the the lesson numbers were lesson 20 for arc length and lesson 21 talked about surface area you're going to want to go to those lessons your corresponding notes the sections from the book in order to find the appropriate formulas for that situation do be aware that you asked to set this thing up so when you set up an integral you do of course need the integral symbol you need to have the the appropriate bounds a and b there needs to be of course a function and that function might be multi-part because of the formulas in play there like for example for surface area in addition to having a square root likely be in the integral you also have a radius to deal with there's also a two pi floating around so look for those but you also need to have your differential that is part of the integral it does represent a geometric part of these problems and so omitting the integral would make it incorrect particularly in these examples right here because the default integral you're actually integrating with respect to ds for which ds can be modified into a dx or a dy that is you can integrate with respect to x or y if you don't specify which one you are using you actually can't necessarily glean from the information what it is it could be dx or dy and therefore you need to be explicitly including that in your calculation there i should also mention that you should simplify these things as appropriate now i don't necessarily require every possible simplification under the sun but you definitely got to get all the low hanging fruit ones and just some natural algebraic simplifications are appropriate here you do not need to evaluate the integral there will be no credit given for that but if you're lacking appropriate and necessary simplifications you could lose some points there as well and that then gets us through the short response section the final section of the exam is the free response section there's four questions here now their points change based upon the perceived difficulty of these questions most of them are in the 12 point range though so question number nine is a 12 point problem and i'll ask you to evaluate an indefinite integral now you are allowed to use any technique you want on this problem but be aware that this question was intended to be a trigonometric substitution so things you can do to show your work because this is a free response question you do need to show your work for full credit there you could say things like what is the trigonometric substitution you're using you can say things like well x equals a times whatever the trig function is let's say it's a tangent substitution i'm not saying that's what the correct substitution is here i'm trying to not give away the answer i want you to work this out on your own but let's say that you think the trig sub is x equals a tangent then you could be like well dx equals a secant squared theta d theta you can then also tell us about the square root the square root of a squared plus x squared is equal to a secant this is all very important information you can then also if you want to you control the triangle not necessary but this is one way to show your work if i have a tangent substitution i might say x over a the square root of x squared plus a squared like so then once that's done of course you can start doing your substitutions there you can be like oh well x squared is a tangent squared now the square root is then a secant again i know i'm not actually doing the wrong problem this is just for illustrative purposes here the dx would be like a secant squared theta d theta then you proceed to simplify this thing compute it but at the end you should probably switch it back into some antiderivative in terms of x unless it's a definite integral if it's a definite integral then of course you can switch the balance and find the number using the trig metric function but most likely this will be an indefinite integral so the variable if you have some function in terms of theta that you would find here that's good partial credit but for full credit you do need to switch it back to the function of x and as this is an indefinite question indefinite integral you need that plus c for full credit there so there's a lot going on because you have to convert from algebra to trig solve the trig antiderivative then switch it back from trig to algebra there's a lot of stuff going on there which is why this is worth so many points 12 points there and also just as a reminder I mean I've already said in this video but in case you missed it before let me just remind you again that the topics of trigometric substitution are discussed in details in lessons 12 through 13 the next question number 10 it's also worth 10 points and it's of comparable difficulty to question number nine the difference now though is that we are given a rational function and so you're going to want to find the antiderivative using the method of partial fraction decomposition as discussed in lessons 14 and 15 and again this one question could cover any of the topics that we've seen using partial fractions so in particular if I was to get started with this there's some things you should look for you might need to use long division that is division of the polynomials this happens in particular if the fraction is improper that is the numerator is bigger than the denominator the example on the screen right now seems to suggest you might need to do that it might also be necessary to complete the square because you do have to sometimes deal with these irreducible quadratics in order to prepare for a tangent typically but it could be a sin or secant as well in order to prepare for a trigometric substitution you often have to complete the square be aware that this is an algebraic skill that might also be necessary on question number nine you might have to complete the square before you can do the substitution clearly factoring might be an important thing that has to happen in this situation can you factor the polynomial denominator I won't make the factorizations too difficult but pulling out things like a greatest common divisor our common factorization forms like uh difference of squares perfect square trinomials maybe like factoring by grouping there are some elementary factoring techniques that uh we're on the on the par of like intermediate algebra math 1010 that would be appropriate and be expected of students to know here so there are some algebraic things you have to do there then of course you're gonna have to do the partial fraction decomposition itself come up with the correct template like we do on question number two from the multiple choice section solve for the coefficients now as you're solving for the coefficients and your partial fraction decomposition of course you can use this technique of annihilation where you choose strategic cool values of x to eliminate everything except for one of the parameters that's acceptable you could set up a system of linear equations and solve it you could do a combination of those two methods however you find the coefficients as long as it's done correctly i don't care how you do it so you'll need to find the partial fraction decomposition and then at the end you do have to integrate the function itself if you've computed the the pfd it shouldn't be too hard to do um you'll have some standard forms there uh typically the anti-derivative involved logarithms and arc tangents that come from u substitutions those are often necessary in the situation um likewise you might have to do some trig subs when you have irreducible quadratics and so just some things to remember here it might be useful to observe that if you have a function of the form du over u this is equal to the natural log of the absolute value of u plus a constant that does show up with enough frequency that it's worth mentioning i should also mention that if you if you have the integral du over a squared plus u squared where u is some function of x a is a constant the anti-derivative in that situation is going to be one over a arc tangent of u over a plus a constant these two anti-derivative forms are of some common frequency when you're dealing with partial fraction decompositions it might be worth you know memorizing these are put in a reference that you can find later on all right let's now move to the last page of the exam so questions 11 and 12 uh question number six sorry question number 11 is worth 16 points um it is the climactic question of this exam and this question is supposed to represent our street fighter integral question that we learned about of course in the lessons here where anything goes on this one as opposed to questions nine and ten where i then i i i i prophesied to you what the topics are going to be you're going to use trig subs on number nine if you do something else you're probably making it harder than needs to be on question number 10 you're going to use partial fraction decomposition of some kind and again if you don't then you're probably doing it wrong not necessarily but you're definitely making it harder if you're not with question 11 anything goes anything goes anything goes so this involves any of the topics that we've learned about now the specific lesson about these about street fighting was lesson 16 which we talked about techniques of integration for which remember those techniques i mean sometimes the right thing to do is some type of algebraic manipulation use algebraic identities to help you out here um that is certainly a possibility here uh but also by the nature of the problem maybe it's trigometric or maybe you do a trig metric substitution um trigometric identities may also be useful in this situation again you might want to go back to lessons 10 and 11 to review some of those topics but like i said you also can use substitution of course the topic of you substitution is perfectly relevant here that actually goes back to lesson two for this for this lecture series here old topics are just as valuable on this example as well so speaking of which we could talk about integration by parts integration by parts was introduced in lessons nine and ten integration by parts might be necessary to work on these type of problems um in addition to trig subs uh excuse me in addition to u-subs you might have to do a trig sub this one i'm looking at i don't see any sums or differences of squares so probably don't need a trig sub but maybe um which of course i've already said it twice but i'll say it one more time trig subs were introduced in lessons 12 and 13 here that might be very very appropriate to use of course methods of partial fraction decomposition might be very relevant for this one which again that was lessons 14 and 15 how do you deal with partial fractions just much like we did in question number 10 there but also an important topic to pay particular attention to in lessons 16 we actually focus a lot of time on a method known as the rationalizing substitution for which you do basically a u-sub that turns your function into well into basically a partial fraction well that is a rational function for to do partial fractions decompositions and this is very useful when you have a substitution like u equals the square root of x um because in that case you get that du is equal to well i'll actually say this way you get that dx is equal to two u du the derivative of the function is actually like similar to the original function itself and that can be very useful this also happens with like exponentials if u equals e to the x then du equals e to the x dx for which that tells us that dx is equal to du over u things like that so this rationalizing substitution is very useful with radicals and exponentials because of the similarities between the functions and its derivative and so when i look at something like this my thoughts are maybe a rationalism substitution could be very useful here but honestly anything can go this is why this question is worth so many points do not give up on this question even if you're stumped that is too many points just to throw away um work at it do what you can play with multiple attempts um you can definitely get some partial credit even if you don't do it all the way um don't forfeit this question it's a very important one and it really much is going to characterize how one does um these techniques of integration the last question uh question number 12 it's only worth 10 points here not as much as the other three certainly um and this is actually a question about improper integrals we have one question the multiple choice there'll be a second one right here in this situation you'll be asked to show that an integral an improper integral either converges or diverges it'll tell you whether it's convergent or divergent um in particular this question this function we chose it in such a way that you will not likely be able to compute an elementary anti derivative of that function um so we can't necessarily apply the fundamental theorem calculus directly instead though what we can use as an a theorem we learned about when we learned about improper integrals this is the actually the so-called comparison test remember the comparison test tells us that if an improper integral is smaller than a convergent integral then it's likewise convergent um but also if an integral is bigger than a divergent improper integral then it also must be divergent so that is we can potentially simplify the function by comparing it to something which you know to be convergent or divergent and then use that make sure you state explicitly when is the comparison test being used and so try to mimic some examples uh that we saw of course in lesson 19 and so that gets us to the end of this exam and we've now talked about all the topics that are covered on this exam uh in addition to of course this video use the exam syllabus to practice exam and other study resources that are available to you on canvas if you have any questions please reach out to me or other appropriate sources to get the help you need as you prepare for this exam i'm quite confident that you will be able to do very well on this exam if you are sufficiently prepared for it and let me help you prepare for the exam if you're not there already