MitPY 6: Integration by Substitution

From frequent commenter, SRK:

A question for commenters: how to explain / teach integration by substitution? To organise discussion, consider the simple case

    \[\boldsymbol{ \int \frac{2x}{1+x^2}dx \,.}\]

Here are some options.

1) Let u = 1 + x^2. This gives \frac{du}{dx} = 2x, hence dx = \frac{du}{2x}. So our integral becomes \int \frac{2x}{u}\times \frac{du}{2x} = \int \frac{1}{u}du. Benefits: the abuse of notation here helps students get their integral in the correct form. Worry: I am uncomfortable with this because students generally just look at this and think “ok, so dy/dx is a fraction cancel top and bottom hey ho away we go”. I’m also unclear on whether, or the extent to which, I should penalise students for using this method in their work.

2)  Let u = 1 + x^2. This gives \frac{du}{dx} = 2x. So our integral becomes \int \frac{1}{u}\times \frac{du}{dx}dx = \int \frac{1}{u}du. Benefit. This last equality can be justified using chain rule. Worry: students find it more difficult to get their integral in the correct form.

3) \frac{2x}{1+x^2} has the form f'(g(x))g'(x) where g(x)=1+x^2 and f'(x) = \frac{1}{x}. Hence, the antiderivative is f(g(x)) = \log (1+x^2). This is just the antidifferentiation version of chain rule.  Benefit. I find this method crystal clear, and – at least conceptually – so do the students. Worry. Students often aren’t able to recognise the correct structure of the functions to make this work.

So I’m curious how other commenters approach this, what they’ve found has been effective / successful, and what other pros / cons there are with various methods.

UPDATE (21/04)

Following on from David’s comment below, and at the risk of splitting the discussion in two, we’ve posted a companion WitCH.

73 Replies to “MitPY 6: Integration by Substitution”

  1. If du/dx isn’t a fraction then what is it? Just notation? Or something else? I’m in year two of uni and am still unclear on what du/dx (for instance) actually is. And I’m worried that if I become some sort of teacher and someone asks me I won’t have any non wishy-washy answer. Maybe, hopefully, the answer just comes from experience, or something.

    1. Hi, Craig. Yes, du/dx is “just notation”, for the derivative function. But the notation is attractive because it is suggestive of a fraction, which in some ways is how derivatives work. It is not mere luck that SRK’s dodgy-looking Method 1 works. The notation is also problematic, however, for the very same reason: it is suggestive of a fraction, which it isn’t.

      As for your general concern, it is part of a teacher’s life to be confronted with questions for which you won’t know the answer. I’ve been caught out plenty of times. That can be embarrassing (if you should know the answer), but it can also be indicative of a thoughtful student. I always try to be honest, praise the question and the questioner, and if i can’t figure it out on the spot then I promise to come back with the best answer I can next lesson/lecture. (It is of course better, however, to sort out the ideas as clearly as practicable before teaching.)

    2. Thanks Craig. If you are a second year uni student and this is confusing you, then I feel vindicated in trying to get my students to think carefully about what is going on here.

  2. I don’t think Method 1 abused the notation. The properties of differentials applied, even though the concept of differentials are not in the course.

    1. Hi, anonymous. However the argument may in future be justified by “differentials”, Method 1 is most definitely an abuse of notation.

        1. Hi, Anon. You mean solving (separable) differential equations of the form dy/dx = f(y)g(x) etc? And by “you” do you mean me (rather than a general “you”)?

          Anyway, the answer for me is “yes-ish”.

          1. Sorry, I meant general ‘you’. If the notation is used in solving differential equations, why we can’t use it to interpret Method 1?

            1. Do you give it meaning? If you don’t, then I don’t think it should be used. If you do, then it certainly doesn’t mean “the top or bottom bit of a piece of other bad notation”, which is an interpretation you won’t be able to get away from.

              You could give it the meaning that it is simply shorthand for other notation, but then why not just use that notation in the first place?

              You’re vexing me today Marty!

              1. The fractional property of dy/dx can be explained by the linear property of the difference $\Delta {x}. But it goes too far, it’s not suitable to teach at secondary level.

                    1. Hi anonymous, I wasn’t thrilled with the responses to the questions in those links, but they correctly point out that dy/dx is most definitely not a fraction.

        2. The method I encourage my students to use when solving separable DEs is the following: (Note that they already know that \int f(u)\frac{du}{dx}dx = \int f(u)du from our work on integration by substitution)

          Given \frac{dy}{dx} = f(x)g(y) then, (with the required provisos) \frac{1}{g(y)}\frac{dy}{dx} = f(x) \Rightarrow \int \frac{1}{g(y)}\frac{dy}{dx}dx = \int f(x)dx \Rightarrow \int \frac{1}{g(y)}dy = \int f(x)dx.

          It may not be quite as quick as a method involving an abuse of notation (where we “multiply both sides by dx”), but that method involves the confusing idea that we integrate the LHS “with respect to y” but integrate the RHS “with respect to x”, and this is confusing for students because it seems to violate principles of equality.

          1. SRK, that is also how I teach separation of variables when I need to at first year uni level. Although, I do always use limits.

            1. Thanks Glen.

              Just trying to make sure I understand your approach properly. Suppose one wanted to solve the DE \frac{dy}{dx} = xy. Would you encourage your students to solve this by writing something like \int_{a}^y \frac{1}{t}\,dt = \int_{b}^x s\, ds, and assume that ay > 0 ?

              1. That would be OK. The way I’d teach it (at say 1st year uni level):

                DE:

                    \[y'(x) = x\,y(x)\]

                for a differentiable function y:(a,b)\rightarrow\mathbb{R}.

                Assume y is uniformly positive (for simplicity, sometimes I give questions where there are branching solutions around degenerate points, but usually only one per test, most students don’t get these out).

                Then for each x\in(a,b) we have

                    \[\frac{y'(x)}{y(x)} = x\]

                which we recognise from the chain rule (here is where they can instead use your integration by substitution, I don’t mind) as being

                    \[\big(\log y(x)\big)' = x \,.\]

                Then, integrate both sides with respect to x (which becomes a dummy variable of integration) from say c to z where c, z\in (a,b) (but are otherwise arbitrary, and become new, independent variables):

                    \[\int_c^z \big(\log y(x)\big)' \,dx = \int_c^z x\,dx\]

                We can evaluate the LHS using the FTC (which I teach before I do DEs, so they need to at least state that the hypotheses of the FTC are satisfied, usually it is obvious), and the RHS using known results. We find

                    \[\log y(z) - \log y(c) = \frac{z^2}{2} - \frac{c^2}{2}\]

                for all c,z\in(a,b), which we can simplify to

                    \[y(z) = \Big(y(c)e^{-\frac{c^2}{2}}\Big)\,e^{\frac{z^2}{2}}\,.\]

                At this point they may choose to relabel z as x, and fix c (so that c is no longer an independent variable of the above equation) and set C = \Big(y(c)e^{-\frac{c^2}{2}}\Big).

                Then we have the representation formula

                    \[y(x) = Ce^{\frac{z^2}{2}} \,.\]

                This gives us a one-parameter family of solutions, parametrised by c (note that C = C(c)). One big advantage of being rigorous with this is that it helps when considering problems down the road. For instance, if we know that the midpoint of y is 1 (i.e. y(\frac{a+b}{2}) = 1), then we obtain a unique solution, because we can use C(\frac{a+b}{2}). It also allows us to be more careful about the domain and range of our solutions — note that here we assumed y is uniformly positive, which remains valid in our representation formula. However since C(c) \ne 0 for any c, if (a,b)=\mathbb{R} then there doesn’t exist any uniformly bounded positive (or negative) solution.

                I suppose the reason I go back to the chain rule for this kind of thing is because I don’t want students to become confused with the variable of integration being in the image — they might think that ay > 0 (I’m using the variables as in your comment) means for a in the domain of y, but actually a here is in the image of y. So the restriction ay > 0 is on the values of the function y. That said, I wouldn’t consider your approach wrong in any way.

                I hope that is enough, but not too much, information!

  3. Once, with some colleagues, I wrote a paper entitled “Calculus: A Marxist approach”.

    Click to access TechPaperFahey.pdf

    In teaching himself differential calculus, Marx used the notation 0/0. This looks like the writings of a careless or confused student. But Marx was a clever bloke: why did he do this?

    On the face of it, 0/0 is no more ridiculous than dy/dx. The meaning is well-known by those who know it well.

    Nearly 60 years ago, I was looking at my father’s mathematics books from school at my grandparents’ place; I knew nothing about calculus so I was intrigued to find a book on calculus in the collection. I recall cancelling the d’s in all the problems on derivatives.

    1. I forwarded the Marx article to my philosopher brother. He was reassured that, unlike some modern acolytes, Marx didn’t take the position that learning times tables oppresses the proletariat.

  4. Method 2 is OK, but I am disturbed by your lack of limits. So long as the role that the chain rule plays is made clear, I think you are winning.

    At uni we re-teach integration by substitution many times. Even by third year students often don’t understand the point.

    One suggestion — try using the “prime” notation for derivatives. The fact of the matter is that if you have a function f then f' is the derivative of f with respect to the first slot — this is unambiguous because f only takes one variable. It doesn’t matter what variable you stick in there, it can be f'(x), f'(y)', f'(my_cat) or anything.

    This is really helpful for the chain rule. You can differentiate f first and then evaluate it at a special point called g(x):

        \[f'(g(x))\]

    or you can evaluate the composition and then differentiate:

        \[(f(g(x)))' \,.\]

    I would be so happy if the difference between these two things could be made notationally clear by at least some teachers at high school. I would be even more happy if this kind of discussion was done around the same time as the chain rule, an application of which is integration by substitution.

    (Students become seriously unstuck when they get to multivariable calculus and then PDE at uni. The problem there is made extreme by functions with arbitrarily many variables and complex substitutions + chain rules usually being done implicitly. Abuse of notation is very common. Typically 1/3 to 1/2 of the semester is spent fixing their ideas around what derivatives are and how the chain rules works.)

    1. Hi, Glen. I’m sympathetic about the problem at uni, but I pretty much disagree with your conclusion. I think function notation is way overdone in schools, at least in Victorian VCE. (I strongly approve, however, of your gratuitous use of “cat”).

      1. Oh I’m really interested to hear your reasons for that. (I’m assuming that you mean using the prime to denote derivatives, because that’s what my point is in my post? But even if not, I’m still interested to know what you’re getting at.)

        1. Hi, Glen. Yes I’m voting for the primacy of Leibniz notation to indicate derivatives in VCE (as well as making the more general claim that function notation is overdone in school). I’ll give my reasons, but I’ll first give others a chance to weigh in on the whole topic.

  5. I might also add, the same issues arise when teaching students how to solve separable differential equations. But if students have a good understanding of integration by substitution, then the method for solving separable differential equations is easily justified.

  6. For what it’s worth here is my 2 bob’s worth.

    I would teach SRK’s method 2 initially. Then after we had worked several problems I would say “By the way – here is what the engineers do” and then show method 1, along with a comment along the lines that “dy/dx is not a fraction, but it works like a fraction.” Most students immediately adopted method 1. Nowadays I might say “dy/dx is not a fraction, until you get to advanced model theory”. If a student really wants to think of dy/dx as a fraction, I am not fussed; what looks like a fraction, works like a fraction and smells like a fraction … The student might be the next Robinsohn.

    At Swinburne we had a policy of not prescribing textbooks; we wrote and issued brief summaries with sets of exercises, widely used at other techs. This type of integral was called OPDOOP (One part derivative of another part) which I thought constituted a good guide as to when to try the method. [There was a subset called NIDOD which yielded log answers.] The students joined the fun in finding acronyms for other methods in the course.

  7. Greetings all, for what its worth I am about to teach this concept in the next week or 2 to a class of Specialist Mathematics students via video-link (which is just horrible to attempt by the way, so my sympathies for anyone having to do it at the moment).

    My approach is somewhere between (1) and (2) but much closer to (2).

    I find a few things helpful with this:

    Students are not going to ever be asked to explain why this works, because VCAA won’t be able to assess that. So I leave the ideology out of the lessons as much as I can morally allow myself.
    Process (3) is a very useful one that I do come back to when helping students to “look” for what to substitute but it is not something I explicitly teach.
    I also teach integration by parts even though it is not in the curriculum because I feel it is useful, helpful for some questions and really not beyond a Specialist student to comprehend.

    1. Best of luck with that RF. I am not even going to try to teach my SM class (or any of my classes) by video conference, all my lessons are pre-recorded and students are sent links to videos.

      In response to the substance of your comment, this is definitely the point that troubles me. I suspect my mathematical and educational judgement is not yet good enough to always know exactly when I’ve crossed the line from good practice to dogmatism. While acknowledging that it’s of enormous practical importance to my students that they are well prepared for their exams, I think they receive an impoverished education if I take the content of the course to be strictly limited by what sorts of questions appear on VCE exams. Given that the chain rule plays such an important role in many areas of SM, I believe it is important to be explicit about that in my teaching, even if students will never be asked in an exam to explicitly justify, using the chain rule, the method of integration by substitution – either in general or in a specific case.

      Of course, in this topic, part of the problem is the perennially infuriating problem of not having access to VCAA marking schemes, and hence I do not know whether or to what extent students are penalised for abusing notation as in Method 1.

      1. Hi, SRK. My understanding is that Method 1 is acceptable to VCE examiners, but others would be able to answer more definitively.

  8. Thanks, everyone, for the very interesting discussion. I’ll still hold back my specific thoughts for now (except for teasing Glen). But i have a few points to help clarify the discussion, if only for me.

    0) It is always worth noting that Maths Methods doesn’t include integration by substitution, one of the strongest of the many reasons why this subject is so monumentally stupid. Note that this point is not off-topic, since many Methods teachers, being smarter than the VCAA idiots, teach integration by substitution, at least for linear substitutions.

    1) I think everyone now accepts and understands that dy/dx etc are not fractions and, whatever the arguable merits of Method 1, it is an abuse of notation. For some, that is the end of the discussion of Method 1. But, not for others (including me).

    2) The discussion is branching in two different ways, which is fine, but is worth noting. One’s sense of a satisfactory answer may depend a lot upon which branch one is on.

    The first branching is:

    i) SRK’s original question was about integration by substitution.

    ii) There has been an expansion of SRK’s question to also discuss separable differential equations, dy/dx = f(x)g(y).

    That expansion makes sense, since integration by substitution (or the chain rule, and ultimately the chain rule) underlies the solution method for these DEs, and since there is an analogous debate about notation and setting out.

    If in principle the questions are very similar, however, I’m not sure they necessarily are in practice. Integration by substitution can be very tricky, since the substitution can be difficult to see. In (ii), by contrast, the “substitution” is never a mystery, since it’s always y = y(x), whatever the function y(x) happens to be. Moreover, the subsequent integrals in (ii) are, in practice (i.e. by design), usually pretty easy. These distinctions don’t imply that one’s preferred approach need be different, but it does raise the possibility.

    The second branching is:

    b) SRK was asking about teaching integration by substitution in VCE, at the senior school level.

    c) That has been expanded to included the discussion of teaching substitution (and separable DEs) at, say, 1st year university.

    a) As suggested in (0), it is reasonable to also expand the discussion to include the survival-mode discussion of substitution in Maths Methods.

    Of course (c) can be further branched, or spectrumised, depending upon how cookbooky or rigorous the subject is, and upon the strength of the uni students.

    In fact, as SRK pointed out, (b) is already branched in this manner. Since the VCAA idiots don’t give a damn whether students understand the justification for substitution (or for anything), the teacher must decide how much precious time, if any, to take away from score-maximisation in order to give the topic some mathematical solidity.

    3) As always, it is worth asking, what does Spivak do? (If you don’t know what this means, consider it homework.)

    1. I know lots of famous people do use dx and du willy-nilly in this context and I’m certain that many of them know precisely what they are doing.

      And, maybe they have a way of justifying it — perhaps it is all a mnemonic for what the correct notation is.

      I don’t like it because what often happens is that students end up under a total misapprehension for what integration by substitution (and the chain rule) actually is saying, they get confused about derivatives, and if this isn’t cleared up at university, they will just never know. What’s the point then, of teaching the topic at all? I think it isn’t so hard to teach it without using dx and du.

      I know Marty has more to say, I’m trying to stay quiet, but it’s difficult :D.

        1. The discussion on pp 130–132 is brilliant. I love it.

          In summary, Spivak says:

          (just after explaining the Leibniz notation will not be gone anytime soon; keep in mind this is written in 1967, so he was certainly right)

          “The policy adopted in this book is to disallow Leibnizian notation within the text, but to include it in the Problems; several chapters contain a few (immediately recognizable) problems which are expressly designed to illustrate the vagaries of Leibnizian notation.”

          (any typos are mine)

          I think that is pretty self-explanatory, but anyone who wants more detail, please go and read the indicated pages, Spivak is a joy to read.

          1. Thanks, Glen. So, Spivak’s choice and message is that Leibniz (dy/dx) notation is bad for discussion, for the theory of what’s going on, but that it is good for at least some computation. Note also that by “Leibniz”, Spivak includes du = 2x dx and whatnot.

            So, Glen (or anyone), are you generally in agreement with Spivak? What then does that say for the various Methods, either for what you say to the students or for how you might want them to frame the answers to questions/problems? Of course the answer could vary, depending upon which branch you’re thinking of.

            1. I think I’ve misrepresented Spivak’s position. He writes three pages on the topic and the quote I picked probably wasn’t the best — it’s just the last-ish thing he writes about it. His exercises are included (i) to make sure the reader is familiar with this notation, because it won’t be going anywhere, and (ii) to make sure the reader can clearly see ambiguities that arise when using the notation. So they don’t make mistakes when using it.

              I don’t think Spivak is making the point that it is good for computation. Of course, I might be misreading it, but I thought he was being (extremely polite and) clear.

              Of course, it is possible to still ask the question: is Leibniz notation good for computation? I think absolutely not. The notation confuses what a dummy variable is, what a derivative is, and I’ve witnessed so many students become completely confused with it when trying to do problems. There is no shadow of a doubt in my mind that this notation harms way more than it helps.

              I can see that if you are already used to using this notation, AND you already know what you are doing, it may be that the speed at which you are getting your thoughts onto paper leads to an increase in computational ease.

              Here’s another question related to this topic. What do people tell students the dx in the integrand means? I’m primarily interested in this question when students are first given the integral, so at high school. At uni, I ask this question often. It is one of the very first questions I ask my Calc students, and i ask them again in second year (in Analysis), and then again in third year (in PDE). It’s always a pretty nice discussion.

              1. Hmm. Maybe I misread Spivak. I’ll look again.

                But of course, as you note, the real question is whether Leibniz is indeed useful for computation. (Even if it is, one can argue against it, because of the drawbacks for proof and discussion of ideas. But if it isn’t, that kills the only strong argument for Leibniz, and that’s the end of it.)

                The Evil Mathologer is on my back right now, so back to you tomorrow.

              2. I feel very out of my depth in this current discussion, but can answer the very last part.

                When I teach integration for the first time, I do it as a Riemann sum and therefore the dx is the width of the rectangles (limit of, rather), so an integral becomes the limit of a sum as the width tends to zero.

                But, based on what I’ve been reading here, I may be at best wrong and at worst completely misguided in my own understanding.

                1. Hi, RF. i wouldn’t worry too much. Glen has asked a question that no matter how you reply, he can argue that you’re wrong.

                2. Hi Red Five!

                  If the dx is the limit of the width of the rectangles, and that width is going to zero… I’d be confused as a student thinking that

                      \[\int_a^b u(x)\,dx = \int_a^b u(x)\,0 = 0?\]

                  I think it’s probably better in the derivation of the Riemann integral to use something like a w (or \delta or something) for width of the rectangles, and then you can just say that the whole integral expression is the result of taking the limit of w to zero.

                  Anyway, that’s more beside the point — after the derivation of Riemann integration, we have this notation

                      \[\int_a^b u(x)\,dx\]

                  and in there is that dastardly dx, that is causing so much trouble :).

                  1. Thanks, Glen. It was a very good sub-question. I strongly agree, that using dx both to mean a little bit of x and in the integral notation is a (very popular) recipe for confusion. In this context I always use \boldsymbol{\triangle x} for a little bit of x.

                  2. I’m not saying the width is becoming zero but asking what is the limit as the width tends to zero.

                    To me there is a very big difference. My students, hopefully, pick this up before I give them the “and now here is a rule you can use…” part of the lesson.

                  3. OK… I have not been able to leave this question alone and have gone back to some of my university textbooks and… the mystery deepens.

                    I always assumed that the dx “the measure” was there to indicate the dummy variable being used in the case of an integrand with constants and variables (assuming a Riemann integral). So I guess I saw of it more as a delimiter rather than having special significance.

                    But then when I have dug a bit further into more general Riemann integrals, I see examples of generalisation where the dx has been simply omitted, such as \int (f+g) = \int f + \int g

                    And then I recall VCAA examiner reports which lament the missing dx in students’ work.

                    1. RF, it’s just notation. Mathematicians prefer \int f, since it says everything you need, and there’s no suggestion of an x being more than a dummy variable. The notation \int f(x)\,{\rm d}x is fine and is standard in schools, although VCAA seems to be sliding away from it. The in between \int f(x) is technically wrong, but the only people I know who really care are pedantic VCAA assholes.

                    2. OK, thanks. So… to continue the pedantry… whilst we keep f as a function of no specific variable, it is OK to not have the measure present, but as soon as a variable is mentioned, the measure is required?

                      VCAA are a bunch of mostly faceless /insert negative adjective/ pedants, but the structure I feel is perhaps more to blame than the cogs of the machine. The 1966 leaving certificate exams (for example, I was reading them today) are about as difficult as modern VCAA papers, but the mathematics is unapologetically pure in nature. The current VCE subjects, in my unqualified opinion, seem to struggle with their identity; are they pure or applied? They don’t seem to know nor care.

                    3. Hi RF!

                      There is a lot to this story. I won’t go through everything, but some random thoughts:

                      I like to denote where an integrand ends, and dx (or whatever is being used) does that job.
                      Sometimes there are multiple measures lying around, for example d\mathcal{L}, d\mathcal{H}, and d\mu (or dS) are common (these are Lebesgue, Hausdorff and the induced measure). For example, you might have seen the divergence theorem in multivariable calculus:

                          \[\int_U \text{div }V\,dx = \int_{\partial U} V\cdot\nu\,dS\,;\]

                      here: U\subset\mathbb{R}^n is an open (smooth) domain in Euclidean space, V is a differentiable vector field on U, \partial U is the boundary of U, the \cdot denotes the dot product in \mathbb{R}^n, \nu:\partial S\rightarrow\mathbb{R}^n is the unit normal vector field to \partial S pointing away from the interior of U (let’s assume we are able to identify an interior and exterior), and finally, dS is the induced area measure on \partial U.

                      Note that in \text{all} these cases the actual variables of integration are omitted.

                      Sometimes there are multiple parameters/variables floating about and then \textbf{yes} you are right, we want to be certain about what variable we are integrating with respect to. For instance, this comes up when working with convolutions (in the myriad of contexts in which they are arise): Suppose we have (continuous say) functions (also in L^1(\mathbb{R})) f,g,h:\mathbb{R}\rightarrow\mathbb{R}. Then f is the convolution of g with h if

                          \[f(x) = \int_{\mathbb{R}} g(y)h(x-y)\,dy\,.\]

                      When working with convolutions in more depth, for instance when dealing with Greens functions, it becomes even more important.

                      There are a lot more, but these come to mind as particularly useful, and places where I’ve seen students make mistakes that could have been prevented if they had used the “dx“. 🙂

                      Cheers!
                      Glen

                    4. OK… maybe now is the time to mention I majored in abstract algebra and not calculus… I did study Vector Analysis though, so do remember Green’s theorem and the associated calculus. I didn’t do much in the way of PDEs, so a lot of this calculus I’ve had to try to learn post-graduation with mixed success. I am, however, genuinely interested in knowing the extent to which what I am teaching my Year 12 students is correct… unsurprisingly, Marty is my number 1 fact-checker on these matters!

                    5. Sorry, RF, I should have added that your comment about other variables etc. is definitely relevant. So, at school the {\rm d}x is just convention, but yes if there are other variables and constants flying around then it makes sense to specify the integration variable.

                      Indicating the “measure” is again part of convention and common sense. The basic principle is that if the notion of integration is clear (measure and variable) then \int f works just fine; if it’s not clear then make it clear. It’s basically in the same spirit as USFB.

                    6. “pathetic”? Yes, you could say it’s pathetic. Or, you could say it’s thoughtless idiocy from a bunch of anal-retentive assholes who don’t know the difference between sheeplike rule-following and properly considered rigour. Your choice.

                  1. And thanks Glen. I know of Lebesgue integration but am not so familiar with the others (again, didn’t choose the calculus topics much past second year…) your explanations do make some sense though, so thanks.

  9. I tend to have my Year 12 Specialist Maths students do it the way that it is presented in the “standard” (whatever that means) first year textbooks.

    That is, if we want to determine \int\frac{2x}{1+x^2}\,dx we’d let u=1+x^2 and so \frac{du}{dx}=2x\Rightarrow du=2x\,dx, thus treating \frac{du}{dx} as a “fraction” (which I emphasize to my students it is not, but rather behaves like one). In any case, this fraction-like behaviour is one that we use without (much) hesitation whenever we are solving separable de’s.

    So, the integral becomes \int\frac{1}{u}\,du.

    The benefit of this method, to my mind, is that we are doing all of the rearranging outside the integral. This allows us to move cleanly from an integral involving nothing but x’s and dx, to an integral involving nothing but u’s and du.

    The methods presented in chapter 7 of the “standard” year 12 textbook vary from OK to egregiously horrible (consider the worked examples on pg 303 where we get to see integrals with only the variable u appearing and a dx tacked on the end. In the first example on that page, the (-2) magically represents \frac{du}{dx} – yikes!). Apologies to those reading who don’t have this textbook on their desks, but I suspect the audience of this blog includes a lot who do.

    Furthermore, what if the integral was a definite integral? Say \int_{0}^{2}\frac{2x}{1+x^2}\,dx? Then the method of doing all of the rearranging outside the integral results in a very nice conclusion:

    Let u=1+x^2 and so \frac{du}{dx}=2x\Rightarrow du=2x\,dx. When x=0, u=1 and when x=2, u=5. Then \int_{0}^{2}\frac{2x}{1+x^2}\,dx=\int_{1}^{5}\frac{1}{u}\,du.

    If you are running a mix of x’s and u’s in your integral, what do you do about the terminals? At what point do they change? Students don’t always know the answer to this and I guess this may be why I see plenty of students messing up simple definite integration problems (with substitution).

    1. Hi, David, and welcome to the Dark Side.

      You haven’t said everything there is to say, but I agree with everything you’ve said. As for the “standard” textbook you mention, it is one big Everest of WitCHery. I’ll look at the part you mention and maybe throw up a companion witCH for this MitPY.

    2. I definitely don’t agree, but I’ve already made all the same points elsewhere on other comments to Marty’s post. So I’ll just register my disagreement with… most? of that, and otherwise be quiet.

    3. Thanks David. The way I deal with your last question would be to tell students that \int_{0}^2 \frac{1}{1+x^2}\frac{du}{dx}dx is ok (since \frac{du}{dx} is just notation for some function of x) but that \int_{0}^2 \frac{1}{u}du or \int_{0}^2 \frac{1}{1+x^2}du is not (or at least, they are not equal to the original definite integral)

      The general thing I tell my students is something like: “When you’ve got a dx at the end of your integral, your limits should be values of x; when you’ve got a du at the end of your integral, your limits should be values of u”.

      I agree with the general point that it’s best to FIRST deal with the change of variables / limits without writing each line as a definite integral, and then writing down the correct definite integral with all variables changed.

  10. The report for 2019 Specialist Exam 1 has been released. https://www.vcaa.vic.edu.au/Documents/exams/mathematics/2019/SM1_examrep.pdf

    Pertinent to this thread is that the examiners suggest that students should be familiar with Method 3 given in the original post, and should not “waste time” by writing out “u = … , etc.”, but just recognise the anti-derivative by “reverse chain ruling”.

    Oh, and in relation to another matter raised on this blog: the distribution of marks for that diabolical Question 10 is interesting. An average mark of 3.2 / 5 marks – despite the question requiring a wall of algebra / arithmetic to entirely complete – suggests to me that only the 5th mark was awarded for doing all that arithmetic, and that the 3rd / 4th marks were for the substitution and knowing exact values for cos and sin.

    1. Thanks, SRK. I haven’t had time to go properly through the report. I think they’re advocating less than Method 3. The integrals in both Q1 and Q8 are of the form \int f'/f, and I’m interpreting that the examiners were just permitting/expecting students to recognise that antiderivative as log(f) + c. Others can say whether that’s reasonable in the context of expected grading and other reports.

        1. Sure. I just don’t think the comment in the report was general advice. Unless they’re indicating all future integrals will be trivial …

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