photo (1)

 

I’ve never taught Calculus, and I find myself struggling with this mistake…

…in a good way. Here’s what’s want through my head:

  • Oh, God, integration.
  • How would I solve it? Is there any way that I can avoid integrating by parts? (Yeah: transform the numerator to (x-1)+1.)
  • Oh, shoot, how do you integrate by parts anyway? It’s vdu – uduvuvvvvvuuu or whaaaaa???
  • [Googles integration by parts]
  • [Solves it using integration by parts]
  • Good Lord I don’t like integrating by parts.
  • What did this kid do wrong?
  • It looks like they ended up with a square root instead of a square?
  • That makes sense. People often make association mistakes when they’re working on a problem where they have to keep in mind a bunch of moving parts.

I think that working through math mistakes like this one would be a great way to prepare for a new course.

(Hey! Someone should start collecting and organizing these so that they’re available to new teachers…)

  • suevanhattum

    I can’t understand why anyone would use integration by parts for this problem, unless they’re just flailing. (Your first method is exactly what I’d do.) Ok, I just tried integration by parts, and I can’t see how to get an answer that way. Either way you split it up into u and dv, you get something worse for vdu. Did you really get a solution with integration by parts? I’m so curious how you did it. (Btw, I like integration by parts. I teach Calc II often. My least favorite part is parametric and polar. I moved that sooner in the course this semester, so that I’d do it justice.)

    • Time for some honesty: I didn’t actually solve it by integrating by parts.

  • A very strange way to do this by parts: set u=x/(x-1) and dv=dx and arrive at

    x^2/(x-1) + Integral x/(x-1)^2 dx

    Integrate this latter by parts by setting u=x and dv = 1/(x-1)^2 dx to arrive at

    x^2/(x-1) – x/(x-1) + Integral 1/(x-1) dx

    This last integral is then ln|x-1| + C, and you’re done.

    This is extremely bad form: typically doing two integrations by parts with the v from the first integration used as the u in the second integration just gives you back your original integral — the second integration by parts undid the first one. The reason it happens to work here is because of the big simplification in the derivative of u in the first integration by parts. When you integrate it up in the second integration, you get your original integral minus a constant. In the end, the two integrations by parts just amount to Michael’s observation that x/(x-1) = 1 + 1/(x-1)

    On the other hand, this paragraph reveals something to me that I have often thought but never vocalized — I always try to make rules to make integration more systematic, but there always seem to be exceptions. Like, “never do two successive integrations by parts, with the u from the second equal to the v from the first”. I still don’t recommend this for the above integral, but now I wonder if there is a more sophisticated problem where this actually would be the easiest way to find a solution.

    The above problem would not be put in the “integration by parts” section of any calculus book, but rather in the u-sub section. Instead of the algebraic simplification Michael suggested, one could also set u=(x-1), du=dx in the original integral to obtain the partially substituted integral:

    Integral x/u du

    Finish the substitution by observing that x=u+1, so we arrive at

    Integral (u+1)/u = Integral 1 + 1/u = u + ln|u| + C = x + ln|x-1| + C

    Part of the huge difficulty of integration is developing the intuition for which techniques are beneficial when, and this student demonstrates a need to further develop this intuition. But before you get to that, you need to master the actual execution of techniques. It is extremely difficult to develop higher-level understanding, like whether an integration by parts has actually made progress towards an answer, until one can almost always differentiate the u and integrate the dv correctly.

    The second integration in the student’s work shows that his/her focus should be on execution. In taking the derivative of u, he/she actually integrates it. Again, a common enough mistake with lots of “moving parts”, but the integral also shows the student assuming that Integral of 1/(something) = ln|something| holds generally.

    • Hao

      From what I remember of calc…, the typical examples for double integration by parts were trig expressions, and the second integration by parts got you something similar to the original expression (but usually of the opposite sign), which you could then combine and simplify. (e.g. sin -> cos -> -sin).

      However, the most recent time such a problem showed up for me was to integrate e^(ax) * sin(bx), and I ended up using the substitution sin(bx) = (e^(ibx) – e^(-ibx))/(2i) instead, which worked out pretty well. 😉

  • Owen Thomas

    i found this in the “squares instead of square roots” category.
    and have thus far skipped looking at the calculations.
    because this confuse-rooting-with-powering bug is *fierce*.
    and looks like it might be worth discussing.

    it might be some sort of locus classicus of some identifiable
    meta-bug… “avoid careful distinctions about vocabulary”
    or something. many beginners act like they believe
    a “definition” is pretty precisely the *opposite* of what-
    ever *i* seem to believe it is… or “should” be… and so
    will say that any-old-thing is true “by definition” with
    no hint what defining property of what allegedly-
    -defined object is being applied…
    anyhow.
    i’ll’ve encountered the root-and-powers thing
    hundreds of times by now. if time allows, i’ll
    usually try to get some kind of discussion
    going about it, too (“rooting and powering
    are a so-called “inverse pair” like adding-
    -and-subtracting or multiplying-and-dividing…
    and we can use ’em in much the same way
    when we need to ‘move things around’
    in equations; let’s look at some examples…”).
    in other than one-to-one… or, anyhow, small
    group… interactions, one can more or less
    count on having the target audience zone out
    pretty quick, though, alas.
    in some cases, now that i think about it, students
    will have been correctly informed that “rooting
    is just a certain kind of powering” (in some such
    language), just as “subtracting is a certain kind
    of adding”. but without the student’s will to look
    carefully into just *what* kind of “powering”,
    such philosophical remarks run the danger
    of being treated as somehow *better than
    the actual definition* of the concept at hand
    (because, say, allegedly closer to the “intuition”).
    this is *damage*, of course.
    and very like the “read for keywords” bug discussed
    around here on some other thread.
    keep ’em coming.