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I've TAed and tutored calculus for years and of the hundreds of students I've interacted with, it is always a shock when I tell them to change the limits of integration when they do substitutions. When I tell them to do that, they are always confused and act like they never heard about that. I remember being taught that. I might think that one or two teachers might not cover it but it's standard in calculus. Even students in Calculus 3 are shocked.

Why do students not know this?

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    $\begingroup$ @DanChristensen Link to what? $\endgroup$ – user5108 Feb 9 '18 at 3:47
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    $\begingroup$ An example of "changing the limits of integration when doing substitutions." I can only guess at what you are talking about, but I suspect it is a problem of notation. $\endgroup$ – Dan Christensen Feb 9 '18 at 4:21
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    $\begingroup$ @DanChristensen: For examples, see OpenStax Calculus I, under "Substitution for Definite Integrals": cnx.org/contents/i4nRcikn@2.72:wxH1chTc@2/Substitution $\endgroup$ – Daniel R. Collins Feb 9 '18 at 4:30
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    $\begingroup$ Not an answer, but I suspect it's because they're just following the steps of a recipe and don't have the foggiest iota of understanding as to the significance of any of the steps. $\endgroup$ – shoover Feb 9 '18 at 16:51
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    $\begingroup$ More on topic for a comment on MESE, this question could be improved by modifying it to ask something like "How can we teach students how to do integration by substitution with definite integrals so that they will reliably match the limits to the variable?" This would allow for either of the methods in Mike Pierce's answer below. $\endgroup$ – shoover Feb 9 '18 at 16:56
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Some students are instead taught to change the substitution variable back into the original variable before evaluating the antiderivative at the bounds.

$$\int\limits_0^2 2x\cos(x^2) \;\mathrm{d}x = \int\limits_{x=0}^{x=2} \cos(u) \;\mathrm{d}u = \sin(u) \Big|_{x=0}^{x=2} = \sin(x^2) \Big|_{x=0}^{x=2} = \sin(4) \\\text{versus}\\ \int\limits_0^2 2x\cos(x^2) \;\mathrm{d}x = \int\limits_{u=0}^{u=4} \cos(u) \;\mathrm{d}u = \sin(u) \Big|_{u=0}^{u=4} = \sin(4) $$

I recall doing it the former way the first time I was taught calculus, so I'm not sure how standard changing the bounds is. I think whether or not an instructor does this or instead changes the bounds is just a matter of preference (although I do worry that instructors gloss over the fact that if you don't change the bounds that they are still $x$-bounds, and so students don't really appreciate that and just change the variable back to $x$ because that's the procedure they've been taught). Now whether it is better to teach one or the other, or whether we should teach both, may be an interesting discussion.

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    $\begingroup$ Perhaps a better way of writing the former is to find an indefinite integral (substituting, then bringing back $x$; so that no bounds appear at all), and then separately substitute the $x$-bounds back into that. This appears in one example of Stein/Barcellos, before making the simultaneous bounds-change is introduced. $\endgroup$ – Daniel R. Collins Feb 9 '18 at 4:26
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    $\begingroup$ @DanielR.Collins Organizing the calculation that way would worry me a bit. Writing down the former method that way, I can easily see myself, after figuring out the antiderivative, forgetting that I was originally looking at an indefinite integral. I think the $x = \dots$ bounds are nice (even if they are a bit awkward looking) because they serve as a reminder to that you're dealing with an indefinite integral, and a reminder that you need to make sure the variables in your bound and integrand match. $\endgroup$ – Mike Pierce Feb 9 '18 at 4:48
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    $\begingroup$ I always show both, but I strongly advocate changing the bounds. The reason is that if you do multiple substitutions in the same problem, you will end up with a ton of algebra (easy to make mistakes) when converting everything back to the original variable at the end. Changing the bounds as you go can save some real headaches. $\endgroup$ – Steven Gubkin Feb 9 '18 at 14:48
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    $\begingroup$ For definite integrals worked in class, where you have a full blackboard to work with and also in the case of elementary calculus (rather than an advanced undergraduate course), I usually found the indefinite integral in terms of the original variable and then plugged in the limits. However, for formal writing (solutions to tests, publications, internet posts, etc.), I often find it best to explicitly indicate the variable that the constants substitute for, like Mike Pierce has done. $\endgroup$ – Dave L Renfro Feb 9 '18 at 16:15
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    $\begingroup$ Changing the limits of integration is definitely part of the high school AP Calculus curriculum. $\endgroup$ – gparyani Feb 10 '18 at 17:48
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Zach:

It is a standard part of how the subject is taught and is in every textbook that I have seen. It's also an easy thing to miss. Unless the students are well drilled, they won't internalize it.

Instead of being exasperated, just see it as one of those little things that people need reminding of. (Look at first year rookies in the NFL who forget you have to be "touched down" unlike the college game. Even though told it, they don't always internalize it.)

Also, realize that you as a TA are in the right end of the bell curve of calculus prowess. Your trainees will generally not be as strong as you are. Have some empathy and just help them get better. Don't be surprised they are not already better.

Nothing has changed. It was always a tricky thing and will always be. Every new batch of students brings new ignorance. That is how teaching and coaching are.

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    $\begingroup$ "Every new batch of students brings new ignorance": I wonder how much the overall dissatisfaction teachers seem to have toward their students is explained by the fact we feel as if we always had the same students, and thus depressed by seeing freshmen not progress from year to year. $\endgroup$ – Benoît Kloeckner Feb 9 '18 at 12:32
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    $\begingroup$ @BenoîtKloeckner: Alternatively, it might be the experience that the more the teacher works/simplifies things, the corresponding less effort students put into school work: mitpressjournals.org/doi/abs/10.1162/rest_a_00013 $\endgroup$ – Daniel R. Collins Feb 9 '18 at 22:22
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    $\begingroup$ @DanielR.Collins: the paper is behind a paywall (and some people still wonder why OA matters...), but do I understand correctly that trying too hard as a teacher might makes thing simpler for student, who would thus try less hard, and would thus learn less? $\endgroup$ – Benoît Kloeckner Feb 10 '18 at 13:29
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    $\begingroup$ @BenoîtKloeckner: Yes, that's one of the findings: "the role of the school effort... affects negatively the effort exerted by children but not that exerted by parents". Based on NCDS data in the UK, N ~ 12K. $\endgroup$ – Daniel R. Collins Feb 10 '18 at 15:36
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    $\begingroup$ "Instead of being exasperated" was this really a necessary comment? $\endgroup$ – user5108 Feb 10 '18 at 17:26
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My guess is this may be due to practicing indefinite integration before definite integration, where such a thing does not come up. I imagine most people learn definite integration as "indefinite integration, followed by plugging in numbers", which lacks a substitution step.

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  • $\begingroup$ I think you are right. I always found indefinite integrals to be confusing and unsatisfactory. $\endgroup$ – copper.hat Feb 10 '18 at 2:38
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At my institution (and I think this is common in semester systems, including the calculus AP curriculum), substitution of definite integrals is the very last subject of the first semester of calculus, so it's both the thing most likely to get shortchanged if the course runs behind schedule, and there's just less time to practice. When students pick up in the next semester, they usually jump in with "methods of integration"---integration by parts, trig substitution, and partial fractions---which reinforce the method of u-substitution, but are less likely to reinforce thinking about the bounds in definite integrals.

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  • $\begingroup$ I think this is quite plausibly a factor. It made me realize I favored indefinite integrals on quizzes etc. because I can reasonably expect students to do more indefinite integrals in a timed setting. $\endgroup$ – pjs36 Feb 10 '18 at 0:35
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I suspect that students don't learn to change bounds of integration well because they have no motivation to do so. Whether we agree with them or not, it is evidently simpler for them conceptually to evaluate a definite integral by first evaluating the corresponding indefinite integral and then applying the fundamental theorem to that result.

As a teacher, there are a couple of things that I do to combat this. First, I devise problems where the whole point of the problem is to change bounds of integration in the first place. For example:

Express $\displaystyle \int_{-2}^{3} f(2x+1)\,dx$ as a definite integral involving $f(x)$.

Or,

Evaluate $\displaystyle \int_0^{\pi} \sin\left(e^{\sin(x)}\right)\cos(x)\,dx$.

Second, I try to impress upon them that there are plenty of practical situations where this translation from one definite integral to another arises. For example, when exploring probability theory (a natural application of integration in Calc II), we often need to translate a normal integral to a standard normal integral. This amounts to showing that $$\int_a^b e^{-(x-\mu)^2/(2\sigma^2)} dx = \int_{(a+\mu)/\sigma}^{(b+\mu)/\sigma} e^{-x^2/2} dx.$$ Thus, a rule that many of them learn in elementary statistics arises as a change of bounds in a definite integral.

As another example, that's nice if you do numeric integration, you might show that $$\int_0^1 \frac{\sin(1/x)}{x} \, dx = \int_1^{\infty} \frac{\sin(x)}{x} \, dx.$$ The point is that the second integral is more palatable from the point of view of numerical integration (rather than from the point of view of symbolic integration, as calc students are accustomed to). The reason is that the higher derivatives of the integrand don't explode over the domain of integration the way they do for the first. You can even illustrate this using a little sage code:

print numerical_integral(sin(1/x)/x, 0,1)
print numerical_integral(sin(x)/x, 1,1000)

# Out: 
# (-1.9426263726635902, 42.80850345186512)
# (0.6241500516015874, 5.034861416675085e-14)

The first computation is completely wrong and has an absurd error bound. The second computation is (not surprisingly) correct to three decimal places.

Note that, in all four of these examples, the integrand cannot be integrated symbolically; a change of bounds is the whole point.

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  • $\begingroup$ How do your students respond to that second example of evaluating $\int_0^{\pi} \sin\left(e^{\sin(x)}\right)\cos(x)\,\mathrm{d}x$? Or maybe more importantly how do you expect them to respond? The students I've TAed (intro to calculus) wouldn't have a clue how to handle that correctly. $\endgroup$ – Mike Pierce Feb 12 '18 at 1:22
  • $\begingroup$ @MikePierce Well, you need to expose them to similar examples. When I teach integration, I spend plenty of time on geometry. They know that the integral of an odd function over an interval centered on the origin is zero. They know that the integral of a constant is the length of the interval times the constant. They most certainly know that $$\int_0^0\sin(e^u)du=0,$$ which is what that integral transforms into. When I teach Calc II, more than half would get that right. But then, I make them change bounds of integration from day one. $\endgroup$ – Mark McClure Feb 12 '18 at 1:43
  • $\begingroup$ Making that substitution $u$ for $\sin(x)$ doesn't seem valid since $\sin(x)$ isn't injective on $(0,\pi)$. I don't think you can just "fold" the domain of integration like that. Here's a silly example: $$\text{Letting } u = x^2,\quad \int_{-1}^1 x^2 \;\mathrm{d}x \;=\; 2\!\int_1^1 \sqrt{u} \;\mathrm{d}u \;=\;0\,.$$ It only works out to be true for $\int_0^{\pi} \sin\left(e^{\sin(x)}\right)\cos(x)\,\mathrm{d}x$ because of the symmetry of the function about $\pi/2$. Doing the substitution $u = x-\pi/2$, the integrand becomes and odd function. $\endgroup$ – Mike Pierce Feb 12 '18 at 2:03
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    $\begingroup$ @MikePierce Thanks for the fun example! In your translation, though, I suspect you replaced $x$ with $\sqrt{u}$, which is valid for only half of the interval. No such mistake is necessary for my integral. Regardless, in the identity $$\int_{\varphi(a)}^{\varphi(b)}f(x)dx = \int_a^b f(\varphi(t))\varphi'(t) dt,$$ there is no assumption of monotonicity on $\varphi$. $\endgroup$ – Mark McClure Feb 12 '18 at 2:30
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    $\begingroup$ It's killing me that you're right, because my intuition is still screaming otherwise. I think my intuition based only on how I originally learned to evaluate integrals via substitution is a bit off then, because I'm struggling to assimilate the statement and conditions of that identity into how I currently think about "doing $u$-substitution". $\endgroup$ – Mike Pierce Feb 12 '18 at 3:10
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Following on Mike's example, you might insist on explicitly specifying the variables being integrated over in the limits as well as the substitution used:

$\int\limits_{x=0}^{x=2} 2x\cos(x^2) \;\mathrm{d}x = \int\limits_{u=0}^{u=4} \cos(u) \;\mathrm{d}u = \sin(u) \Big|_{u=0}^{u=4} = \sin(4) \space\space$ where $u=x^2, \space du=2x\space dx$

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I'm surprised this hasn't been mentioned yet, but it seems likely that the OPs experience is due in part to selection bias. That is: the students that one encounters when tutoring, or in office hours, are more likely to be precisely the students that are struggling to understand something. There are probably lots and lots of students who are learning this skill when it is taught to them; you aren't seeing them, because they don't need to come to you for help.

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I found that it was safer for weaker students to evaluate the indefinite interval, back substitute, and the evaluate the limits once. Multiple evaluations just gave them additional additional steps in which to make errors and didn't really improve their somewhat limited grasp of what it all meant.

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    $\begingroup$ As in another comment of mine, this appears in one example of Stein/Barcellos (immediately prior to showing the simultaneous change in bounds technique), and it look rigorous to my eye. $\endgroup$ – Daniel R. Collins Feb 10 '18 at 23:02

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