5
$\begingroup$

We have theoretical questions in our exams (often in ABCD format). For example, I can state the question:

For an indefinite integral, does it hold that $\int f(g(x)) \cdot g'(x) \, dx = \int f(x) dx $?

or should I state the question:

Let $f, g: \mathbb{R} \rightarrow \mathbb{R}$ be two strictly monotonic differentiable functions with continuous derivatives. Does it hold that $\int f(g(x)) \cdot g'(x) \, dx = \int f(x) dx $?

Surely the 2nd one is more correct. But it gets even worse with 2-variable calculus, where one often has to put a bunch of conditions on the most basic theorems. I feel that exams get bloated with conditions, but on the other hand, we should make the questions precise if we grade them. What is your opinion on this?

$\endgroup$
14
  • 14
    $\begingroup$ Why are you choosing those conditions in the second statement? It is false both ways, and adding/removing conditions does not change the fact that it is false. If you changed the integrals to definite integrals (from $a$ to $b$ and $g(a)$ to $g(b)$ respectively) it would be true as long as $f$ is continuous and $g$ is continuously differentiable, with no monotonicity assumptions. $\endgroup$ Aug 24, 2023 at 8:43
  • 5
    $\begingroup$ If multiple choice questions (I'm guessing this is what you mean by "ABCD format"), then you could simply ask something like the following: "Given suitable assumptions on the functions $f$ and $g,$ which of the following equalities holds?" Incidentally, I don't see any reason you need to write the words "indefinite integral". In fact, your wording might not be technically correct since there are two indefinite integrals that follow "For an indefinite integral", unless by "indefinite integral" the reference is to the operation itself, in which case it should be "For the indefinite integral". $\endgroup$ Aug 24, 2023 at 9:30
  • 2
    $\begingroup$ Also, too many conditions obscure the content of the theorem and can confuse students who, otherwise, have a decent working knowledge of what is going on. $\endgroup$
    – Adam
    Aug 24, 2023 at 12:19
  • 12
    $\begingroup$ Yes, be rigorous. If you're sloppy, how are you going to teach the students to be rigorous? Also, I agree with JochenGlueck and StevenGubkin. If I was a student and I was faced with a question like your indefinite integral, and asked simply for true or false, I wouldn't be able to answer, because I wouldn't be able to mindread the teacher and guess what unspoken assumptions they had and what the integral bounds were on this indefinite integral. In general, if the students need mindreading to understand the exam questions, then the exam questions are bad. $\endgroup$
    – Stef
    Aug 24, 2023 at 18:21
  • 2
    $\begingroup$ @JakeB. This indicates, I think, a major lack of calculus knowledge on your part. Try $f(x) = x$ and $g(x) = 2x$ for example. $\int f(g(x))g'(x) dx = \int 4x dx = 2x^2 + C$. $\int f(x) dx = \int x dx = \frac{1}{2}x^2 + C$. These indefinite integrals are not equal. I am concerned about your ability to teach a calculus class. The theorem you are trying to quote only applies to definite integrals. $\endgroup$ Aug 24, 2023 at 22:10

1 Answer 1

12
$\begingroup$

On the one hand, the assumptions are an indispensable part of any theorem and many mistakes (well beyond the student homework) have been made exactly because somebody didn't understand that the conclusion of a certain result holds only under certain conditions which might be well-hidden in the sentences "assuming that $f$, $g$ are as in Lemma 5". So remembering formulae without clear understanding what classes of functions they can be applied to is not enough.

On the other hand, a lot of times we start teaching results in simple situations that are very far from full generality in which they really hold, so it is important to emphasize that some conditions are imposed just because of the students' limited knowledge, not because they are really needed.

The best practice is to bite a bit from both ends, i.e., to state and prove the theorem under assumptions that are easy to understand and check, but to show some really pathological counterexamples too so that the students would know that without assuming anything at all things can go awry.

If it were a decent analysis course, I would phrase your particular test question as follows.

Assuming that $f$ and $g$ are defined on the entire real line, $g$ is non-decreasing, and $a,b\in\mathbb R$, the change of variable formula $$ \int_a^b f(g(x))g'(x)dx=\int_{g(a)}^{g(b)} f(y)dy $$ holds for the Riemann integral

A) For arbitrary locally Riemann integrable $f$ and differentiable $g$

B) For arbitrary continuous $f$ and continuously differentiable $g$

C) For arbitrary locally Riemann integrable $f$ and differentiable $g$ with locally Riemann integrable derivative.

D) For arbitrary locally Riemann integrable $f$ and continuously differentiable $g$

Choose the most general option that is correct.

In calculus you, probably, do not want to go that far, but you can make a similar list with options that are more obviously true or false.

$\endgroup$
7
  • $\begingroup$ I am embarrassed to say that I am not sure whether $(C)$ is correct. I cannot think of a counterexample, but a proof seems tough. My usual tricks for proving different variants of this theorem are not strong enough. Care to enlighten me? $\endgroup$ Aug 25, 2023 at 11:31
  • $\begingroup$ @StevenGubkin Note that I imposed the condition that $g$ is non-decreasing, so $g'$ is non-negative. Now, $f$ (as a Riemann integrable function) can be squeezed between two piecewise constant functions with almost the same Riemann integral, so it is enough to check that both sides make sense and are equal when $f$ is a characteristic function of an interval. I'll let you take it from there :-) $\endgroup$
    – fedja
    Aug 25, 2023 at 17:17
  • $\begingroup$ Oh, I missed your non-decreasing condition. I can certainly prove it in this case. Do you have a proof or counterexample of (C) without this condition? $\endgroup$ Aug 25, 2023 at 18:14
  • 1
    $\begingroup$ Too difficult. Probably in the lecture some theorem was shown, but not the most general one. The test shd check that the student can see why if one of the conditions is dropped (not just weakened) it wouldnt work anymore. Especially if time allocated for multiple choice is tight (usually less than 2 Minutes, often less than 1minute) $\endgroup$
    – lalala
    Aug 25, 2023 at 19:39
  • 1
    $\begingroup$ @fedja I posted an MSE question: math.stackexchange.com/q/4759152/34287 $\endgroup$ Aug 26, 2023 at 22:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.