# Calculus II Professor will not accept my correct integral evaluation that uses a different method, should I bring this up further?

I am a freshman enrolled at an American University. Recently, I took an examination in which the following problem appeared:

Evaluate the following integral:

$$\int_0^4\sqrt{16-x^2}dx$$

My answer: 4$$\pi$$, was correct. However, I received reduced credit for this answer because I did not solve it correctly (according to the professor). The exams are time-limited and have a fair amount of content, so when I saw this problem, I noticed it was the equation of the top half of a circle centered at (0, 0) and with radius 4. Knowing this, and my knowledge of the integral indicating the signed area under a curve, I merely took the area of a quarter-circle of radius 4, $$\frac{1}{4}\pir^2$$ and wrote my answer of 4$$\pi$$.

The context of the test was surrounding our unit on inverse trigonometry and integration by parts. This section of the test did not list any other instructions besides evaluating the definite integrals. I've talked to my professor about it and his only response was that I solved it wrong:

To receive full credit, you would have had to evaluate an integral, as the instructions indicated.

Is my interpretation of evaluating the integral different? Does the instruction "Find the antiderivative and then evaluate" not need to exist for that to be required?

Thank you.

• you profesor in mixed up – Mikey Spivak Mar 15 at 0:23
• can you post the exact problem question? – Mikey Spivak Mar 15 at 0:27
• @MikeySpivak The question sheet is no longer in my possession, but I remember exactly what it said "Evaluate the following integrals:" (there were several). I wrote the integral exactly as it appeared. – user146073 Mar 15 at 0:31
• Did you obtain the correct answer via a logically sound argument? If so (as you did here), then you solved the problem. Of course, the professor can use whatever arbitrary grading metric he wants (and if he meant "evaluate the integral by computing an antiderivative for some reason, then he probably should have said so); but as far as the underlying math goes, recognizing that an integral corresponds to some other quantity that's easier to compute via a different method is completely legit. – anomaly Mar 15 at 0:56
• It's a common misconception, especially for U.S. freshmen, that they can ignore all the material taught in a course and pass exams with outside, prior knowledge. Related: academia.stackexchange.com/questions/80898/… – Daniel R. Collins Mar 15 at 4:29

I've done some undergraduate teaching and my policy is always if you get the correct answer by any means then you get full credit, but others have different policies and it's really up to them.

You could argue your case. Your professor could argue back that solving the integral by trig substitution does not require the formula $$A = \pi r^2$$, and he did not permit the use of that formula. He could argue that using that formula entails circular reasoning (the formula for the area of a circle has to be gotten by some limiting or integration method equivalent to evaluating $$\int_{-r}^r \sqrt{r^2-x^2}dx$$).

It could go either way for you. But I think it would be a waste of your and your professor's time.

• +1 for "circular reasoning". – JonathanZ Mar 15 at 1:49
• Got eem $\space$ – D_S Mar 15 at 2:12
• Thank you for the response, fair enough, I don't think I'll pursue further. This made a lot of sense. – user146073 Mar 15 at 2:15
• Even if you lost points, good on you for noticing a slick way to solve the problem instead of mindlessly grinding out a trig substitution – D_S Mar 15 at 2:20
• @user146073 I don't think this counts as circular reasoning. You are using a specialized version of a more general result. The general result is provable without the question in the OP. Even in undergraduate level, the area of a circle should be a well-established fact. If you want to really grind those points, see if your textbooks include an exercise where you prove the general result. – WorldSEnder Mar 15 at 5:30

An argument could be made that you should include a proof that the integral evaluates the area of a half-disk, rather than just asserting the answer.

Whether you “should” have gotten full points is more a matter of pedagogy than of mathematics, but as a practical tip: using (correct) method Y to solve a problem with instructions to use method X (especially in an intro class and when you are not familiar with the instructor and their teaching philosophy) is always a gamble.

Normally I would have just made a brief comment to the effect that I totally agree with you and this represents an extremely poor response by your professor, but this is simply too egregious to ignore. Assuming you are giving an accurate description of the events (yes, students sometimes embellish things, even lie) and assuming you are not omitting any relevant context (such as a note on the blackboard after the test began supplementing the directions, due to an oversight in the directions not realized until shortly after the tests were handed out --- I've done this several times), then I find this professor’s behavior very puzzling, even surprising, at least if you pointed this out shortly after receiving your graded test back.

Incidentally, if this is something you waited several hours, even days, before telling your professor about, then you have to realize that he/she might consider it possible that you copied the answer from another student's paper, and then later realized (or were told by someone else) how the evaluation could be done in without paper and pencil computations. In what follows, I'm going to assume that you brought this to the attention of your professor soon enough after the tests were handed out so that he/she could reasonably dismiss the possibility that you retroactively came up with an explanation for why no work was shown.

I find it surprising that praise and encouragement were not given for pointing out a neat way to evaluate the integral. I also find it surprising that the professor appears not to have anticipated this approach.

The onus for correctly writing questions and instructions lies with the professor, who has a much broader knowledge of the subject than his/her students, and who has far more experience in precise and correct and appropriate writing than his/her students. Surely the professor knows that many calculus texts and courses in the past 20-30 or so years have been emphasizing such methods (i.e. what you did) in evaluating definite integrals at the beginning of their study? Surely the professor knows that such methods were emphasized in the various calculus reforms that were all the rage in the 1990s (such as this) and whose innovations continue to be felt in present-day teaching? Surely the professor knows that during this time such methods for evaluating definite integrals have become increasingly taught in AP calculus classes and their presence on AP tests has increased?

Well, maybe not. But what bothers me even more is this apparent Dunning–Kruger effect on the part of someone who should know better. Namely, someone NOT aware of the things I’ve pointed out but who nevertheless thinks saying something like "To receive full credit, you would have had to evaluate an integral, as the instructions indicated" is an appropriate response to your method.

OK, enough of what we can’t fix. Let’s focus on what you should do on future tests in this class and in other classes. Unless you are absolutely sure that work is not being considered (such as a multiple choice question, but even these might be designed so that partial credit is possible), I would advise that you always give at least a brief explanation for answers. In fact, I consider this to be a useful skill to develop (i.e. something whose importance goes well beyond simply providing evidence that you did not copy from someone else’s test), even outside of mathematics, and as a (former) teacher this is something I often tried to encourage. For example, in the case of your integral evaluation, something like the following: The integral represents the first quadrant area under $$y = \sqrt{16 - x^2},$$ and hence is $$\frac{1}{4}$$ of the area of the circle $$y^2 = 16 – x^2$$ of radius $$4,$$ and thus is equal to $$\frac{1}{4}(\pi \cdot 4^2) = 4\pi.$$ Also, in the future if you think a certain method is clever and might perhaps be an unintended way to solve a problem, then (if you have time) I would advise giving the clever method as an alternative way of solving the problem, or (if you do not have time and the professor is proctoring the test) silently describe the alternative way to your professor and ask whether this would be sufficient work for the problem.

Regarding how to provide sufficient work in situations such as this, the various comments and answers to A student is cheating and I don't know how might give you some insight from a teacher’s point of view. Incidentally, here is the picture that for some reason was removed from the question, a picture that is needed to fully understand the comments and answers there. Maybe it was removed due to handwriting privacy issues, I don’t know.

I noticed it was the equation of the top half of a circle centered at (0, 0) and with radius 4. Knowing this, and my knowledge of the integral indicating the signed area under a curve, I merely took the area of a quarter-circle of radius 4, $$\frac{1}{4}\pir^2$$ and wrote my answer of 4$$\pi$$.

Did you write this clearly in your test (as you did here)? If not, it is fair to give reduced points. One should always explain where answers come from. If yes, proceed reading.

To receive full credit, you would have had to evaluate an integral, as the instructions indicated.

Do the instructions clearly disallow your solution? If not, it was not fair to you, and you should insist on it. If yes, read further.

Were these instructions available a priori, or they were included in the test itself? If available a priori, you should have complained about them before the test. If not, the instructions are unfair, and you should try to insist about it as well.

Also, recall that most of this is up to the professor, so you might be with bad luck, sadly.

As a student who has had a similar thing happen and heard of it happening to others, my personal recommendation would be to not bring it up again, it will probably be a losing battle. :(

But I would not consider your method "wrong" or "incorrect." There are many ways to solve a problem, you simply just made use of one of them, that wasn't the desired one.

As long as you explained how you came to your answer, the reason why your professor $$\textit{probably}$$ marked you down is that based on the class and section the test covered, the question was designed so that you would display and make use of your knowledge of trig substitution to solve the problem.

I would hope your professor didn't take too many points off (since the method does work), but in these classes you will usually be expected to give the professor a specific method that they are looking for.

I just wanted to add to the chorus of voices proclaiming that you're right and that your professor is completely wrong.

The evaluation of an integral should be taken as valid, as long as valid mathematical techniques are employed and shown to be employed. This holds as long as the question did not explicitly stipulate or forbid certain method(s). Just haing the test on a topic titled "Trigonometric integrals" is not sufficient. I'll expand on this point in a bit.

To see why this professor's "logic" leads to a slippery slope, consider this question:

"Evaluate the integral $$\displaystyle \int_0^{\frac{\pi}{2}} \sin^2 x dx$$".

Perhaps most students might evaluate this in the most obvious way, using a double angle cosine identity and plodding through.

But suppose a bright student does this:

"Let $$I = \displaystyle \int_0^{\frac{\pi}{2}} \sin^2 x dx$$.

By substituting $$y = \frac{\pi}{2} - x$$, we can see that $$I = \displaystyle \int_0^{\frac{\pi}{2}} \cos^2 y dy = \displaystyle \int_0^{\frac{\pi}{2}} \cos^2 x dx$$ since the variable of integration in a single variable definite integral is a dummy variable.

Hence $$2I = \displaystyle \int_0^{\frac{\pi}{2}} \sin^2 x + \cos^2 x dx = \int_0^{\frac{\pi}{2}} 1dx = \frac{\pi}{2}$$.

Therefore $$I = \frac{\pi}{4}$$."

Now this is a perfectly valid and highly elegant solution. It even uses "calculus-ey" working rather than a graphical shortcut. But something tells me your professor may not be the sort to accept this sort of creative solution, either, and his reasons for the rejection may be equally arbitrary.

We should never aim to discourage creativity in mathematics, only temper it with rigour.

Now, getting back to your situation, there are ways the professor could've avoided any ambiguity about how he wanted the question solved. For example:

"By using an appropriate trigonometric substitution, evaluate $$\displaystyle \int_0^4 \sqrt{16-x^2}dx$$",

in which case I would be on the side of your professor if you'd use the circular area to work out your integral.

Finally, a word of encouragement. As a fairly good student of math and physics, I've encountered rather poor behaviour from educators. I've had a math teacher marking my perfectly valid working wrong because of her own lack of knowledge. I've had a physics instructor ordering me to show my solution to an Olympiad prep question on the blackboard in front of the class, derisively refusing to accept my insistence that the question itself was incorrect, sending me back to my seat in embarrassment, and then later circulating a memo to the students in the class confirming exactly what I'd said (that the question was wrong), but not even acknowledging my existence, let alone rendering an apology for his conduct.

I've had some great teachers too, so it's not that I have a chip on my shoulders about teachers in general or anything. But we must acknowledge that teachers are human, with human foibles. And if we were to become teachers ourselves, we must try not to make the mistakes we've observed in others.

• Thumbs up for it all, especially the final two paragraphs – Allawonder 2 days ago

I suspect that the problem here is related to the fact that the context was:

"inverse trigonometry and integration by parts"

and that the professor thus wanted an answer that used methods relating to those. In this case, that would mean using a trigonometric substitution, such as $$x = 4 \sin u$$.

However, I would say that, although it is highly likely that this is what the professor in question wanted, he was not engaging in what I'd consider good practice by taking marks off an otherwise-valid answer because someone did not get his innuendo that these were the only acceptable methods. To me, if there is going to be an additional requirement as to the method of solution that would exclude other valid methods, the professor (and if I were one, I would) should state explicitly that that is what they want - i.e. "Solve the following integrals by trigonometric substitution". Such is even standard language in many textbook exercises, and there is no reason it could not be used on an exam. Moreover, this situation is made even worse by the fact of his response, which, if taken at its face value, should permit your solution because the "instructions" did not "indicate" that such a method was not allowed since they said nothing to prohibit it.

It is a matter of expectations and key concepts and methods discussed on the course (in lecture/seminar sessions). The instructor must have expected the demonstration of trigonometric substitution for full credit. For any other valid method, the mark is reduced.

In order to avoid such conflicts usually exam questions specify which method to use for full credit.

You are lucky to get this simple integral, however, your method might not work for harder integral and you need to show the knowledge and skill of using more universal method such as the trigonometric substitution.

You may want to ask model (detailed) answers/solutions from your instructor and discuss with other students before going to higher officers. Good luck.

Your method is completely correct then in my opinion! Perhaps you can explain to him that Bernhard Riemann defined the integral as a sequence of of rectangular areas, thus, your solution, if anything, is a more concise solution!

This is exactly what I think is great about math, regardless of who you are, if your solution is valid it doesn’t matter if Einstein himself walks out of his grave and tells you you’re wrong; your answer is still right!

It’s possible the professor just made a mistake or was tired, I’m sure that they would actually appreciate it if you can go their office hours and discussed it with him.

I love your solution! It reminds me of the following question. What is $${{\int}_0}^{2\pi}{\rm{cos}^2}x{\rm{d}}x$$? Answer: $$\pi$$. The reasoning is that as $${\rm{cos}^2}x + {\rm{sin}^2}x = 1$$, as the areas under $${\rm{sin}^2}x$$ and $${\rm{cos}^2}x$$ in the interval $$[0,2{\pi}]$$ are equal, and as $${{\int}_0}^{2\pi}{{\rm d}}x = 2{\pi}$$, we conclude that $${{\int}_0}^{2\pi}{cos^2}x{\rm{d}}x ={{\int}_0}^{2\pi}{sin^2}x{\rm{d}}x = {\pi}$$ The whole point of mathematics to me is beautiful pieces of inspiration like the one you had on your exam. You should get full credit.

• I don't think $\cos^2x+\cos^2x=1.$ Take $x=0,$ for example. – Allawonder 2 days ago
• Thanks! It was a typo. I meant to say "${\rm{cos}^2}x + {\rm{sin}^2}x = 1$", and corrected my post. Many thanks for pointing out my mistake. – student yesterday