### Find the errors!

Using methods discussed in class this term, find the mean value over $[-\pi,\pi]$ of the function $f(x) = \sin(2x) \cdot \exp [1 - \cos (2x)]$.

The conceptual parts of the question are (based on the syllabus of the course)

1. Connecting “mean value of a continuous function over an interval” with “integration”, an application of calculus to probability theory and statistics.
2. Evaluating an integral by substitutions/change of variables.
3. Familiarity with the trigonometric functions $\sin, \cos$ and their properties (periodicity, derivative relations, etc).

I was told to grade with an emphasis on the above, so I prepared a grading rubric such that the above three key ideas gave most of the points. Here’s an otherwise reasonable answer that unfortunately does not use the methods discussed in class and so would receive (close to) zero credit (luckily no students turned in an answer like this):

The function $f(x)$ satisfies $f(x) = - f(-x)$, i.e. it is odd. So the average $(f(x) + f(-x))/2 = 0$. Since for every $x\in [-\pi,\pi]$, we also have $-x \in [-\pi,\pi]$, the mean value of $f(x)$ over that interval must be zero.

Here are some responses that can get quite a good number of points* (at least more than the above answer) based on the grading rubric (I guess it means I wasn’t imaginative enough in coming up with possible student errors). (I took the liberty of combining some of the most awful bits from different answers; the vast majority of the students’ answers are not nearly that horrible**, though only one student remembered that when changing variables one also needs to change the limits of integration.) Since most students who made any reasonable attempt on the question successfully wrote down the integral

$\displaystyle \mu = \frac{1}{\pi - (-\pi)} \int_{-\pi}^\pi f(y)~\mathrm{d}y$

(which is not to say no unreasonable attempts were made: just ask the poor bloke who decided that the Mean Value Theorem must play a role in this question), I will start from there. All mistakes below are intentional on my part. What amazed me most is how many students were able to get to the correct mean value…

Response 1
Integrate by substitution: we set $u = 1-\cos 2x$ then $\mathrm{d}u = 2\sin(2x) \mathrm{d}x$, so
$\displaystyle \mu = \frac{1}{2\pi} \int_{-\pi}^{\pi} 2 e^u \mathrm{d}u = \frac{1}{\pi} \frac{e^{u+1}}{u+1} |_{-\pi}^\pi$.
Plugging back in we have
$\displaystyle \mu = \frac{1}{\pi} \left( \frac{e^{2 - \cos 2\pi}}{2 - \cos 2\pi} - \frac{e^{2 - \cos -2\pi}}{2 - \cos -2\pi} \right)$.
Since $\cos 2\pi = \cos -2\pi = \cos \pi = -1$, we have that $\mu = \frac{1}{3\pi} ( e^3 - e^3) = 0$.

Response 2
Integrate by substitution: we set $u = \exp (1 - \cos 2x)$ then $\mathrm{d}u = \sin(2x) \mathrm{d}x$. So
$\displaystyle \mu = \frac{1}{2\pi} \int_{-\pi}^\pi u \mathrm{d}u = \frac{1}{4\pi} u^2 |^{\pi}_{-\pi} = \frac{1}{4\pi}(\pi^2 - \pi^2) = 0$

Response 3 (this one receives fewer points because it didn’t integrate by substitution)
By the product rule of integration***
$\displaystyle \int_{-\pi}^\pi \sin(2x) e^{1-\cos 2x} \mathrm{d}x = \int_{-\pi}^\pi \sin(2x)\mathrm{d}x \int_{-\pi}^\pi e^{1-\cos 2x} \mathrm{d}x$
We evaluate the first factor
$\displaystyle \int_{-\pi}^\pi \sin(2x)\mathrm{d}x = \cos(2x) |_{-\pi}^\pi = 0$
so $\mu = 0$.

Response 4 (getting even more ridiculous)
Since $\frac{\mathrm{d}}{\mathrm{d}x} e^{g(x)} = g'(x) e^{g(x)}$, by the rules of integration,
$\displaystyle \int_{-\pi}^\pi \sin(2x) e^{1-\cos 2x} \mathrm{d}x = \frac{\sin 2x}{x - \frac12 \sin 2x} e^{1 - \cos 2x} |_{-\pi}^\pi$.
Since $\sin(2\pi) = \sin(-2\pi) = 0$ we have that $\mu = 0 - 0 = 0$.

* A side effect of modern exam grading which emphasize fairness is the “no double jeopardy rule”. That is, if a mistake is made in one computation, and if the remainder of the computation is correct assuming the mistake, no more points should be deducted. Hence something like in response 2 would only get deducted some small amount of points for incorrectly computing $\mathrm{d}u$ and some small amount for writing down the wrong integration limits. The cynical side of me wonders if some students are not in fact exploiting this rule intentionally to great effect.

** In fact, somewhere around 10 or 15 percent of the students did the problem perfectly (as long as we overlook the problem with limits of integration), with another 10 or 15 percent missing out on “perfect” only because they did not remember the values of $\cos(2\pi)$ and $\cos(-2\pi)$ and didn’t want to risk guessing incorrectly.

*** The so called “product rule of integration” appeared uncharacteristically often during the midterm exam. Since clearly the lecturer would not have mentioned such nonsense, the fact that so many students “learned” this rule seems to indicate that when students study together, factoids can be easily mistaken for facts in a sort of crowd mentality. That this appears again on the final exam despite the lecturer having most likely clarified the issue in class maybe due to the “cheat sheet” system. For a lot of classes offered to non-majors, the instructors allow the students to prepare for themselves one or several sheets of paper, which can contain useful information and formulae, so long as the sheets are hand-written. The intention I assume is that the process of producing a cheat sheet would help with rote learning, since it involve looking up relevant information and copying it down by hand. The downside, however, is that once a mistake makes its way onto the cheat sheet, the student probably won’t remember to remove it even after the mistake was pointed out to him…