Some Special Madness

Our second post on the 2017 VCE exam madness concerns a question on the first Specialist Mathematics exam. Typically Specialist exams, particularly the first exams, don’t go too far off the rails; it’s usually more “meh” than madness. (Not that “meh” is an overwhelming endorsement of what is nominally a special mathematics subject.) This year, however, the Specialist exams have some notably Methodsy bits. The following nonsense was pointed out to us by John, a friend and colleague.

The final question, Question 10, on the first Specialist exam concerns the function \boldsymbol{f(x) = \sqrt{\arccos(x/2)}}, on its maximal domain [-2,2]. In part (c), students are asked to determine the volume of the solid of revolution formed when the region under the graph of f is rotated around the x-axis. This leads to the integral

    \[V \ = \ \pi \int\limits_{-2}^{2}  \arccos(x/2)\, {\rm d}x\,.\]

Students don’t have their stupifying CAS machines in this first exam, so how to do the integral? It is natural to consider integration by parts, but unfortunately this standard and powerful technique is no longer part of the VCE curriculum. (Why not? You’ll have to ask the clowns at ACARA and the VCAA.)

No matter. The VCAA examiners love to have the students to go through a faux-parts computation. So, in part (a) of the question, students are asked to check the derivative of \boldsymbol{x\arccos(x/a)}. Setting a = 2 in the resulting equation, this gives

    \[ \frac{{\rm d}\phantom{x}}{{\rm d}{x}}\left(x\arccos(x/2)\right)= \arccos(x/2) - \dfrac{x}{\sqrt{4-x^2}}\,.\]

We can now integrate and rearrange, giving

    \[ \aligned V \ &= \ \pi\left\Big[\!x\arccos(x/2)\!\!\right\Big]\limits_{-2}^{2} \quad +\quad \pi \int\limits_{-2}^{2} \dfrac{x}{\sqrt{4-x^2}}\, {\rm d}x\\[2\jot] \ &= \ 2\pi^2\quad +\quad \pi \int\limits_{-2}^{2} \dfrac{x}{\sqrt{4-x^2}}\, {\rm d}x\,.\endaligned\]

So, all that remains is to do that last integral, and … uh oh.

It is easy to integrate \boldsymbol{x/\sqrt{4-x^2}} indefinitely by substitution, but the problem is that our definite(ish) integral is improper at both endpoints. And, unfortunately, improper integrals are not part of the VCE curriculum. (Why not? You’ll have to ask the clowns at ACARA and the VCAA.) Moreover, even if improper integrals were available, the double improperness is fiddly: we are not permitted to simply integrate from some –b to b and then let b tend to 2.

So, what is a Specialist student to do? One can hope to argue that the integral is zero by odd symmetry, but the improperness is again an issue. As an example indicating the difficulty, the integral \boldsymbol{\int\limits_{-2}^2 x/(4-x^2)\,{\rm d}x} is not equal to 0. (The TI Inspire falsely computes the integral to be 0, which is less than inspiring.) Any argument which arrives at the answer 0 for integrating \boldsymbol{x/(4-x^2)} is invalid, and is thus prima facie invalid for integrating \boldsymbol{x/\sqrt{4-x^2}} as well.

Now, in fact \boldsymbol{\int\limits_{-2}^2 x/\sqrt{4-x^2}\,{\rm d}x} is equal to zero, and so \boldsymbol{V = 2\pi^2}. In particular, it is possible to argue that the fatal problem with \boldsymbol{x/(4-x^2)} does not occur for our integral, and so both the substitution and symmetry approaches can be made to work. The argument, however, is subtle, well beyond what is expected in a Specialist course.

Note also that this improperness could have been avoided, with no harm to the question, simply by taking the original domain to be, for example, [-1,1]. Which was exactly the approach taken on Question 5 of the 2017 Northern Hemisphere Specialist Exam 1. God knows why it wasn’t done here, but it wasn’t and the consequently the examiners have trouble ahead.

The blunt fact is, Specialist students cannot validly compute \boldsymbol{\int\limits_{-2}^2 x/\sqrt{4-x^2}\,{\rm d}x} with any technique they would have seen in a standard Specialist class. They must either argue incompletely by symmetry or ride roughshod over the improperness. The Examiners’ Report will be a while coming out, though presumably the examiners will accept either argument. But here is a safe prediction: the Report will either contain mealy-mouthed nonsense or blatant mathematical falsehoods. The only alternative is for the examiners to make a clear admission that they stuffed up. Which won’t happen.

Finally, the irony. Look again at the original integral for V. Though this integral arose in the calculation of a volume, it can still be interpreted as the area under the graph of the function y = arccos(x/2):

But now we can consider the corresponding area under the inverse function y = 2cos(x):

It follows that

    \[V \ = \  \pi \int\limits_{-2}^{2}  \arccos(x/2)\, {\rm d}x \ = \ \pi \int\limits_{0}^{\pi} \left[  2\cos(x) - (-2)\right]  \, {\rm d}x \ = \ 2\pi^2\,.\]

Done.

This inverse function trick is standard for Specialist (and Methods) students, and so the students can readily calculate the volume V in this manner. True, reinterpreting the integral for V as an area is a sharp conceptual shift, but with appropriate wording it could have made for a very good Specialist question.

In summary, the Specialist Examiners guided the students to calculate V with a jerry-built technique, leading to an integral that the students cannot validly compute, all the while avoiding a simpler approach well within the students’ grasp. Well played, Examiners, well played.