College Math Teaching

December 13, 2012

Domains and Anti Derivatives (Indefinite Integration)

Grading student exams sometimes inspires me to revisit elementary topics. For example, I recently spoke about some unusual (but mostly correct) integration techniques used by students on a final exam.

I’ll recap (and adjust the example slightly): on a recent exam, a student encountered $\int \frac{2}{1-x^2} dx$. I had expected the student to use the usual partial fractions expansion to obtain $\int \frac{1}{1+x} dx + \int \frac{1}{1-x} dx = ln|1+x| - ln|1-x| + C$ which is valid when $x \ne \pm 1$. I admit to being a bad professor and not being picky about domains.

But one student noticed the $1 - x^2$ in the denominator of the fraction and so used the trig substitution $x = sin(\theta), dx = cos(\theta) d\theta$ which leads to the following integral: $\int \frac{2}{cos(\theta)} d\theta = 2ln|sec(\theta) + tan(\theta)| + C$ which leads to $2ln|\frac{1}{\sqrt{1-x^2}} + \frac{x}{\sqrt{1-x^2}}| + C = 2ln|\frac{1+x}{\sqrt{1-x^2}}| = ln|1+x| - ln|1-x| + C$ for $x \in (-1,1)$. Note that, strictly speaking, the “final answer” is really defined for all $x \ne \pm 1$ though the equalities do not hold outside of the domain for $x$ used in the original trig substitution.

And yes, I was a bad professor; I gave full credit to this answer even though we “lost domain” during the string of equalities.

But that got me to wondering: is there a trig substitution that works for $|x| > 1$? Answer: of course:

$\int \frac{2}{1-x^2} dx = -\int \frac{2}{x^2 -1} dx$. Now use $x = sec(\theta), dx = sec(\theta) tan(\theta) d\theta$ which leads to $-2\int csc(\theta) d\theta = 2ln|csc(\theta) + cot(\theta)| + C = 2ln|\frac{x}{\sqrt{x^2-1}} + \frac{1}{\sqrt{x^2 -1}}| + C$ which leads us to our ultimate solution for $|x| > 1$

So, if one REALLY wanted to use trig substituions for this problem, one could and do it in a way to cover the entire domain.

But…as our existence and uniqueness theorems imply, once we get a candidate for an anti-derivative that “works” or the domain, it really doesn’t matter if we did “illegal” steps to get it; we need only show that it is an anti derivative and is valid for the entire domain for the integrand.

Now if one wants a more detailed discussion on domain issues for anti-derivatives, I can recommend the article The Importance of Being Continuous by D. J. JEFFREY which appeared in Mathematics Magazine, Vol 67, pp 294 – 300. (reprint can be found here, scroll down a bit; this mathematician has written quite a bit!). Note: I can recommend this little paper as it talks about the domains of the anti derivatives themselves and not just the domains assumed in doing the calculations along the way or the domains of validity of the substitutions. Note: integral tables and computer algebra systems don’t always give the anti derivative with the “largest” possible domain. One has to watch for that.

December 1, 2012

One challenge of teaching “brief calculus” (“business calculus”, “applied calculus”, etc.)

Today’s exam covered elementary integrals and partial derivatives; in our course we usually mention two variable functions and show how to calculate some “easy” partial derivatives.

So today’s exam saw a D/F student show up late (as usual); keep in mind this is an 8 am class (no class prior to it). He, as usual, got little or nothing correct. Of course we had the usual $\int \frac{1}{x^2} dx = ln(x^x) + C, \int^1_0 3e^{5x}dx = (15e^5 -15) + C$, etc.

But there was this too: note that we had barely discussed partial derivatives and how to calculate them “by the formula”. But I did give the following bonus question: “is it possible to have a function $f(x,y)$ where $f_x = x^3 + y^3$ and $f_y = 3xy$? Yes, this is a common question in multivariable calculus (e. g., “is this vector field conservative?”) but remember this is a “brief calculus” course.

A few students took the challenge; some computed $\int(x^3 + y^3)dx = \frac{x^4}{4}+ xy^3 + C, \int (3xy^2)dy = \frac{3}{2}xy^2+C$ and noted that the two functions cannot be made to match (I didn’t expect them to recognize that functions of one variable alone represents constants of integration). Some took the second partials and noted $f_{xy} = 3y^2, f_{yx} = 3y$ and that these don’t match. Again, this was NOT a problem that we practiced.

Another instance: given the ideal gas law $PV = nRT$ I challenged them to show $\frac{\partial P}{\partial V}\frac{\partial V}{\partial T}\frac{\partial T}{\partial P} = -1$ and someone got it!

Bottom line: in one course, we have some bright, interested students who enjoy thinking and we have some who either don’t or can’t. This makes teaching difficult; if one tries to “teach to the mean” one is teaching to the empty set. It is almost: either bore half the class, or blow away half the class.