'Rigor Mathis' - A Calculus Paradox or...

Update: I've added another 'paradox' in the comments. With the AP Calculus (BC) Exam looming, AP teachers may want to share this with their students for review.

[The following AP level question is designed for upper level students.]

Why do mathematicians have to be so rigid, um, I mean, rigorous?

Here's an AP Calculus problem brought to my attention this morning by one of our outstanding Calculus teachers, Mr. D. He found this in an AP Review book.

[Rather than play with the symbols, I'll 'write it out']:

The definite integral of sec²(x) from x = 0 to x = 3pi/4 is?
It was multiple choice and the answer given was -1.

I shared this problem with my AP group later on in the morning and I asked them why that answer makes no sense. R.J. immediately replied, "The answer can't be negative since sec²(x) is never negative."
Of course, but let's work it out!

[I intentionally did it incorrectly at first]:
By the Fundamental Theorem, the integral equals tan(3pi/4) - tan(0) = -1!

What's going on here! I was gratified that one of my students recognized that the function sec²(x) has an infinite discontinuity at x = pi/2, so the original integral is improper. When we integrate from 0 to pi/2, then from pi/2 to 3pi/4, and apply the rigorous limit definition of an improper integral, we see that the integral diverges! If anything, the 'area' is infinite or unbounded.

Using these kinds of 'paradoxical' examples and asking students to 'FIND THE ERROR' is a wonderful device many educators use to deepen student understanding of mathematics and demonstrate the need to be rigorous!

Now why isn't the definite integral of 1/x from -1 to 1 equal to zero, since the region in the first quadrant 'clearly cancels' the part in the 3rd quadrant?? Hmmm... I'll bet some of you could explain this and find many other such 'paradoxes'!