Nope, you fell for the intuitive trap! Infinity is not a concrete number, and "one infinity" may not equal the "other" < for example x - x^2 , lim x --> infinity = -infinity>. **Since** sine is cyclic, how do you know where to terminate the point of integration? Is sin(infinity) =0, 1, -1, 1/sqrt[2], ...?

You *don't* terminate it and you *don't* evaluate it at infinity. The total change of the sin x function *is* infinity. Good reasoning, bad wording. That's what I get for trying to do calculus when I'm playing WoW.

You think the integral of sin(x) from -infinity to infinity is 0, or am I misinterpreting you? By total change you mean absolute value?

This is exactly the intuitive trap I set. The curve |sin(x)| is mostly above the x axis, but is always either above or touching the x axis. Also sin(x) is a periodic function, (e.g. it does not decrease in magnitude at an increasing rate as we approach infinity). Therefore logically, and (rigorously enough IMO although some pure mathematicians would disagree), the integral of |sin(x)| from -infinity to infinity is infinite.

Now here is your

**intuitive error**:

*by removing the absolute value sign, we have an equal number of identically shaped curve segments above and below the x axis from -infinity to infinity* (false premise), therefore

the integral is zero.

Since you're approaching infinity, you are not terminating your evaluation at any specific x value. Since sin(x) (and cos(x)) do not converge to any point as x --> infinity, you do not know precisely what you are integrating, but you do know that there may not be an equal area to the right of the y axis as there is to the left of it, and that the net area on either side of the y axis may not be zero.

Here is a slightly more rudimentary question that may help you see my point:

What is the integral of sin(x) from x = 0 to infinity?

To be more explicit, Integral[sin(x)] (-infinity,infinity) = -cos(infinity) + cos(infinity) = ? + ? = undefined, since trig function(infinity) does not converge to any specific point.

Learn to use MathType ...

LaTeX is so much nicer, and regular text is so much lazier

.

This is why intuition must be so carefully applied. If we establish using the FTC is not acceptable, then we cannot use it to justify a non-rigorous answer. There is no obvious reason that the integral should converge to 2 other than 1.999999999999999999999999. If FTC is the reason, then FTC should be explained, and the Riemann sum calculations would not be valid support.

Are you suggesting that the teacher expects students in Algebra II to use something taught to Calculus students half way through the year? To me, it seems that the Reimann Sum solution is much more reasonable, especially since using FTC2 can be used to prove it (without actually proving why FTC2 is true).

I expect that he should be able to derive the tools necessary to solve the problem exactly, using riemann sums. It is not a small task for someone who has just taken algebra, although it is very simple and short in my opinion, and does not use any mathematics he is unfamiliar with. If he doesn't want to start from just algebra though, taking the definition of a derivative (which is really easy to show) as a given, makes the whole problem very easy.

He could simply calculate a few Riemann sums and prove that his estimation is better as the number of Riemann sums used increases, but just saying it obviously converges to 2 isn't good math or good logic.