# Improper integrals and convergence/divergence, help please

I thought I understood what I was doing, but my last homework grade says otherwise. It was a simple assignment, just check what would prove a given integral converges or diverges, or be able to tell from a graph.
(just work with me on the notation here, hard to get right on the board)

Here’s one question -

``````

∞
∫f(x) dx
3

``````

Which of the following can be used to prove [emphasis theirs] that the integral diverges?

f(x) approaches 0 as x approaches infinity.
f(x) approaches 5 as x approaches infinity.
For all b ≥ 3, [the same integral above with b substituted for infinity] = ln|b| - ln(3)
For all x ≥ 3, 0 < f(x) < 1/(sqrt(x))
none of these

Now, the way I understand it, the first two can’t be used to prove anything, neither convergence nor divergence of the integral (though, again, from how I understand it, it would prove the integral diverged if f(x) went to infinity at x = 5 or some other constant though I’m not wholly sure). This would be due to the fact that say, 1/x and 1/x^3 both act the same way near infinity (that is to say, they’re both zero), but the integral of one converges and the other diverges.

The next two confuse me. I figured that for option 3, since the lim (b -> infinity) would come out to infinity, it doesn’t converge to a definite point => divergence.

And for option 4 I thought did nothing because divergence is proven by comparison if 0 <= g(x) <= f(x), where g(x) diverges (whereas convergence is proven by 0 <= f(x) <= g(x), where g(x) converges). Where did my thinking go wrong on these options?

I can probably figure out the other one (with convergence on a different integral) I missed with the help from this question (if not I’ll post it). But need help on the graph:

http://img230.imageshack.us/img230/6006/53284.jpg

The question is basically, is the area between 1/x^2 and 1/x [though they could have been any two arbitrary functions, labeled or not, so I need to be able to figure it out from the graph, not the function itself, which is what I did to get it right] infinite or finite? And then a bunch of do h/k/g/f(x) converge, diverge, or is it impossible to tell? I was totally lost on this one and mostly just guessed.
[[And sorry, I’m tacking on a mini-rant because I can]]
Sorry for unloading basic math questions, I have an instructor who I’m sure knows what he’s doing… but he certainly can’t teach it (and, sadly, I’m not a very good book learner, I need the instructor portion). For reference, on the last test I got a 65/100 and it was still the fourth highest grade in a class of 35 people (I wouldn’t complain if this was an upper-division course, but it’s a 100 level course). Last semester I never got below an 87. Oh, and he cribbed questions from other instructor’s tests, so it wasn’t just a hard test. And he takes off half or so of the points for writing down a four instead of a two in your final answer because you brain-stupided for a second, even if you got the actual concept and calculus right. Luckily I think I have the current section (areas and volumes of shapes/solids by integral) down, hopefully.

The first choice is a necessary condition for convergence, but it is not sufficient. It doesn’t prove anything.
f(x) must approach 0 as x approaches infinity in order for that improper integral to converge so the second choice indeed proves that it diverges.
The third choice, as b goes to infinity, ln(b) - ln(3) goes to infinity, so the integral diverges.
I’m not quite sure what you mean by your last question, so I’ll leave you with this:

integral of 1/x^p converges for p>1

The area between 1/x^2 and 1/x is infinite. The reason is, as gladtobeglazed pointed out, that the area under 1/x is infinite and the area under 1/x^2 is finite. Taking away a finite area from an infinite area will leave an infinite area.

This tells you nothing. Both f(x) = 1/x and f(x) = 1/x^2 satisfy this condition, but one converges and the other diverges.

Yes, 1 and 4 give no information, while 2 and 3 guarantee divergence. For the last two, the reason is that the integral doesn’t approach a limit (more precisely, goes to infinity).