# Homework Help: Calculus II improper integrals

1. Aug 18, 2011

### GreenPrint

Hi,

I was wondering if it was really necessary to evaluate improper integrals with limits? Could anyone really say I was wrong if I did something like

find the area bounded by the region y=1/x^2, x=2, and the x-axis

integral[1,inf] dx/x^2 = (-1/x)|[2,inf] = (-0)-(-1/2)=1/2

Like I don't really see the need to remove infinity from my work and plug in some variable and evaluate using limits when I can just plug in infinity if that makes sense and was wondering if anyone would consider my work to be wrong because my text book insists that I evaluate improper integrals with limits but I see no real reason to

also if there was a problem were limits were necessary and I would get a undefined answer I evaluate improper integrals with like 3^+ or 3^- to indicate I was evaluating the integral from the right of 3 or to the left of 3

Last edited: Aug 18, 2011
2. Aug 18, 2011

### dynamicsolo

If you were in an upper-level undergraduate or graduate physics course, you might get away with writing things that way (physicists write improper integration like that all the time). But you should be aware that it is a "shorthand" for indicating the limit. "Infinity" is not a number*, and one attempts to treat it as one at their peril. In improper integration of many kinds of functions, simply "plugging in infinity" will leave you with an expression you won't begin to know how to approach without keeping in mind where the limits are operating.

I can tell you that most graders would probably dock you a point or two on an exam if you wrote your evaluation of the integral that way. What you do in your own personal work is, of course, up to you...

*EDIT: Well, yeah, I mean in standard analysis; after all, in projective geometry, they put a point "at infinity" and make it a location...

Last edited: Aug 18, 2011
3. Aug 18, 2011

### GreenPrint

Hm interesting thanks for the advice. So I take it if I "plugged in" inf^-1 instead it would still be considered a improper way of evaluating a integral as well because it's not a number neither...

I never thought of it that way, infinity is not a number, so you must integrate evaluating limits...

would one however say it's improper to integrate from like c^- or c^+ were c is a actual number, like for example 1 or 2?

4. Aug 18, 2011

### vela

Staff Emeritus
Yes, in the sense that that's how the improper integral is defined. I think your textbook, by requiring you to write out the steps explicitly, is simply trying to pound it into you that this is what it really means when you have an infinite limit.

The question of whether you will always need to write those steps out depends on your graders. For your current course, it would probably be safest to always write it out because it's one of the concepts you're learning in the course. In later courses, it's probably fine to omit those steps. Not writing the limit explicitly is pretty common, so it's arguable whether a grader would ding you for doing that. Simply put, you need to know your audience. In other words, just ask the grader if he or she wants to see those steps or not.

5. Aug 18, 2011

### obafgkmrns

You might be interested in looking into "non-standard analysis". It sets forth a rigorous approach to infinitesimals and their inverses without introducing limits. The blanket statement that infinity is not a number is not quite correct in that context. Wikipedia has a tolerably correct description.

6. Aug 18, 2011

### GreenPrint

is this notation ok though

integral[a^+/-,b^+/-] f(x)dx

were a and b are actual numbers and not infinity, is this short hand acceptable, were a^- is a from the left, and a^+ is a from the right?

7. Aug 18, 2011

### GreenPrint

hmmm interesting... I'll look into that topic

8. Aug 18, 2011

### Staff: Mentor

No, it's not. Here's an integral similar to what you're indicating above.
$$\int_0^1 \frac{dx}{x(x - 1)}$$

This integral is improper, because the integrand function is undefined at both limits of integration. The right way to do this is to break the integral into two improper integrals, like so:
$$\lim_{a \to 0^+}\int_a^{.5} \frac{dx}{x(x - 1)} + \lim_{b \to 1^-}\int_{.5}^b \frac{dx}{x(x - 1)}$$

At this point, you find an antiderivative for each integral, and evaluate it at the two endpoints, and then take the limit of each.

BTW, I chose .5 arbitrarily. Any number (strictly) between 0 and 1 would do.

9. Aug 18, 2011

### dynamicsolo

Of course, to manage that requires some significant changes in the way the structure of number sets are viewed, since the hyperreals are a bit more involved than just the real number line with infinitesimals jammed in everywhere. I suspect we couldn't have gotten to that idea without passing through consideration of a rigorous definition for limits first...

BTW, obafgkmrns, have you considered appending the brown-dwarf spectral types to your name?

10. Aug 18, 2011

### obafgkmrns

"BTW, obafgkmrns, have you considered appending the brown-dwarf spectral types to your name?"