improper integrals, convergent or divergent? please help!

Limits, differentiation, related rates, integration, trig integrals, etc.
mike09
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improper integrals, convergent or divergent? please help!

Hi, I'm having a terribly difficult time deciphering whether an improper integral is divergent or convergent.

From what I notice, is it divergent whenever the answer is infinity? and convergent whenever the answer equals a specific number?

I just got an answer that is:

the limit as a approaches negative infinity -1/2a^2 + the limit as b approaches infinity 1/2b^2

is the answer infinity because if you plug any number into a it gets infinitely smaller and if you plug any number into b it gets infinitely bigger? is that explanation correct?

I also just got another answer that is:
the limit as b approaches negative infinity (1/2-1/2e^2b)=1/2 ............how does 1/2e^2b as b is approaching negative infinty equal zero for that to be true? I don't see that at all. Thanks in advance for any help.

nona.m.nona
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Re: improper integrals, convergent or divergent? please help

...is it divergent whenever the answer is infinity? and convergent whenever the answer equals a specific number?
To be convergent, an integral must have a finite value. If the value is not finite, then the integral, by definition, does not converge.
the limit as a approaches negative infinity -1/2a^2 + the limit as b approaches infinity 1/2b^2
Your formatting is unclear. Is the sum either of the following?

$\mbox{1: }\, \lim_{a \rightarrow -\infty}\, -\frac{1}{2a^2}\, +\, \lim_{b \rightarrow \infty}\, \frac{1}{2b^2}$

$\mbox{2: }\, \lim_{a \rightarrow -\infty}\, -\frac{1}{2}a^2\, +\, \lim_{b \rightarrow \infty}\, \frac{1}{2}b^2$

I suspect the former is intended.

Are you given any relationship between "a" and "b"?
the limit as b approaches negative infinity (1/2-1/2e^2b)=1/2 ............how does 1/2e^2b as b is approaching negative infinty equal zero for that to be true?
Is the expression meant to be as follows?

$\lim_{b \rightarrow -\infty}\, \left(\frac{1}{2}\, -\, \frac{1}{2e^2 b}\right)$

If so, then: What expression did you obtain after converting to the common denominator, combining, and then dividing the numerator and denominator by e2b?

mike09
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Joined: Mon Oct 15, 2012 9:32 pm
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Re: improper integrals, convergent or divergent? please help

hey sorry about the formatting, I was able to figure it out though today in class. Appreciate ya trying to help.

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