Improper integral, divergence/convergence

In summary, the conversation is about using the comparison theorem to evaluate if the integral \int^\infty_2 \frac{dx}{\sqrt{x^3+1}} diverges or converges. The person asking for help has tried using simpler functions as a comparison but is unsure if their reasoning is correct. They also mention that they have had trouble with similar attempts. Another person suggests using x^{-3/2} as a comparison, which the original person agrees with and expresses relief that they didn't make a mistake.
  • #1
usn7564
63
0

Homework Statement


Evaluate if the integral diverges or converges using the comparison theorem.

[tex]\int^ \infty_2 \frac{dx}{\sqrt{x^3+1}}[/tex]

Having trouble with this question, the exercises I have managed I generally guessed if it was convergent/divergent, and then found a smaller of bigger function at the interval and went by that. Here I assumed it was convergent and used [tex]\frac{1}{\sqrt{x^2+1}}[/tex], though that was divergent which says absolutely nothing. Ran into the same issue with other attempts.

Is my reasoning the issue or have I just not found the correct manipulation? Not really approaching with anything more involved than "Find something bigger and hope it converges".
Any help would be appreciated, thanks.
 
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  • #2
It seems to me that a simpler comparison is [itex]x^3< x^3+ 1[/itex] so that [itex]\frac{1}{\sqrt{x^3+ 1}}< \frac{1}{\sqrt{x^3}}= x^{-3/2}[/itex].
 
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  • #3
I feel slightly moronic now, was too focused on manipulating the x term for whatever reason.

Thanks, glad I didn't get it all backwards at least.
 

1. What is an improper integral?

An improper integral is an integral where one or both of the limits of integration are infinite or where the integrand function is unbounded over the interval of integration.

2. How do you determine if an improper integral converges or diverges?

To determine if an improper integral converges or diverges, you can use one of several convergence tests such as the Comparison Test, Limit Comparison Test, or the Integral Test.

3. What is the difference between a convergent and divergent improper integral?

A convergent improper integral is one where the limit of the integral exists and is a finite value. A divergent improper integral is one where the limit does not exist or approaches infinity.

4. Can an improper integral converge even if the integrand is unbounded?

Yes, an improper integral can still converge even if the integrand function is unbounded. This can happen if the unbounded portion of the integrand has a small enough contribution to the overall integral.

5. How can we use improper integrals to solve real-world problems?

Improper integrals can be used to solve various real-world problems such as finding the area under a curve, calculating the volume of a solid with a curved boundary, and determining the average value of a function over an infinite interval.

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