Improper Integral using Comparison to determine Convergence/Divergence

In summary, the problem is asking for a comparison to determine if the improper integral converges or diverges, and if it converges, to give an upper bound for the value. The suggested comparison is θ3 + θ > θ for θ > 1, and the goal is to manipulate the left side to be equal to the integrand 1/√(θ3 + θ) in order to apply the comparison theorem.
  • #1
MiDnite89
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Homework Statement



Use a comparison to determine if the improper integral converges or diverges. If the integral converges, give an upper bound for the value.

Integral of d(theta) / (theta^3 + theta)^1/2 from 1 to infinity

Homework Equations



N/A

The Attempt at a Solution



I'm not sure which function would be a good comparison to use to determine convergence or divergence. Earlier in the assignment, I ran across an equation that was dx/(9 - x^2)^1/2 which was arcsin(x/3), and since this is +, it'd be arccos - however that was with x^2 and this is theta^3, I'm not sure if that would be a direct comparison or not.

Any help would be appreciated, thanks.
 
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  • #2
∫ dx/√(9 - x2) is arcsin(x/3) + C, but a + between the two terms in the denominator would not involve arccosine in the integral but instead, arctangent. In fact, that integral could also be written as -arccos(x/3) + C. Most integrals involving arcsine or arccosine like this one can differ only by a sign and constant.
For your problem, you don't have to actually find the integral, just find another integral with which to compare it.

Start with the comparison θ3 + θ > θ (for θ > 1) and work with it until you can make the left side look like your integrand 1/√(θ3 + θ)
 
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FAQ: Improper Integral using Comparison to determine Convergence/Divergence

1. What is an improper integral?

An improper integral is an integral with one or both limits being infinite or the integrand having a discontinuity at some point in the interval of integration. It is used to find the area under a curve that is not bounded within a finite interval.

2. How is comparison used to determine the convergence or divergence of an improper integral?

Comparison is a method used to determine the convergence or divergence of an improper integral by comparing it to a known integral that is either convergent or divergent. If the unknown integral is smaller or equal to the known integral, it will have the same convergence or divergence behavior.

3. What criteria can be used to determine which integral to use for comparison?

When using comparison to determine the convergence or divergence of an improper integral, the integrands of the known and unknown integral must have the same behavior at infinity. This can be determined by looking at the highest power of x in the integrand or by taking the limit of the integrand as x approaches infinity.

4. Can comparison always be used to determine the convergence or divergence of an improper integral?

No, comparison can only be used to determine the convergence or divergence of an improper integral if the integrands of the known and unknown integral have the same behavior at infinity. If they do not, other methods such as the limit comparison test or the ratio test must be used.

5. Can the comparison method be used for integrals with infinite limits in both directions?

Yes, the comparison method can be used for integrals with infinite limits in both directions. In this case, two separate comparisons must be made, one for the integral with the upper limit being infinite and one for the integral with the lower limit being infinite.

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