Determine if the improper integral converges or diverges

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Homework Help Overview

The discussion revolves around determining the convergence or divergence of the improper integral of the function \( \frac{2}{(x-2)^{8/3}} \) over the interval from 1 to 3. Participants are exploring the implications of discontinuities and the behavior of the function near the singularity at \( x=2 \).

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are questioning the nature of the discontinuity at \( x=2 \) and whether the absence of discontinuities in the endpoints implies divergence. There is discussion about how to approach the integral by splitting it at the point of discontinuity and considering limits.

Discussion Status

Some participants have suggested methods for evaluating the integral by considering limits as they approach the discontinuity. There is an ongoing exploration of whether the integral converges or diverges based on the behavior of the function near the singularity.

Contextual Notes

Participants are discussing the implications of the function approaching infinity at \( x=2 \) and how this affects the overall convergence of the integral. There is a mention of the need to verify calculations and the importance of understanding the nature of singularities in integrals.

physics=world
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1. integrate from (1 to 3) of function (2) / (x-2)^(8/3)

Can someone explain why this diverges. i do not understand it. when i plugged in the numbers there are no discontinuities and this is where i am stuck at. If there are no discontinuity does that means that it diverges?


Homework Equations


The Attempt at a Solution

 
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physics=world said:
1. integrate from (1 to 3) of function (2) / (x-2)^(8/3)

Can someone explain why this diverges. i do not understand it. when i plugged in the numbers there are no discontinuities and this is where i am stuck at. If there are no discontinuity does that means that it diverges?
What happens at ##x=2##?
 
oh. it would equal to zero. so does that mean that it is continuous on the interval [1,3] except at 2? if so, do i proceed with solving it from 1,2 to 2,3 ?
 
Solve it from 1 to t, t to 3 and do the limit as t approaches 2 from the right and left.
 
i got the answer -12/5. Since its negative does that means that it diverges?
 
Graph the function in your head...as it approaches 2 the denominator (x-2) term goes to zero, so the function goes to infinity. Hence the area under the curve also goes to infininity (diverges).
 
lordsurya08 said:
Graph the function in your head...as it approaches 2 the denominator (x-2) term goes to zero, so the function goes to infinity. Hence the area under the curve also goes to infininity (diverges).
That's not automatically true. For example, ##1/|x|^{1/2}## diverges to infinity as ##x \rightarrow 0##, but the function has a finite (improper) integral over any finite-length interval even if the interval includes 0.

In general, whether the integral diverges or not at a singularity depends on how "wide" the singularity is: the integral of ##1/|x|^p## over an interval including 0 will converge or diverge depending on the value of ##p##. Larger ##p## = wider singularity.
 
Last edited:
physics=world said:
i got the answer -12/5. Since its negative does that means that it diverges?
If you got a finite answer (positive or negative), that would mean the integral converges. However, please check your work or post it here. -12/5 is incorrect.
 

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