Determine if the improper integral converges or diverges

In summary, the integral from 1 to 3 of the function 2/(x-2)^(8/3) diverges because as x approaches 2, the denominator becomes 0 and the function goes to infinity. Whether the integral converges or diverges at a singularity depends on the "width" of the singularity, with wider singularities leading to divergence and narrower ones leading to convergence.
  • #1
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?


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  • #2
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##?
 
  • #3
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 ?
 
  • #4
Solve it from 1 to t, t to 3 and do the limit as t approaches 2 from the right and left.
 
  • #5
i got the answer -12/5. Since its negative does that means that it diverges?
 
  • #6
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).
 
  • #7
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.
 
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  • #8
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.
 

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 function being integrated is not defined at one or more points within the interval of integration.

How do I determine if an improper integral converges or diverges?

To determine if an improper integral converges or diverges, you can use one of the following methods: 1) Use the limit comparison test, 2) Use the direct comparison test, 3) Use the integral test, 4) Use the p-series test, or 5) Use the alternating series test.

What is the limit comparison test?

The limit comparison test is a method used to determine if an improper integral converges or diverges by comparing it to a known convergent or divergent integral using the limit of the ratio of the two integrals.

What is the direct comparison test?

The direct comparison test is a method used to determine if an improper integral converges or diverges by comparing it to a known convergent or divergent integral, and then using the comparison to make a conclusion about the original integral.

What is the integral test?

The integral test is a method used to determine if an improper integral converges or diverges by comparing it to a corresponding infinite series and using the limit of the partial sums of the series to make a conclusion about the integral.

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