# Determine if the improper integral converges or diverges

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?

## The Attempt at a Solution

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jbunniii
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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?
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).

jbunniii
Homework Helper
Gold Member
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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jbunniii