# Determine if the improper integral converges or diverges

by physics=world
Tags: converges, determine, diverges, improper, integral
 P: 104 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? 2. Relevant equations 3. The attempt at a solution
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P: 2,933
 Quote by physics=world 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##?
 P: 104 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 ?
P: 542

## Determine if the improper integral converges or diverges

Solve it from 1 to t, t to 3 and do the limit as t approaches 2 from the right and left.
 P: 104 i got the answer -12/5. Since its negative does that means that it diverges?
 P: 4 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).
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P: 2,933
 Quote by lordsurya08 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.