# Improper Integral discontinuity

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• October 1st 2008, 03:39 PM
Retromingent
Improper Integral discontinuity
How is this improper? Plug in 1 and it's continuous. Plug in 3 and it's still continuous.

$\int^3_1 \displaystyle\frac{2} {(x-2)^{8/3}} dx$
• October 1st 2008, 03:42 PM
Krizalid
Having $1\le x\le3$ then $x\in[1,3],$ but in this interval, 2 lies there, and $1\le2\le3,$ so there's the discontinuity.