How is this improper? Plug in 1 and it's continuous. Plug in 3 and it's still continuous.

$\displaystyle \int^3_1 \displaystyle\frac{2} {(x-2)^{8/3}} dx$

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- Oct 1st 2008, 03:39 PMRetromingentImproper Integral discontinuity
How is this improper? Plug in 1 and it's continuous. Plug in 3 and it's still continuous.

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