Hi,

$\displaystyle \int_{-1}^8 \frac{1}{x^{2/3}} \, dx$

How is this even possible since the integrand isn't even defined for values less than 0?...

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- Oct 11th 2011, 09:55 AMsjmillerImproper Integral
Hi,

$\displaystyle \int_{-1}^8 \frac{1}{x^{2/3}} \, dx$

How is this even possible since the integrand isn't even defined for values less than 0?... - Oct 11th 2011, 10:03 AMPlatoRe: Improper Integral