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Q. 2TF

Expert-verifiedFound in: Page 617

Book edition
1st

Author(s)
Peter Kohn, Laura Taalman

Pages
1155 pages

ISBN
9781429241861

Improper Integrals: Determine whether the following improper integrals converge or diverge.

${\int}_{1}^{\infty}\frac{1}{{x}^{2}}dx.$

The series is a convergent series.

$Givenanintegral:{\int}_{1}^{\infty}\frac{1}{{x}^{2}}dx.$

${\int}_{1}^{\infty}\frac{1}{{x}^{2}}dx\phantom{\rule{0ex}{0ex}}={\left[\frac{-1}{x}\right]}_{1}^{\infty}\phantom{\rule{0ex}{0ex}}=\left[\frac{-1}{\infty}-\left(\frac{-1}{1}\right)\right]\phantom{\rule{0ex}{0ex}}=\left[0+1\right]=1.\phantom{\rule{0ex}{0ex}}Therefore,theseriesisaconvergentseries.$

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