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Q. 2TF

Expert-verified

Calculus

Found in: Page 617

Book edition 1st

Author(s) Peter Kohn, Laura Taalman

Pages 1155 pages

ISBN 9781429241861

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Short Answer

Improper Integrals: Determine whether the following improper integrals converge or diverge.
$\int_{1}^{\infty} \frac{1}{x^{2}} d x .$

The series is a convergent series.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Given Information.

$G i v e n a n i n t e g r a l : \int_{1}^{\infty} \frac{1}{x^{2}} d x .$

Step 2. Solving the integral.

$\int_{1}^{\infty} \frac{1}{x^{2}} d x = {[\frac{- 1}{x}]}_{1}^{\infty} = [\frac{- 1}{\infty} - (\frac{- 1}{1})] = [0 + 1] = 1 . T h e r e f o r e, t h e s e r i e s i s a c o n v e r g e n t s e r i e s .$

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