Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 7 - Techniques of Integration - 7.8 Improper Integrals - 7.8 Exercises: 3

Answer

T = $\infty$........... A =0.5

Work Step by Step

$\int1/x^{3}$ = $-x^{-2}/2$ = $-1/(2x^{2})$ $-1/(2*10^{2})$ - $-1/(2*1^{2})$ = 0.495 $-1/(2*100^{2})$ - $-1/(2*1^{2})$ = 0.49995 $-1/(2*1000^{2})$ - $-1/(2*1^{2})$ = 0.4999995 as x approaches infinity the number gets closer and closer to 0.5 meaning $\lim\limits_{x \to \infty}$ $-1/(2*x^{2})$ - $-1/(2*1^{2})$ = 0.5
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