University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Section 9.2 - Infinite Series - Exercises - Page 499: 92


$8 m^2$

Work Step by Step

The area, A, can be found as: $A=2^2+(\sqrt 2)^2+(1)^2+(\dfrac{1}{\sqrt 2})^2+....=4+2+1+\dfrac{1}{2}$ or, $A=\dfrac{4}{1-\dfrac{1}{2}}=8 m^2$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.