Calculus 8th Edition

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

Chapter 11 - Infinite Sequences and Series - 11.9 Representations of Functions as Power Series - 11.9 Exercises - Page 798: 32



Work Step by Step

$\int_{0}^{0.3}\frac{x^{2}}{1+x^{4}}dx=\int_{0}^{0.3}\frac{x^{2}}{1-(-x^{4})}dx$ $=\int_{0}^{0.3}x^{2}\Sigma_{0}^{\infty}(-1)^{n}(x^{4n})dx$ $=\frac{(0.3)^{3}}{3}-\frac{(0.3)^{7}}{7}+...$ $=0.008969$
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