Answer
$0.008969$
Work Step by Step
$\int_{0}^{0.3}\frac{x^{2}}{1+x^{4}}dx=\int_{0}^{0.3}\frac{x^{2}}{1-(-x^{4})}dx$
$=\int_{0}^{0.3}x^{2}\Sigma_{0}^{\infty}(-1)^{n}(x^{4n})dx$
$=\frac{(0.3)^{3}}{3}-\frac{(0.3)^{7}}{7}+...$
$=0.008969$
Multivariable Calculus, 7th Edition
by Stewart, James
Published by
Brooks Cole
ISBN 10:
0-53849-787-4
ISBN 13:
978-0-53849-787-9
Chapter 11 - Infinite Sequences and Series - 11.9 Exercises - Page 776: 32
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