Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 11 - Section 11.10 - Taylor and Maclaurin Series - 11.10 Exercises - Page 772: 60



Work Step by Step

$x^{2}e^{-x^{2}}=\Sigma_{n=0}^{\infty}\dfrac{(-1)^{n}x^{2n+2}}{(n)!}$ $\int_{0}^{0.5}x^{2}e^{-x^{2}}dx=\int_{0}^{0.5}\Sigma_{n=0}^{\infty}\dfrac{(-1)^{n}x^{2n+2}}{(n)!}$ Then $\approx \Sigma_{n=0}^{1}\dfrac{(-1)^{n}(0.5)^{2n+3}}{(2n+3)(n)!}$ $\approx 0.035$
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.