Calculus: Early Transcendentals 8th Edition

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

Chapter 5 - Review - Exercises: 17

Answer

$5.25$

Work Step by Step

$\int^1_{0}y(y^2+1)^5dy$ $=\frac{1}{2}\times\frac{1}{6}(y^2+1)^6$ $=\frac{1}{12}(y^2+1)^6$ $=[\frac{1}{12}(y^2+1)^6] ^1_{0}$. Sub in the bounds and subtract the lower bound (0) from the upper bound (1): $=[\frac{1}{12}((1)^2+1)^6]-[\frac{1}{12}((0)^2+1)^6]$ $=(\frac{1}{12}\times64)-\frac{1}{12}$ $=5.25$.
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