Calculus 8th Edition

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

Chapter 5 - Applications of Integration - 5.5 Average Value of a Function - 5.5 Exercises - Page 391: 5



Work Step by Step

Given $$f(t)=t^2(1+t^3)^4,\ \ \ [0,2] $$ Then \begin{align*} f_{\text{ave}}&=\frac{1}{b-a}\int_{a}^{b}f(t)dt \\ &=\frac{1}{2-0}\int_{0}^{2}t^2(1+t^3)^4dt\\ &=\frac{1}{6} \int_{0}^{2}3t^2(1+t^3)^4dt\\ &=\frac{1}{30}(1+t^3)^{5}\bigg|_{0}^{2}\\ &=\frac{1}{30}(59049-1)\\ &=\frac{59048}{30} \end{align*}
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.