Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 5: Integrals - Practice Exercises - Page 308: 67


a) b ; b) b

Work Step by Step

a. The average value is defined as: $\dfrac{1}{2} \int_{-1}^1 (mx+b ) dx$ $\dfrac{1}{2} [(m)\dfrac{x^{1+1}}{1+1}+bx]_{-1}^1=\dfrac{1}{2} [m(\dfrac{1}{2} -\dfrac{1}{2} )+b(1+1)]=b$ b. The average value is defined as $\dfrac{1}{2k} \int_{-k}^k (mx+b ) dx$ Then, $\dfrac{1}{2k} [(m)\dfrac{x^{1+1}}{1+1}+bx]_{-k}^k1=\dfrac{1}{2k} [m(\dfrac{k}{2}-\dfrac{k}{2})+b(k+k)]=(\dfrac{1}{2k}) (2bk)=b$
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.