University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Section 5.3 - The Definite Integral - Exercises: 54

Answer

$\frac{b^2}{4}+b$

Work Step by Step

$\int_{0}^{b} (\frac{x}{2}+1)dx=[\frac{x^2}{4}+x]_{0}^{b}=[\frac{b^2}{4}+b]-[\frac{0^2}{4}+0]=[\frac{b^2}{4}+b]-0=\frac{b^2}{4}+b$
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.