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 - Page 309: 54



Work Step by Step

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