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: 46



Work Step by Step

$\int^{0}_{3}(2z-3)dz=2\int^{0}_{3}zdz-3\int^{0}_{3}dz $ $=2(\frac{0^{2}}{2}-\frac{3^{2}}{2})-3(0-3)$ $=-9+9=0$
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