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: 44



Work Step by Step

$\int^{\sqrt 2}_{0}(t-\sqrt {2})dt=\int^{\sqrt 2}_{0}tdt- \sqrt {2}\int^{\sqrt 2} _{0}dt$ $=\frac{(\sqrt {2})^{2}}{2}-\sqrt {2}(\sqrt {2}-0)= \frac{2}{2}-2= -1$
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