Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 5 - The Integral - 5.4 The Fundamental Theorem of Calculus, Part I - Exercises - Page 257: 8



Work Step by Step

$\int^{2}_{-3}u^{2}du= [\frac{u^{3}}{3}]^{2}_{-3}= \frac{2^{3}}{3}-\frac{-3^{3}}{3}=\frac{35}{3}$
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