Calculus (3rd Edition)

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

Chapter 5 - The Integral - 5.3 The Indefinite Integral - Exercises - Page 252: 15


$\int(18t^5 -10t^4 -28t)dt= 3t^6-2t^5 -14t^2+C$

Work Step by Step

$\int(18t^5 -10t^4 -28t)dt= 18\times\frac{1}{6}t^6 - 10\times\frac{1}{5}t^5 -28\times \frac{1}{2}t^2 +C= 3t^6-2t^5 -14t^2+C$
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