Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 7 - Integration - 7.1 Antiderivatives - 7.1 Exercises - Page 366: 5

Answer

$6k+C$

Work Step by Step

Since $k^0=1$, we can rewrite $6$ as $6k^0$ and use the Power Rule: $\int 6dk$ $=\int 6k^0dk$ $=\frac{6k^1}{1}+C$ $=6k+C$
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