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: 9

Answer

$2t^3-4t^2+7t+C$

Work Step by Step

$\int(6t^2-8t+7)dt$ $=\int 6t^2dt-\int 8tdt+\int 7dt$ $=6\int t^2dt-8\int t^1dt+7\int t^0dt$ $=6*\frac{t^3}{3}-8*\frac{t^2}{2}+7*\frac{t^1}{1}+C$ $=2t^3-4t^2+7t+C$
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