Answer
$2t^3-4t^2+7t+C$
Work Step by Step
$\int(6t^2-8t+7)dt$
$=\int 6t^2dt-\int 8tdt+\int 7dt$
$=6\int t^2dt-8\int t^1dt+7\int t^0dt$
$=6*\frac{t^3}{3}-8*\frac{t^2}{2}+7*\frac{t^1}{1}+C$
$=2t^3-4t^2+7t+C$
Calculus with Applications (10th Edition)
by Lial, Margaret L.; Greenwell, Raymond N.; Ritchey, Nathan P.
Published by
Pearson
ISBN 10:
0321749006
ISBN 13:
978-0-32174-900-0
Chapter 7 - Integration - 7.1 Antiderivatives - 7.1 Exercises - Page 366: 9
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