Answer
$\dfrac{t^3}{6}+t^4+C$
Work Step by Step
Since, we have $\int (\dfrac{t^2}{2}+4t^3) dt=\dfrac{t^3}{6}+t^4+C$
and
$\dfrac{d}{dt}(\dfrac{t^3}{6}+t^4+C)=\dfrac{t^2}{2}+4t^3$
University Calculus: Early Transcendentals (3rd Edition)
by Hass, Joel R.; Weir, Maurice D.; Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0321999584
ISBN 13:
978-0-32199-958-0
Chapter 4 - Section 4.8 - Antiderivatives - Exercises - Page 272: 28
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