Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 4: Applications of Derivatives - Section 4.7 - Antiderivatives - Exercises 4.7 - Page 239: 35

Answer

$-2 \sin t +C$

Work Step by Step

Simpify. $\int (-2 \cos t) dt$ Thus, $\int (-2 \cos t) dt=-2 \sin t +C$ Also, $\dfrac{d}{dt}(-2 \sin t +C)=-2 \cos t$
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