Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 6 - Section 6.1 - Areas Between Curves - 6.1 Exercises - Page 434: 15

Answer

$6\sqrt3$

Work Step by Step

$\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}(8cos(x)-sec^2(x))dx= [8\sin(x)-\tan(x)]_{-\frac{\pi}{3}}^{\frac{\pi}{3}}=6\sqrt3$
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