Calculus: Early Transcendentals 8th Edition

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

Chapter 5 - Section 5.3 - The Fundamental Theorem of Calculus - 5.3 Exercises: 39

Answer

$= \frac{4\pi}{3}$

Work Step by Step

$\int^{\sqrt 3}_{\frac{1}{\sqrt 3}} \frac{8}{1+x^{2}} dx$ $= 8\int^{\sqrt 3}_{\frac{1}{\sqrt 3}} \frac{1}{1+x^{2}} dx$ $= 8 \arctan x |^{\sqrt 3}_{\frac{1}{\sqrt 3}}$ $= 8[ (\arctan \sqrt 3) - (\arctan \frac{1}{\sqrt 3})]$ $= 8[ (\frac{\pi}{3}) - (\frac{\pi}{6})]$ $= 8[ (\frac{2\pi}{6}) - (\frac{\pi}{6})]$ $= 8(\frac{\pi}{6})$ $= (\frac{8\pi}{6})$ $= \frac{4\pi}{3}$
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