University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Section 5.4 - The Fundamental Theorem of Calculus - Exercises - Page 320: 32



Work Step by Step

$\int_{0}^{1/\sqrt 3} \frac{dx}{1+4x^2}=\frac{1}{2}\cdot\int_{0}^{1/\sqrt 3} \frac{2dx}{1+(2x)^2}=\frac{1}{2}\cdot[arctan(2x)]_{0}^{1/\sqrt 3}=\frac{1}{2}\cdot[arctan(\frac{2\cdot1}{\sqrt 3})]-\frac{1}{2}\cdot[arctan({2\cdot0})]\approx0.4285-0\approx0.4285$
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