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: 37


$\sqrt {26}-\sqrt 5\approx2.863$

Work Step by Step

$\int_{2}^{5} \frac{x\cdot dx}{\sqrt {1+x^2}}=[\sqrt {1+x^2}]_{2}^{5}=[\sqrt {1+5^2}]-[\sqrt {1+2^2}]=\sqrt {26}-\sqrt 5\approx2.863$
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