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 321: 70

Answer

$y=\int_{1}^{x} \sqrt {1+t^2}dt -2$

Work Step by Step

We are given that $\dfrac{dy}{dx}=\sqrt {1+t^2}$ $\dfrac{d}{dx}[\int_{1}^{x} \sqrt {1+t^2} dt -2]=\sqrt {1+x^2}$ This implies that $y(1)=\int_{1}^{1} \sqrt {1+x^2} dx-2 =-2$ Hence, $y=\int_{1}^{x} \sqrt {1+t^2}dt -2$
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