Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 5: Integrals - Section 5.4 - The Fundamental Theorem of Calculus - Exercises 5.4 - Page 287: 60


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

Work Step by Step

Given: $\dfrac{dy}{dx}=\sqrt {1+t^2}$ $\implies \dfrac{d}{dx} [\int_{1}^{x} \sqrt {1+t^2} dt] -2=\sqrt {1+t^2}$ $\implies y(1)=\int_{1}^{1} \sqrt {1+t^2} dt $ $\implies y(1)=-2$ Hence, $y=\int_{1}^{x} \sqrt {1+t^2}dt -2$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.