Thomas' Calculus 13th Edition

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

Chapter 4: Applications of Derivatives - Section 4.7 - Antiderivatives - Exercises 4.7 - Page 240: 73



Work Step by Step

The anti-derivative for $\dfrac{dy}{dx}=\dfrac{1}{x^2}+x$ is $y=\dfrac{-1}{x}+\dfrac{x^2}{2}+C$ Apply Initial conditions $y(2)=1$.. This implies $\dfrac{-1}{2}+\dfrac{4}{2}+C=1$ or, $ C=-\dfrac{1}{2}$ Hence, $y=\dfrac{-1}{x}+\dfrac{x^2}{2}-\dfrac{1}{2}$
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.