Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 5 - The Integral - Chapter Review Exercises - Page 278: 33


$$y= 2\sqrt{x}-1 $$

Work Step by Step

Given$$ \frac{d y}{d x}=x^{-1 / 2}, \quad y(1)=1$$ Since :\begin{align*} y&= \int (x^{-1 / 2}) dx\\ &= 2\sqrt{x} +c \end{align*} at $x=1,\ \ \ y= 1 $, then $c= -1$, hence $$y= 2\sqrt{x}-1 $$
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