University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Practice Exercises - Page 342: 36


See below.

Work Step by Step

We know that $\int x^{n} dx=\dfrac{x^{n+1}}{n+1}+C$ where $C$ is a constant of proportionality. $y=\int_{0}^{x} (1+2 \sqrt {\sec t}) dt$ or, $y'=1+2 \sqrt {\sec x}$ and $y''=2 (1/2) (\sec x)^{-1/2} (\sec x\tan x)=(\sec x)^{-1/2+1} \tan x=\sqrt {\sec x} \tan x$ Now, $y(0)=0$ and $y'(0)=1+2 \sqrt {\sec 0}=3$
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