University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 4 - Practice Exercises - Page 278: 104


$\sqrt {7+\theta^2}+c$

Work Step by Step

Since, we know $\int x^{n} dx=\dfrac{x^{n+1}}{n+1}+c$ where $c$ is a constant of proportionality. Substitute $k=7+\theta^2 \implies 2\theta d\theta =dk$ Thus, $\dfrac{1}{2} \int \dfrac{1}{\sqrt k} dk=\dfrac{1}{2} \int k^{-1/2} dk$ or, $=\dfrac{1}{2} [\dfrac{k^{-1/2+1}}{-1/2+1}]+c$ or, $=k^{1/2}+c$ plug back in: $k=7+\theta^2 $, we have $=\sqrt {7+\theta^2}+c$
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