University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 8 - Section 8.7 - Improper Integrals - Exercises - Page 471: 15


$\sqrt 3$

Work Step by Step

Re-arrange the given integral as follows: $\int_{0}^{1} \dfrac{\theta +1}{\sqrt {\theta^2+2\theta}}=\lim\limits_{p \to 0^{+}}\int_{p}^{1} \dfrac{\theta +1}{\sqrt {\theta^2+2\theta}}$ ....(1) Plug $u(\theta)=\theta^2+2\theta \implies \dfrac{du}{2}=(\theta+1) d\theta$ and $u(p)=p^2+2b; u(1)=3$ Then, we have $\lim\limits_{p \to 0^{+}}(\dfrac{1}{2}) \int_{p^2+2p}^{3} \dfrac{du}{\sqrt u} d\theta=[\sqrt u]_{p^2+2p}^{3} $ Thus, equation (1) becomes $\lim\limits_{p \to 0^{+}}[\sqrt 3-\sqrt{p^2+2p}]=\sqrt 3$
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