Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 7 - Section 7.8 - Improper Integrals - 7.8 Exercises - Page 534: 25

Answer

$2$ (converges)

Work Step by Step

$\displaystyle \int e^{-\sqrt{y}}dy=\left[\begin{array}{ll} u=y^{1/2} & \\ du=\frac{1}{2}y^{-1/2}du & dy=2udu \end{array}\right]=\displaystyle \int e^{-u}(2udu)$ Table integral 96: $\displaystyle \int ue^{au}du=\frac{1}{a^{2}}(au-1)e^{au}+C$ $2\displaystyle \int ue^{-u}du=2(u-1)e^{u}=2(\sqrt{y}-1)e^{-\sqrt{y}}$ The integral is a type I improper integral, so $I=\displaystyle \int_{0}^{\infty}e^{-\sqrt{y}}dy=\lim_{t\rightarrow\infty}\int_{0}^{t}e^{-\sqrt{y}}dy=\lim_{t\rightarrow\infty}\left[2(\sqrt{y}-1)e^{-\sqrt{y}}\right]_{0}^{t}$ $=\displaystyle \lim_{t\rightarrow\infty}\left[\frac{2(\sqrt{y}-1)}{e^{\sqrt{y}}}\right]_{0}^{t}$ $=\displaystyle \lim_{t\rightarrow\infty}\left[\frac{2(\sqrt{t}-1)}{e^{\sqrt{t}}}\right]-\lim_{t\rightarrow\infty}\left[\frac{-2}{1}\right]$ ...L'Hospital's rule $\displaystyle \lim_{t\rightarrow\infty}\frac{\sqrt{t}-1}{e^{\sqrt{t}}} =\displaystyle \lim_{t\rightarrow\infty}\frac{\frac{1}{2\sqrt{t}}}{e^{\sqrt{t}}\cdot\frac{1}{2\sqrt{t}}}=\lim_{t\rightarrow\infty}\frac{1}{e^{\sqrt{t}}}=0$ $I=0+2$ $=2$ (converges)
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