University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 5 - Practice Exercises - Page 343: 77


$$\int^{4}_1\frac{dt}{t\sqrt t}=1$$

Work Step by Step

$$A=\int^{4}_1\frac{dt}{t\sqrt t}=\int^{4}_1\frac{dt}{t^{3/2}}$$ $$A=\int^{4}_1t^{-3/2}dt$$ $$A=\Big(\frac{t^{-1/2}}{-\frac{1}{2}}\Big)\Big]^{4}_1=\Big(-\frac{2}{\sqrt t}\Big)\Big]^{4}_1$$ $$A=-\frac{2}{\sqrt4}-\Big(-\frac{2}{\sqrt1}\Big)$$ $$A=-\frac{2}{2}+\frac{2}{1}=-1+2$$ $$A=1$$
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