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: 78


$$\int^{4}_1\frac{(1+\sqrt u)^{1/2}}{\sqrt u}du=\frac{4}{3}(3\sqrt3-2\sqrt2)$$

Work Step by Step

$$A=\int^{4}_1\frac{(1+\sqrt u)^{1/2}}{\sqrt u}du$$ We set $a=1+\sqrt u$, which means $$da=\frac{du}{2\sqrt u}$$ $$\frac{du}{\sqrt u}=2da$$ For $u=1$, $a=2$ and for $u=4$, $a=3$ Therefore, $$A=2\int^{3}_2a^{1/2}da$$ $$A=\Big(\frac{2a^{3/2}}{\frac{3}{2}}\Big)\Big]^{3}_2=\Big(\frac{4a^{3/2}}{3}\Big)\Big]^{3}_2$$ $$A=\frac{4}{3}(3^{3/2}-2^{3/2})$$ $$A=\frac{4}{3}(3\sqrt3-2\sqrt2)$$
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