Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 5: Integrals - Practice Exercises - Page 307: 19



Work Step by Step

Given $ x=2y^2$ Re-write as: $y=\dfrac{\sqrt x}{\sqrt 2} \implies \dfrac{\sqrt x}{\sqrt 2}=3 $ Consider $f(x)= 3-\dfrac{\sqrt x}{\sqrt 2}$ Use formula such as follows: $\int x^{n} dx=\dfrac{x^{n+1}}{n+1}+C$ This implies that $[3x-\dfrac{1}{\sqrt 2}(\dfrac{x^{(3/2)}}{\dfrac{1}{(3/2)}})]_0^{18}=3(18-0)-\dfrac{2}{3\sqrt 2}( 18^{(\frac{3}{2})}-0)$ Hence, $f(x)=\int_0^{18} (3-\dfrac{\sqrt x}{\sqrt 2})dx=18$
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