Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 4 - Section 4.9 - Antiderivatives - 4.9 Exercises - Page 362: 12

Answer

$H\left( z \right) = \frac{5}{3}{z^{1.8}} - \frac{{2{z^{ - 1.5}}}}{3} + C$

Work Step by Step

$$\eqalign{ & h\left( z \right) = 3{z^{0.8}} + {z^{ - 2.5}} \cr & {\text{Using the formulas in Table 2 }}\left( {{\text{see page 358}}} \right){\text{ }} \cr & {\text{Function: }}cf\left( x \right) \to {\text{Particular antiderivative:}}cF\left( x \right) \cr & {\text{Function: }}{x^n}\left( {n \ne - 1} \right) \to {\text{Particular antiderivative: }}cF\left( x \right) \cr & {\text{Function: }}f\left( x \right) + g\left( x \right) \to {\text{Particular antiderivative: }}f\left( x \right) + g\left( x \right) \cr & {\text{And applying the Theorem 1}},{\text{ we obtain}} \cr & H\left( z \right) = 3\left( {\frac{{{z^{0.8 + 1}}}}{{0.8 + 1}}} \right) + \frac{{{z^{ - 2.5 + 1}}}}{{ - 2.5 + 1}} + C \cr & H\left( z \right) = 3\left( {\frac{{{z^{1.8}}}}{{1.8}}} \right) + \frac{{{z^{ - 1.5}}}}{{ - 1.5}} + C \cr & H\left( z \right) = \frac{5}{3}{z^{1.8}} - \frac{{2{z^{ - 1.5}}}}{3} + C \cr} $$
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