Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 4 - Applications of the Derivative - 4.9 Antiderivatives - 4.9 Exercises: 62

Answer

$$ - \frac{8}{{\sqrt x }} - \frac{4}{{{x^{3/2}}}} + 16$$

Work Step by Step

$$\eqalign{ & f\left( x \right) = \left( {4\sqrt x + 6/\sqrt x } \right)/{x^2} \cr & f\left( x \right) = \frac{{4\sqrt x }}{{{x^2}}} + \frac{6}{{\sqrt x }}\left( {\frac{1}{{{x^2}}}} \right) \cr & f\left( x \right) = \frac{4}{{{x^{3/2}}}} + \frac{6}{{{x^{5/2}}}} \cr & f\left( x \right) = 4{x^{ - 3/2}} + 6{x^{ - 5/2}} \cr & {\text{find an antiderivative of }}f\left( x \right),{\text{ use power rule}} \cr & F\left( x \right) = 4\left( {\frac{{{x^{ - 1/2}}}}{{ - 1/2}}} \right) + 6\left( {\frac{{{x^{ - 3/2}}}}{{ - 3/2}}} \right) + C \cr & F\left( x \right) = - 8{x^{ - 1/2}} - 4{x^{ - 3/2}} + C \cr & {\text{using the initial condition }}F\left( 1 \right) = 4 \cr & 4 = - 8{\left( 1 \right)^{ - 1/2}} - 4{\left( 1 \right)^{ - 3/2}} + C \cr & {\text{then}} \cr & 4 = - 8 - 4 + C \cr & C = 16 \cr & {\text{so}}{\text{,}} \cr & = - 8{x^{ - 1/2}} - 4{x^{ - 3/2}} + 16 \cr & = - \frac{8}{{\sqrt x }} - \frac{4}{{{x^{3/2}}}} + 16 \cr} $$
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