## Calculus: Early Transcendentals (2nd Edition)

Published by Pearson

# Chapter 4 - Applications of the Derivative - 4.9 Antiderivatives - 4.9 Exercises: 76

#### Answer

$$v\left( x \right) = 3{x^{4/3}} + 3{x^{2/3}} - 20$$

#### Work Step by Step

\eqalign{ & v'\left( x \right) = 4{x^{1/3}} + 2{x^{ - 1/3}} \cr & v\left( x \right) = \int {v'\left( x \right)} dx \cr & then \cr & v\left( x \right) = \int {\left( {4{x^{1/3}} + 2{x^{ - 1/3}}} \right)} dx \cr & find{\text{ the general solution by the power rule}} \cr & v\left( x \right) = 4\left( {\frac{{{x^{4/3}}}}{{4/3}}} \right) + 2\left( {\frac{{{x^{2/3}}}}{{2/3}}} \right) + C \cr & v\left( x \right) = 3{x^{4/3}} + 3{x^{2/3}} + C \cr & {\text{using the initial condition }}v\left( 8 \right) = 40 \cr & 40 = 3{\left( 8 \right)^{4/3}} + 3{\left( 8 \right)^{2/3}} + C \cr & 40 = 3\left( {16} \right) + 3\left( 4 \right) + C \cr & C = - 20 \cr & {\text{the solution to the initial value problem is}} \cr & v\left( x \right) = 3{x^{4/3}} + 3{x^{2/3}} - 20 \cr}

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