Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 5 - The Integral - 5.3 The Indefinite Integral - Exercises - Page 253: 53



Work Step by Step

$\frac{dy}{dx}=(3x+2)^3;$ $y(0)=1$ $dy=(3x+2)^3dx$ $\int dy=\int(3x+2)^3dx$ $y+C=\frac{1}{3}\int3(3x+2)^3dx$ $y=\frac{1}{3}(\frac{{(3x+2)}^4}{4})+C$ Substitute in initial condition to solve for $C$ $1=\frac{4}{3}+C$ $C=-\frac{1}{3}$ $y=\frac{{(3x+2)}^4}{12}-\frac{1}{3}$
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