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 252: 49



Work Step by Step

$\frac{dy}{dt}=2t+9t^2;$ $y(1) = 2$ $dy=(2t+9t^2)dt$ $\int dy=\int(2t+9t^2)dt$ $y=t^2+3t^3+C$ $2=1^2+3(1^3)+C$ $C=-2$ $y=3t^3+t^2-2$
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