Answer
$y=2x^4+x^3-40$
Work Step by Step
$\frac{dy}{dx}=8x^3+3x^2;$ $y(2)=0$
$dy=(8x^3+3x^2)dx$
$\int dy=\int(8x^3+3x^2)dx$
$y=2x^4+x^3+C$
Substitute in initial condition to solve for $C$
$0=2(16)+(8)+C$
$C=-40$
$y=2x^4+x^3-40$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
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: 50
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