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: 65



Work Step by Step

$f’’(x)=x^3-2x+1,$ $f’(0)=1,$ $f(0)=0$ $f’(x)=\int(x^3-2x+1)dx$ $f’(x)=\frac{x^4}{4}-x^2+x+C_1$ Substitute in $f’(0)=1$ to solve for $C_1$ $1=0-0+0+C_1$ $C_1=1$ $f’(x)=\frac{x^4}{4}-x^2+x+1$ $f(x)=\int(\frac{x^4}{4}-x^2+x+1)dx$ $f(x)=\frac{x^5}{20}-\frac{x^3}{3}+\frac{x^2}{2}+x+C_2$ Substitute in $f(0)=0$ to solve for $C_2$ $0=0-0+0+C_2$ $C_2=0$ $f(x)=\frac{x^5}{20}-\frac{x^3}{3}+\frac{x^2}{2}+x$
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