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: 64



Work Step by Step

$f’’(x)=x^3-2x,$ $f’(1)=0,$ $f(1)=2$ $f’(x)=\int(x^3-2x)dx$ $f’(x)=\frac{x^4}{4}-x^2+C_1$ Substitute in $f’(1)=0$ to solve for $C_1$ $0=\frac{1}{4}-1+C_1$ $C_1=\frac{3}{4}$ $f’(x)=\frac{x^4}{4}-x^2+\frac{3}{4}$ $f(x)=\int(\frac{x^4}{4}-x^2+\frac{3}{4})dx$ $f(x)=\frac{x^5}{20}-\frac{x^3}{3}+\frac{3x}{4}+C_2$ Substitute in $f(1)=2$ to solve for $C_2$ $2=\frac{1}{20}-\frac{1}{3}+\frac{3}{4}+C_2$ $C_2=\frac{23}{15}$ $f(x)=\frac{x^5}{20}-\frac{x^3}{3}+\frac{3x}{4}+\frac{23}{15}$
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