Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 4 - Integration - 4.2 The Indefinite Integral - Exercises Set 4.2 - Page 279: 46


$$f(x)= \frac{1}{6}x^3-\cos x+2x+1$$

Work Step by Step

Given $$ f''(x) = x + \cos x,\ \ \ f(0)=1,\ \ f'(0)=2$$ Since \begin{align*} f'(x)&=\int f''(x)dx\\ &=\int (x+\cos x)dx\\ &=\frac{1}{2}x^2+\sin x+c \end{align*} Since $f'(0)=2$ , then $c=2$ and \begin{align*} f(x)&=\int f'(x)dx\\ &=\int (\frac{1}{2}x^2+\sin x+2)dx\\ &= \frac{1}{6}x^3-\cos x+2x+c_1 \end{align*} Since $f'(0)=0$ , then $c_1=1$ and $$f(x)= \frac{1}{6}x^3-\cos x+2x+1$$
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