Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 5 - The Integral - Chapter Review Exercises - Page 278: 37


$$f(t) =\frac{1}{2}t^2- \frac{1}{3}t^3-t+2 $$

Work Step by Step

Given$$ f^{\prime \prime}(t)=1-2 t, f(0)=2, \text { and } f^{\prime}(0)=-1 $$ Then \begin{align*} f'(t) &= \int (1-2t)dt\\ &= t-t^2+c \end{align*} At $t= 0$, $f'(0)=-1,$ then $c= -1$, hence $$f'(t)= t-t^2-1 $$ \begin{align*} f(t) &= \int ( t-t^2-1 )dt\\ &= \frac{1}{2}t^2- \frac{1}{3}t^3-t+c_1 \end{align*} At $t=0$ $f=2$, then $c_1= c_1=2 $, hence $$f(t) =\frac{1}{2}t^2- \frac{1}{3}t^3-t+2 $$
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