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: 67



Work Step by Step

$f’’(t)=t^{-\frac{3}{2}},$ $f’(4)=1,$ $f(4)=4$ $f’(t)=\int t^{-\frac{3}{2}}dt$ $f’(t)=-2t^{-\frac{1}{2}}+C_1$ Substitute in $f’(4)=1$ to solve for $C_1$ $1=-2\sqrt{\frac{1}{4}}+C_1$ $C_1=2$ $f’(t)=-2t^{-\frac{1}{2}}+2$ $f(t)=\int(-2t^{-\frac{1}{2}}+2)dt$ $f(t)=-4t^{\frac{1}{2}}+2t+C_2$ Substitute in $f(4)=4$ to solve for $C_2$ $4=-4\sqrt4+8+C_2$ $C_2=4$ $f(t)=-4t^{\frac{1}{2}}+2t+4$
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