Answer
$s(t)=\frac{t^3}{3}-2t\sqrt{t}+\frac{8}{3}$
Work Step by Step
Note that $s(t)=\int v(t) \ dt \Rightarrow s(t)= \int t^2-3t^{\frac{1}{2}} dt.$
$$=t^3-2t\sqrt{t}+C.$$ since $s(4)=8$, then
$$8=\frac{64}{3}-16 +C \Rightarrow C=\frac{8}{3}$$.
Calculus: Early Transcendentals 9th Edition
by Stewart, James
Published by
Cengage Learning
ISBN 10:
1337613924
ISBN 13:
978-1-33761-392-7
Chapter 4 - Section 4.9 - Antiderivatives - 4.9 Exercises - Page 363: 66
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