Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 4 - Integrals - 4.3 The Fundamental Theorem of Calculus - 4.3 Exercises - Page 327: 9


$g'(s) = (s-s^2)^8$

Work Step by Step

We are told to find the derivative using Part 1 of the Fundamental Theorem of Calculus given the expression: $g(s) = \int ^{s}_{5} (t-t^2)^8dt$ Since $g(s) = \int^{b}_{a}f(t)dt$ $g'(s) = f(s)$ Thus, $g'(s) = (s-s^2)^8$
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