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

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

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