Chapter 5 - Section 5.3 - The Fundamental Theorem of Calculus - 5.3 Exercises - Page 406: 5

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

$$g(s) = \int\limits_5^s{(t - t^2)^8}dt$$ Plug upper bound into t and multiply by the derivative of the upper bound (chain rule). $$g'(s) = ((s)-(s)^2)^8 \times (s)'$$ Simplify. $$g'(s) = (s-s^2)^8$$