Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 5 - Section 5.3 - The Fundamental Theorem of Calculus - 5.3 Exercises: 9

Answer

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

Work Step by Step

$$g(s) = \int\limits_5^s{(t - t^2)^8}dt$$ 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. $$g'(s) = (s-s^2)^8$$
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