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: 7

Answer

$$g'(x) = \sqrt{x + x^3}$$

Work Step by Step

$$g(x) = \int\limits_0^x{\sqrt{t + t^3}}dt$$ Plug in the upper bound into t and take the derivative of the upper bound. $$g'(x) = \sqrt{(x) + (x)^3} \times (x)'$$ Simplify $$g'(x) = \sqrt{x + x^3}$$
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