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: 10

Answer

$$h'(u) = \frac{\sqrt{u}}{u+1}$$

Work Step by Step

$$h(u) = \int\limits_0^u {\frac{\sqrt{t}}{t+1}}dt$$ Plug in the upper bound into t and multiply by the derivative of the upper bound due to chain rule. $$h'(u) = \frac{\sqrt{u}}{u+1} \times (u)'$$ Simplify. $$h'(u) = \frac{\sqrt{u}}{u+1}$$
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