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: 11

Answer

$$F'(x) = -\sqrt{1 + sec(x)}$$

Work Step by Step

$$F(x) = \int\limits_x^0{\sqrt{1+sec(t)}}dt$$ Switch the bounds of the integral (This will make the equation negative). $$F(x) = - \int\limits_0^x{\sqrt{1+sec(t)}}dt$$ Plug the upper bound into t and multiply by the derivative of the upper bound. $$F'(x) = - \sqrt{1 + sec(x)} \times (x)'$$ Simplify. $$F'(x) = - \sqrt{1 + sec(x)}$$
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