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: 12

Answer

$$R'(y) = -y^{3}sin(y)$$

Work Step by Step

$$R(y) = \int\limits_y^2{t^3}sin(t)dt$$ Switch the bounds of the integral making it negative. $$R(y) = -\int\limits_w^y{t^3sin(t)}dt$$ Use FTC 1 by plugging the upper bound into t and multiply by the derivative of the upper bound (chain rule). $$R'(y) = -y^3sin(y) \times(y)'$$ Simplify. $$R'(y) = -y^3sin(y)$$
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