Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 4 - Integrals - 4.3 The Fundamental Theorem of Calculus - 4.3 Exercises: 12

Answer

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

Work Step by Step

We are told to find the derivative using Part 1 of the Fundamental Theorem of Calculus given the expression: $R(y) = \int ^{2}_{y}t^3sin(t)dt$ Swap upper and lower bounds: (this makes the expression negative) $R(y) = -\int ^{y}_{2}t^3sin(t)dt $ Since $R(y) = \int^{x}_{0}f(t)dt$ $R'(y) = f(y)$ Thus, $R'(y) = -y^3sin(y)$
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