Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 5: Integrals - Section 5.4 - The Fundamental Theorem of Calculus - Exercises 5.4 - Page 287: 64


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Work Step by Step

Let us consider $\int_0^{3}[2- \dfrac{2}{(x+1)^2}] dx=[2x+\dfrac{2}{x+1}]_0^3$ $\implies [2x+\dfrac{2}{x+1}]_0^3=6+\dfrac{2}{4}-0-2=4.5$ Hence, we have $ \$ 4500$
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