Answer
$ \$ 4500$
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$
Thomas' Calculus 13th Edition
by Thomas Jr., George B.
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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