Answer
$\$4500$
Work Step by Step
Here, we have $\int_0^{3}2- \dfrac{2}{(x+1)^2} dx$
or, $\int_0^{3}2- \dfrac{2}{(x+1)^2} dx=[2x+\dfrac{2}{x+1}]_0^3$
Thus, $6+\dfrac{2}{4}-0-2=4.5$
Hence, $\$4500$
University Calculus: Early Transcendentals (3rd Edition)
by Hass, Joel R.; Weir, Maurice D.; Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0321999584
ISBN 13:
978-0-32199-958-0
Chapter 5 - Section 5.4 - The Fundamental Theorem of Calculus - Exercises - Page 321: 74
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