Answer
$9$
Work Step by Step
Here, we have $f(x)=2x$
Then, $\Sigma_{i=1}^n (\dfrac{6}{n}) c_i=(\dfrac{1}{n}) \Sigma_{i=1}^n \dfrac{18i}{n^2}$
or,
$\dfrac{18n(n+2)}{2n^2})=\dfrac{9(n+1)}{n}=9$
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 5.2 - Sigma Notation and Limits of Finite Sums - Exercises - Page 300: 40
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