University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Section 5.3 - The Definite Integral - Exercises - Page 311: 87


$U-L \lt \epsilon(b-a)$

Work Step by Step

Here, we have $U-L=\Sigma_{i=1}^n \triangle x_i \cdot M_i-\Sigma_{i=1}^n \triangle x_i \cdot m_i $ This implies that $U-L=\Sigma_{i=1}^n ( M_i- x_i)\triangle x_i \lt \Sigma_{i=1}^n \epsilon \triangle x_i $; for $i=1,2,3,...n$ Because $\Sigma_{i=1}^n \epsilon \triangle x_i=\epsilon\Sigma_{i=1}^n \triangle x_i=\epsilon(b-a)$ Thus, we have $U-L \lt \epsilon(b-a)$
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