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

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

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