Answer
$U-L \lt \epsilon(b-a)$
Work Step by Step
Let us consider $U-L=\Sigma_{i=1}^n \triangle x_i \cdot M_i-\Sigma_{i=1}^n \triangle x_i \cdot m_i $
and
$U-L=\Sigma_{i=1}^n ( M_i- x_i) \cdot \triangle x_i \lt \Sigma_{i=1}^n \epsilon \cdot \triangle x_i $; (for $i=1,2,3,...n)$
Since, $\Sigma_{i=1}^n \epsilon \cdot \triangle x_i=(\epsilon) \cdot \Sigma_{i=1}^n \triangle x_i=\epsilon \cdot (b-a)$
Then, we get $U-L \lt \epsilon (b-a)$
