Answer
$M_{10} \leq 0.01125$
Work Step by Step
We have the error
$(M_N) \leq \dfrac{k_2(b-a)^3}{24N^2}$
For, $M=10$, we have:
Error $M_{10} \leq \dfrac{(1)(4-1)^3}{24(10)^2} \\ \leq \dfrac{(1)(27)}{24(100)}\\ \leq \dfrac{9}{800} \\\leq 0.01125$
