Answer
Error $(T_{10})\leq 0.045$
Work Step by Step
We have the error
$(T_N) \leq \dfrac{k_2(b-a)^3}{12N^2}$
For, $N=10$, we have:
Error $(T_{10}) \leq \dfrac{2(4-1)^3}{12(10)^2} \\ \leq \dfrac{2(27)}{12(100)}\\ \leq \dfrac{9}{200} \\\leq 0.045$
