## Calculus (3rd Edition)

$(T_{20}) \leq 0.000104167$
We have the error $(T_N) \leq \dfrac{k_2(b-a)^3}{12N^2}$ For, $N=20$, we have: Error $(T_{20}) \leq \dfrac{(\dfrac{1}{16})(2-0)^3}{12(20)^2} \\ \leq \dfrac{\dfrac{1}{16}(8)}{12(400)}\\ \leq \dfrac{1}{9600} \\\leq 0.000104167$