Answer
Error $(M_{20}) \leq 5.04659 \times 10^{-5}$
Work Step by Step
Error $(M_N) \leq \dfrac{k_2(b-a)^3}{24N^2}$
For, $M=20$, we have:
Error $M_{20} \leq \dfrac{(1)(\dfrac{\pi}{4}-0)^3}{24(20)^2} \\ \leq \dfrac{(1)(\dfrac{\pi^3}{64})}{24(400)}\\ \leq \dfrac{\pi^3}{614400} \\\leq 5.04659 \times 10^{-5}$
