Science:Math Exam Resources/Courses/MATH101 B/April 2025/Question 07 (b)/Solution 1

To calculate the error, we will use $M=64\cdot 4800$ and hence, we want $\text{Error}\le \frac{M}{12}\frac{(b-a{)}^3}{n^2}\le \frac{64\cdot 4800}{12}\frac{4^3}{n^2}\le 64\cdot 400\frac{64}{n^2}\le 64.$ This means we need $n^2\ge 64\cdot 400$, so $n\ge 8\cdot 20=160$. Therefore, we can take $n=160$.