Science:Math Exam Resources/Courses/MATH101 C/April 2024/Question 11/Solution 1

Let $S_{n}$ denote the Simpson approximation with $n$ subintervals. The inequality in the statement supports the choice $L=9$ in the well-known error estimate formula. This leads to

$${\begin{aligned}|I-S_{n}|\leq {\frac {L}{180}}{\frac {(b-a)^{5}}{n^{4}}}={\frac {9}{180}}{\frac {(5)^{5}}{n^{4}}}.\end{aligned}}$$

To use this, we chain the desired tolerance onto the right side of the inequality, obtaining

$${\begin{aligned}{\frac {1}{20}}{\frac {(5)^{5}}{n^{4}}}\leq {\frac {1}{20\cdot 5^{3}}}.\end{aligned}}$$

Multiplying both sides by $20(5)^{3}n^{4}$ gives

$${\begin{aligned}{\frac {5^{5}}{20}}\cdot 20\cdot 5^{3}\leq n^{4},\quad \mathrm {i.e.,} \quad 5^{8}\leq n^{4}.\end{aligned}}$$

This is equivalent to $n\geq 5^{2}=25.$

Since the number of subintervals used in Simpson's Rule must be even, then any even $n\geq 26$ will achieve the specified tolerance.