# Chapter 8 - Techniques of Integration - 8.9 Numerical Integration - Exercises - Page 458: 38

$N \geq 13416.4079$. The actual error is approximately $6 \times 10^{-7}$, which is less than $10^{-6}$.

#### Work Step by Step

Error $(T_N) \leq \dfrac{k_2(b-a)^3}{12N^2} \leq \dfrac{(80)(3-0)^3}{12N^2} \\ \leq \dfrac{180}{N^2}$ We want to choose $N$ such that the error is at most $10^{-6}$. So, we have: $\dfrac{180}{N^2} \leq \dfrac{1}{10^6} \implies N \geq 13416.4079$ The given integral for $N=13417$ can be computed as: $\int_0^3 (5x^4-x^5) dx =[x^5-\dfrac{x^6}{6}]_0^3 =121.5$ By using a calculator, we get $T_{13417} \approx 121.5000006$ The actual error is approximately $[121.5000006-121.5 ] \approx 6 \times 10^{-7}$, which is less than $10^{-6}$.

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.