Answer
$N \geq 13416.4079$.
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 $10^{-6}$.
