## Calculus (3rd Edition)

The actual error is approximately $3.333 \times 10^{-7}$, which is less than $10^{-6}$.
Error $(T_N) \leq \dfrac{k_2(b-a)^3}{24N^2} \leq \dfrac{(12)(1-0)^3}{12N^2} \\ \leq \dfrac{1}{N^2}$ We want to choose $N$ such that the error is at most $10^{-6}$. So, we have: $\dfrac{1}{N^2} \leq \dfrac{1}{10^6} \implies N \geq 10^3$ The given integral for $N=1000$ can be computed as: $\int_0^4 x^4 dx =[\dfrac{x^5}{5}]_0^1 =\dfrac{1}{5}=0.2$ By using a calculator, we get $T_{1000} \approx 0.2000003333$ The actual error is approximately $[ 0.2000003333-0.2 ] \approx 3.333 \times 10^{-7}$, which is less than $10^{-6}$.