## Thomas' Calculus 13th Edition

$n \geq 26$
Consider the error bound for the Trapezoidal Rule as follows: $|E_T| \leq \dfrac{M(b-a)^3}{12n^2}$ The maximum value of $|f^{2}(x)|$ on $[0,1]$, $M=8$ Now, $|E_T| \leq \dfrac{8(1-0)^3}{12n^2}=\dfrac{2}{3n^4}$ We will choose $n$ such that $\dfrac{2}{3n^2} \leq 10^{-3}$ So, $n \geq 25.8 \approx 26$ and $n \geq 26$