 Solution

The global bound requiresSince the strip count must be even, the smallest count certified by this bound is $n=88$.

For the tighter bound, use the bisection method on the continuous extension $B(x)-10^{-10}$ for $x\ge 2$. The initial root bracket $[2,88]$ is valid: $B(2)=1/2880>10^{-10}$, while $B(88)$ is below the global bound and hence below $10^{-10}$. To identify the smallest admissible even count, the monotonicity of $B$ can also be justified.

Put $h=2/x$ andFor $h>0$, differentiation showsThe numerator is nonnegative because $e^h-e^{-h}\ge 2h$: its difference from $2h$ starts at zero and has nonnegative derivative $e^h+e^{-h}-2$, by the arithmetic-geometric mean inequality. Thus $C^{\prime }(h)\ge 1/2$, and $C(0^{+})=e/2-1>0$. Therefore $h^{4}C(h)$ increases with $h$, so $B(x)$ is strictly decreasing with $x$. Bisection locates its unique threshold crossing; then check the neighboring even integers, retaining the strict error inequality.

For example, $B(66)\approx 1.09626\times 10^{-10}$ and $B(68)\approx 9.71694\times 10^{-11}$, so the tighter bound certifies $68$ strips. These are evaluations of the bound, not an assumption that it equals the actual quadrature error.