= Solution
Write $f^{(4)}(x)=24(5-4x)/x^6$. Its <derivative> is $f^{(5)}(x)=240(2x-3)/x^7$, so on $[1,2]$ it decreases to $x=3/2$ and then increases. Its endpoint and turning-point values are
$$
f^{(4)}(1)=24,\qquad
f^{(4)}(3/2)=-\frac{512}{243},\qquad
f^{(4)}(2)=-\frac98.
$$
Thus $M_4=24$, and the preceding bound becomes
$$
\boxed{|I-S_n|\leq\frac{2}{15n^4}\qquad(n\text{ even}).}
$$
For the <rounding> estimate, standard binary64 has spacing $\epsilon=2^{-52}$ above one. Balancing this nominal absolute accuracy with the bound gives
$$
n\approx\left(\frac{2}{15\epsilon}\right)^{1/4}\approx4.95\times10^3.
$$
Thus \b[a few thousand even strips, roughly $5000$, reach the double-precision scale according to this bound]. A scale-aware estimate leads to the same order. Indeed,
$$
I=-\frac{19}{6}-4\log2\approx-5.9392553889
$$
lies in magnitude between $4$ and $8$, where the binary64 spacing is $2^{-50}$. Balancing the error bound against half that spacing gives about $4.16\times10^3$ strips.
These are saturation estimates from <balancing quadrature truncation and round-off error>, not sharp guarantees of the actual <rounding> plateau. The error bound can be conservative, and evaluation and summation errors can limit accuracy earlier. Merely storing the final result does not specify every source of <rounding error>.
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