From the formula on Wikipedia, $\frac{M\left(b-a\right)^2}{2n} = \frac{1}{2 \cdot n} < 0.0001$ and my simplification is due to $M$ being the maximum of the absolute value of the first derivative on the interval (so $\left| - \frac{1}{x^2} \right|$ when $1 \leq x \leq 2$. And, $b-a=2-1=1$
But then I get an "unusually" large number when I solve for $n$. It seems that $ n = 5001$ is the minimum amount since I'd need $n$ to be greater than $5000$. The reason I say unsual is because the trapezoidal, midpoint, and Simpson's are so much smaller that I feel some sort of calculation error. For those, I got $41$, $29$ and $8$
Like I know that the other methods were better but I didn't realize it was soo much better than Riemann sums. (But I can see how Riemann sums aren't to good since their error estimates don't have $n$ to the power of anything, and so you'd need a large $n$ in order to make up for no exponents.)
My question: is n=5001 correct?