Calculate ln2=integral from 1 to ² dx/x using Simpson's
Calculate $\ln2=\displaystyle\int_1^2\frac{dx}{x}$ using Simpson's formula with $2n=10$ subintervals, and estimate the error.
Explanation of Method
Apply Simpson's rule with $h=(b-a)/(2n)=0.1$ and the alternating $1,4,2,4,\dots,4,1$ weight pattern to $f(x)=1/x$ at the 11 equally spaced nodes, then compare against the exact value $\ln2$ and bound the theoretical error using $\max|f^{(4)}|$ on $[1,2]$.
Step 1: Setup
With $2n=10$, $h=(2-1)/10=0.1$, nodes $x_i=1+0.1i$ for $i=0,\dots,10$.
Step 2: Function values and Simpson weights
| $i$ | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| $x_i$ | 1.0 | 1.1 | 1.2 | 1.3 | 1.4 | 1.5 | 1.6 | 1.7 | 1.8 | 1.9 | 2.0 |
| $f(x_i)$ | 1.000000 | 0.909091 | 0.833333 | 0.769231 | 0.714286 | 0.666667 | 0.625000 | 0.588235 | 0.555556 | 0.526316 | 0.500000 |
| weight | 1 | 4 | 2 | 4 | 2 | 4 | 2 | 4 | 2 | 4 | 1 |
Weighted sum: $S=20.794506921$.
Step 3: Simpson's approximation
\[I_{\text{Simpson}}=\frac{h}{3}\,S=\frac{0.1}{3}\times20.794506921=0.693150231.\]
\[\boxed{I_{\text{Simpson}}\approx0.693150231}\]
Step 4: Compare with the exact value
\[\ln2\approx0.693147181,\qquad E_{\text{actual}}=|0.693150231-0.693147181|\approx3.05\times10^{-6}.\]
Step 5: Theoretical error bound
For $f(x)=1/x$: $f^{(4)}(x)=24/x^5$, decreasing on $[1,2]$, so its maximum magnitude on this interval is at $x=1$: $\max|f^{(4)}|=24$. Then
\[|E_{\text{Simpson}}|\le\frac{(b-a)^5}{180(2n)^4}\max|f^{(4)}|=\frac{1}{180\times10^4}\times24=\frac{1}{75{,}000}\approx1.33\times10^{-5}.\]
The actual error $3.05\times10^{-6}$ sits comfortably inside this bound (about $4\times$ smaller), typical of these bounds being conservative worst-case estimates.
Result
Final answer:
\[\boxed{\ln2\approx0.693150231\ (\text{Simpson},\,2n{=}10),\qquad E_{\text{actual}}\approx3.05\times10^{-6}\le1.33\times10^{-5}}\]