Calculate integral from 0 to 1 dx/(1+x) by dividing [0,1]

Calculate $\displaystyle\int_0^1\frac{dx}{1+x}$ by dividing $[0,1]$ into 8 subintervals using (a) the rectangle (midpoint) formula, (b) the trapezoidal formula, (c) Simpson's formula, and compare each with the exact value.

Step 0: Exact value

\[I_{\text{exact}}=\int_0^1\frac{dx}{1+x}=\big[\ln(1+x)\big]_0^1=\ln2\approx0.693147181.\]

Step 1(a): Rectangle (midpoint) method, $n=8$

With $h=0.125$ and midpoints $x_i^*=(i-\tfrac12)h$:

Sum $=5.541284$, so $I_{\text{rect}}=0.125\times5.541284\approx0.692661$.

\[\boxed{I_{\text{rect}}\approx0.692661},\qquad E_{\text{rect}}\approx4.87\times10^{-4}\]

Step 1(b): Trapezoidal method, $n=8$

With nodes $x_i=0.125i$, $i=0,\dots,8$, weighted $1,2,2,\dots,2,1$: weighted sum $=11.105950$, so

\[I_{\text{trap}}=\frac{0.125}{2}\times11.105950\approx0.694122.\]

\[\boxed{I_{\text{trap}}\approx0.694122},\qquad E_{\text{trap}}\approx9.75\times10^{-4}\]

Step 1(c): Simpson's method, $n=8$ ($2n=8$ intervals)

Same nodes, weights $1,4,2,4,2,4,2,4,1$: weighted sum $=16.635709$, so

\[I_{\text{Simpson}}=\frac{0.125}{3}\times16.635709\approx0.693155.\]

\[\boxed{I_{\text{Simpson}}\approx0.693155},\qquad E_{\text{Simpson}}\approx7.35\times10^{-6}\]

Step 2: Comparison

Simpson's rule is roughly $66\times$ more accurate than the midpoint rule and about $133\times$ more accurate than the trapezoidal rule here — consistent with its $O(h^4)$ error order versus $O(h^2)$ for the other two, and with Simpson's rule being exact for cubics while the others are exact only for linear functions.