Proof. The linear interpolant at \(x_0 = a\), \(x_1 = b\) has error \(\frac {f''(\xi (x))}{2}(x-a)(x-b)\). Its integral is \(\frac h2[f(a)+f(b)]\), the area of a trapezoid. The error integral is \(\frac 12\int _a^bf''(\xi (x))(x-a)(x-b)\,dx\); since \((x-a)(x-b)\) does not change sign on \([a,b]\), the Weighted Mean Value Theorem (Theorem 1.1.5) gives \[\frac {f''(\xi )}{2}\int _a^b(x-a)(x-b)\,dx = \frac {f''(\xi )}{2}\left (-\frac {h^3}{6}\right ) = -\frac {h^3}{12}f''(\xi ) .\] □

Proof. Integrating the quadratic interpolant gives the weights \(\frac h3,\frac {4h}3,\frac h3\). For the error, the direct approach gives only \(O(h^4)\), because the cubic \((x-a)(x-x_1)(x-b)\) changes sign. Instead expand \(f\) about \(x_1\): \(f(x) = f(x_1)+f'(x_1)(x-x_1)+\frac {f''(x_1)}{2}(x-x_1)^2+\frac {f'''(x_1)}{6}(x-x_1)^3+\frac {f^{(4)}(\xi (x))}{24}(x-x_1)^4\). Integrating over \([x_1-h,x_1+h]\), the odd powers vanish by symmetry, and the Weighted Mean Value Theorem applied to \((x-x_1)^4\geq 0\) gives \[\int _a^bf = 2hf(x_1)+\frac {h^3}{3}f''(x_1)+\frac {f^{(4)}(\xi _1)}{60}h^5 .\] Replacing \(f''(x_1)\) by the central difference \(\frac {f(a)-2f(x_1)+f(b)}{h^2}-\frac {h^2}{12}f^{(4)}(\xi _2)\) (derived by adding the Taylor expansions of \(f(x_1\pm h)\)) gives \[\int _a^bf = \frac h3[f(a)+4f(x_1)+f(b)]-\frac {h^5}{12}\left [\frac 13f^{(4)}(\xi _2)-\frac 15f^{(4)}(\xi _1)\right ],\] and a more careful argument shows that the bracket can be replaced by \(\frac {2}{15}f^{(4)}(\xi )\), giving \(-\frac {h^5}{90}f^{(4)}(\xi )\). □

Solution.

Both Simpson rules are exact for \(x^2\) (degree of precision \(3\)); neither is exact for \(x^4\). For the trapezoidal rule, \(h = 2\) is far too large; the next subsection shows how to use these rules on small subintervals.