Numerical Quadrature and Error Bounds

Approximate a function locally

With $n$ equal subintervals, $h=(b-a)/n$ and $x_i=a+ih$ , the composite trapezoidal rule joins neighboring samples by straight lines:

T_n=h\left[\frac{f(x_0)+f(x_n)}2+\sum_{i=1}^{n-1}f(x_i)\right].

Composite Simpson's rule fits quadratics across pairs of subintervals. It requires even $n$ :

S_n=\frac h3\left[f(x_0)+f(x_n)+4\sum_{i\text{ odd}}f(x_i)+2\sum_{\substack{i\text{ even}\\0<i<n}}f(x_i)\right].

Samples must correspond to the same equally spaced grid. An odd interval count or inconsistent spacing invalidates these composite Simpson weights.

Visual guide

VISUAL GUIDETrapezoids overestimate this convex curve

x²
Trapezoid tops

For x² on [0, 2], straight chords lie above the curve. Two trapezoids of width 1 give area 3; the exact integral is 8/3. Convexity explains the error direction for this example.

Worked example: a known integral

For $f(x)=x^2$ on $[0,2]$ and $n=2$ , samples are $0,1,4$ and $h=1$ . Thus $T_2=(0+4)/2+1=3$ , while $S_2=(0+4\cdot1+4)/3=8/3$ . The exact integral is $8/3$ . Simpson is exact here because it integrates polynomials of degree at most three exactly; trapezoids overestimate this convex function.

Worked example: choosing resolution

If $|f''|\le M_2$ on $[a,b]$ , then $|I-T_n|\le M_2(b-a)^3/(12n^2)$ . For $f=e^x$ on $[0,1]$ , use $M_2=e$ . To guarantee error below $0.001$ , require $e/(12n^2)<0.001$ , so $n=16$ suffices. This is a sufficient bound, not a claim that $15$ necessarily fails in practice.

If $f$ has a continuous fourth derivative bounded by $M_4$ , Simpson's error satisfies $|I-S_n|\le M_4(b-a)^5/(180n^4)$ . These guarantees depend on derivatives being bounded over the entire interval. They do not apply unchanged to a singular integrand such as $1/\sqrt x$ on $[0,1]$ .

Practice

Compute $T_2$ for $f(x)=x$ on $[0,2]$ .
Compute $S_2$ for $x^3$ on $[0,2]$ .
What happens to the stated trapezoidal error bound when $n$ doubles?

Show worked solutions

Samples $0,1,2$ give $T_2=2$ , exactly the integral.
Samples $0,1,8$ give $S_2=(0+4+8)/3=4$ , exactly the integral.
It becomes one quarter as large. The actual error may behave better, including being zero for a linear function.