Composite trapezoidal rule

Each subinterval forms a trapezoid from two adjacent heights. An interior grid value belongs to two trapezoids and receives the factor 1/2 twice, for total weight 1. The first left and final right values each occur in only one trapezoid and keep weight 1/2. This yields Tn=h(f0/2+f1+…+f(n−1)+fn/2). Equal weights would count the endpoint area incorrectly. With unevenly spaced measurements, calculate each trapezoid using its own interval width. With uniform step h, the familiar endpoint weights are one half. With unequal spacing, each subinterval needs its own width rather than one global h.

No. For a convex function, the secant between two sample points lies above the graph, so the trapezoid overestimates the integral on that interval. For a concave function, the secant lies below and the area is underestimated. If curvature changes, local errors may partly cancel. The total error sign therefore requires additional conditions. For twice continuously differentiable functions, an error bound depends in part on the maximum second derivative and the step size. For x² on [0,1], the secant lies above the graph, so the error is positive. For −x², its sign reverses on the same interval.

Refine the uniform grid by increasing n and reducing h=(b−a)/n. For sufficiently smooth functions, the composite trapezoidal rule’s global error typically scales as h². Halving h therefore reduces the error by about a factor of four once the asymptotic regime is reached, at the cost of roughly twice as many function evaluations. Measurements may not be available at finer spacing; then use uncertainty estimates or a higher-order quadrature method. When possible, compare results at two step sizes rather than inferring accuracy from many displayed decimal places.

If f is twice continuously differentiable and the grid is uniform with step h, then |I−T_n|≤(b−a)h² max|f″|/12. The bound shows how curvature matters: a linear function has f″=0 and is integrated exactly, while stronger curvature allows a larger error. For f(x)=x² on [0,1] with n=2, h=0.5 and max|f″|=2. The bound is 1·0.25·2/12=1/24≈0.0417, equal here to the actual error 0.375−1/3. In applications, max|f″| is often known only roughly, so the bound remains conservative. Since max|f″| is an upper curvature bound, the actual error is often smaller.

For a sufficiently smooth function, the leading trapezoidal error on uniform grids is of order h². Compute T_n with step h and T_2n with h/2; then (4T_2n−T_n)/3 cancels the leading error term. For the integral of x² on [0,1], T₁=0.5 and T₂=0.375, so the combination (4·0.375−0.5)/3=1/3, the exact value. This improvement requires both grids to use the same endpoints and the asymptotic error regime to apply. With discontinuities or highly irregular measured data, the extrapolation is not reliable. It often improves the number of reliable decimal places.