Consider the numerical evaluation:

∫₀^π/2 sin(x) dx

whose exact value is 1. Using the trapezoidal method:

∫₀^π/2 sin(x) dx ≃ (π/2)/2 [sin(0) + sin(π/2)] = π/4 ≈ 0.785398164

This is a poor approximation of the exact value 1, due to approximating sin(x) over [0, π/2] with a degree 1 polynomial.

A better approach is to divide the integration interval [a,b] into n subintervals of length:

h = (b−a)/n

Let the points be x_i, i = 0,1,2,...,n with x₀ = a and xₙ = b. Applying the trapezoidal method on each subinterval [x_i,x_i+1]:

∫_a^b f(x) dx ≃ ∑_i=0^n-1 ∫_xi^x_i+1 f(x) dx ≃ ∑_i=0^n-1 h/2 [f(x_i) + f(x_i+1)]

Summing these contributions, all interior f(x_i) appear twice except the first and last:

∫_a^b f(x) dx ≃ h/2 [ f(x₀) + 2(f(x₁) + f(x₂) + ... + f(x_n-1)) + f(xₙ) ]

This is the composite trapezoidal formula. Regarding the error, each subinterval contributes an error − f″(η_i) / 12 h³. Denoting n = (b−a)/h, the total error becomes:

E ≃ − (b−a) / 12 f″(η) h², η ∈ [a,b]

Although the derivation is not fully rigorous (since each subinterval has a potentially different η_i), applying the mean value theorem gives the global error. The composite trapezoidal method is therefore of order 2.