The Composite Trapezoidal Error Bound

Why the Error Depends on Curvature

The composite trapezoidal rule estimates an integral by dividing an interval into equal panels and approximating the function by a straight line on each panel. A straight line captures the endpoint values exactly, but it does not generally follow the curve between them. The size of that discrepancy depends on how much the function bends: a bound on the magnitude of its second derivative controls the error.

This tutorial derives an error estimate for a fixed interval and a fixed number of equal-width panels. The result is useful because it gives a guaranteed upper bound, not just a numerical approximation. It also explains why doubling the number of panels reduces this bound by a factor of four, provided the same second-derivative bound applies.

Rule and Assumptions

Let \(f\) be twice continuously differentiable on the closed interval \([a,b]\), where \(a<b\). Choose a positive integer \(n\), divide the interval into \(n\) equal panels, and let \(h=(b-a)/n\) be the width of each panel. The grid points are \(x_i=a+ih\) for \(i=0,1,\ldots,n\).

Here \(T_n\) is the composite trapezoidal estimate of \(\int_a^b f(x)\,dx\). The smoothness assumption ensures that the second derivative exists and is continuous throughout the interval. Consequently, its magnitude has a maximum there. Define that maximum by

The bound applies to the error in the whole composite estimate. It requires the stated smoothness on the entire closed interval; it is not automatically applicable if the function or its second derivative is undefined or unbounded at a point in the interval.

Deriving the Bound on One Panel

First consider just one panel \([u,v]\), with width \(h=v-u\). Let \(L(x)\) be the straight-line function that agrees with \(f\) at the two endpoints. The area under \(L\) is the trapezoid area:

To bound the difference between \(f\) and \(L\), we use the linear interpolation remainder. Here is a direct justification. Fix any interior point \(x\in(u,v)\), and set

The definition of \(L\) gives \(g(u)=g(v)=0\), and the choice of \(K\) gives \(g(x)=0\). Rolle’s theorem applied between these three zeros shows that \(g'\) has two zeros, one in \((u,x)\) and one in \((x,v)\). A further application of Rolle’s theorem gives a point \(\xi\in(u,v)\) where \(g''(\xi)=0\). Since \(L\) is linear and the second derivative of \((t-u)(t-v)\) is \(2\), this means

Therefore \(K=f''(\xi)/2\). Substituting the definition of \(K\) gives the remainder formula at the chosen point:

Because \(|f''(\xi)|\le M_2\) and \((x-u)(x-v)\le 0\), it follows that

This also holds at the endpoints, where both sides are zero. Integrate the inequality across the panel. With \(s=x-u\), the distance \(v-x\) is \(h-s\), so

Thus one panel contributes an error of at most \(M_2h^3/12\). This local estimate is the ingredient needed for the composite bound.

Adding the Panel Errors

On panel \(i\), from \(x_i\) to \(x_{i+1}\), the trapezoid estimate is \(h(f(x_i)+f(x_{i+1}))/2\). Summing these panel areas produces \(T_n\): each interior grid value occurs in two adjacent panels, with a combined coefficient of \(h\), while each endpoint occurs once with coefficient \(h/2\).

The integral over \([a,b]\) is likewise the sum of the integrals over the \(n\) panels. Applying the triangle inequality to the difference between these sums, and then applying the one-panel estimate to every panel, gives

Since \(h=(b-a)/n\), substituting this width gives the theorem:

The inverse-square dependence follows from adding \(n\) local errors, each proportional to \(h^3\), while the width \(h\) itself shrinks like \(1/n\). In symbols, \(n h^3=n((b-a)/n)^3=(b-a)^3/n^2\).

Worked Examples

A Common Error: Forgetting to Add All Panels

A one-panel bound is not yet a bound for the whole composite rule. For example, with \(f(x)=x^2\) on \([0,1]\) and \(n=4\), the local bound on one panel is

It would be incorrect to claim that \(1/384\) bounds the total error: the actual total error is \(1/96\), which is larger. There are four panels, so adding their local bounds gives

The correction is to sum the panel bounds before substituting \(h=(b-a)/n\). That missing factor of \(n\) is precisely why the composite bound has \(n^2\) in the denominator rather than \(n^3\).

Using the Estimate Carefully

The bound is a guarantee when its assumptions hold, but it is not generally the actual error. It may be exact, as in the quadratic example, or larger than the error, as in the cubic and sine examples. A smaller bound is more informative, but the theorem itself does not assert that the error equals its right-hand side.

To apply the result, one must have a valid upper bound for \(|f''(x)|\) across the full interval. The theorem does not supply that bound automatically. If \(f\) is not twice continuously differentiable on the closed interval, or if no finite bound on \(|f''|\) is available there, this particular estimate cannot be invoked as stated. In particular, applying it across an endpoint singularity without checking the assumptions is not justified.

The factor \((b-a)^3\) also matters: the estimate depends on the length of the interval as well as the panel count. For a fixed interval and a fixed \(M_2\), replacing \(n\) by \(2n\) changes the bound by the factor \(n^2/(2n)^2=1/4\). This describes the change in the guaranteed error bound; it is not a claim that every actual error must decrease by exactly that factor.