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Apply the midpoint quadrature rule on each of $n$ equal panels of a partition of an interval, with $h=(b-a)/n$:Each quadrature node is a panel midpoint and each quadrature weight is $h$. For $f\in C^2[a,b]$, put $K={\max }_{x\in [a,b]}|f^{\prime \prime }(x)|$. Summing the panel quadrature errors givesfor some $\xi \in (a,b)$. Each panel contributes at most $K{h}^3/24$; there are $n=(b-a)/h$ panels, explaining the composite midpoint rule error of order $O(h^2)$ in big O notation. On a nonuniform partition of an interval with panel widths $h_j$ and largest width $H$, the corresponding bound is $K{\sum }_j{h}_j^3/24\le K(b-a)H^2/24$.

For $f(x)=x^2$ on $[0,1]$, summing the sampled polynomial exactly yieldsDoubling $n$ divides the absolute error by exactly four. For a general $C^2$ function, halving $h$ divides the displayed error bound by four, but the actual absolute error need not follow that ratio: cancellation can make it smaller or zero. Continuous functions still give convergence as the largest panel width tends to zero, by uniform continuity; without the bounded second derivative, the second-order rate is not guaranteed. Noise sensitivity of a quadrature rule explains a separate limitation when samples are inexact.

On an equal-panel partition of an interval, with $h=(b-a)/n$, sum the trapezoidal rule over adjacent panels:There are $n+1$ distinct quadrature nodes. Each interior sample belongs to two panels and therefore has quadrature weight $h$, while each endpoint has quadrature weight $h/2$. Equivalently, the composite trapezoidal rule integrates the piecewise linear function obtained by linear interpolation exactly.^[ref]

For $f\in C^2[a,b]$ and $K={\max }_{x\in [a,b]}|f^{\prime \prime }(x)|$, summing panel quadrature errors givesfor some $\xi \in (a,b)$. Thus panel quadrature errors of order $O(h^3)$ accumulate to order $O(h^2)$, using big O notation.^[ref] On a nonuniform partition of an interval with widths $h_j$ and largest width $H$, the bound is $K{\sum }_j{h}_j^3/12\le K(b-a)H^2/12$.

For the polynomial $f(x)=x^2$ on $[0,1]$, direct summation yieldsBoth quadrature errors shrink by exactly four when $n$ doubles, with opposite signs. This exact relation belongs to this polynomial example; the trapezoidal rule is not always twice as inaccurate as the midpoint quadrature rule.

The left plot uses exact polynomial sums; the right adds a common sample bias $\delta =0.002$ to the composite trapezoidal rule. That mathematical model converges to $1/3+\delta$, showing why smaller panels alone do not ensure accuracy for noisy samples. It illustrates noise sensitivity of a quadrature rule, rather than measured data.

Suppose $f$ has a fourth derivative that is a continuous function on the interval $[a,b]$, and $|f^{(4)}(x)|\le M$ there. With $I={\int }_a^{b}f(x)dx$, the signed quadrature error here is $I-S_n$. Then, for some point $\xi \in (a,b)$,In particular, a strictly positive fourth derivative makes $I-S_n<0$: composite Simpson's rule overestimates the integral.^[ref]

The bound's scale follows by adding the panel errors. A panel of width $2h$ has signed quadrature error $-h^5{f}^{(4)}({\xi }_j)/90$ and absolute error at most $M{h}^5/90$. There are $(b-a)/(2h)$ panels, so the triangle inequality gives $M(b-a)h^4/180$. The intermediate value theorem for the continuous fourth derivative replaces the average of the panel values $f^{(4)}({\xi }_j)$ by a single value $f^{(4)}(\xi )$, giving the signed formula.

The underlying single-panel calculation can be seen through the Peano kernel theorem. On the interval $[0,2]$, define the raw Peano kernel $K(t)=L[(x-t{)}_{+}^3]$, where $L$ is integral minus Simpson's rule and $(x-t{)}_{+}=\max (x-t,0)$ is the positive part. Then $K(t)=-t^3(4-3t)/12$ for $0\le t\le 1$ and $K(t)=-(2-t{)}^3(3t-2)/12$ for $1\le t\le 2$. It is strictly negative inside the interval and has integral $-1/15$. ThusThe uniform norm $\Vert f^{(4)}{\Vert }_{\infty }$ bounds the fourth derivative. The polynomial $f(x)=x^4/24$ attains equality. Scaling to endpoint spacing $h$ multiplies the coefficient by $h^5$, giving the sharp single-panel bound $h^5\Vert f^{(4)}{\Vert }_{\infty }/90$. The normalized Peano kernel is $K/3!$, so the external factorial must be omitted when using that normalization.

For the exponential function on $[0,1]$, $f^{(4)}(x)=e^x$, so $M=e$ and the bound is $e/(180n^4)$. Its leading actual absolute error is $(e-1)/(180n^4)$, smaller than the bound because a uniform upper bound on the fourth derivative does not use its variation across the interval. The theorem bounds truncation error with exact function evaluations; sample errors and roundoff error are additional contributions, as described by noise sensitivity of a quadrature rule.

Numerical integration approximates a definite integral using finitely many sampled function values. For an integrand $f$ on a finite interval $[a,b]$, $a<b$, a quadrature rule has the formThe quadrature nodes $x_j$ tell us where to evaluate $f$; the quadrature weights $w_j$ turn those values into an approximation to the accumulated signed area. The exact definite integral depends on the integrand throughout the interval, whereas the quadrature rule sees only these samples. Integrating a chosen interpolant exactly still generally approximates the original integrand's integral.^[ref]

The fundamental theorem of calculus evaluates $I(f)$ as $F(b)-F(a)$ when an antiderivative $F$ is available. Numerical integration is useful when no convenient antiderivative is known, or each evaluation of the integrand requires an expensive computation. The task then becomes choosing informative quadrature nodes and controlling the quadrature error with as few evaluations as possible. Fixed data also permit a quadrature rule, but refinement cannot supply new information between observations.^[ref]

For an exact comparison, take the polynomial $f(x)=x^2$ on $[0,1]$. Its antiderivative $x^3/3$ gives $I(f)=1/3$. On a partition of an interval into $n\ge 1$ equal panels, the composite midpoint quadrature rule uses $n$ midpoint samples, each with quadrature weight $1/n$; the composite trapezoidal rule uses $n+1$ endpoint samples, with outer quadrature weights $1/(2n)$ and interior quadrature weights $1/n$. The existing derivations under these two quadrature rules giveIn particular, $M_1=1/4$ and $T_1=1/2$ already bracket $1/3$; two panels give $M_2=5/16$ and $T_2=3/8$. The convex function $x^2$ explains the opposite signs. A midpoint rectangle can cross the curve within its panel, so its total signed area, rather than pointwise ordering, determines the underestimate.

Define the signed approximation error as $E_n=Q_n(f)-I(f)$. In this example the absolute errors are $1/(12n^2)$ and $1/(6n^2)$: doubling $n$ reduces each by exactly four. A plot with both axes on a logarithmic scale therefore has slope $-2$, expressing order two with respect to panel refinement. This is an exact calculation for this polynomial, including $n=1$, rather than an observed rate extrapolated from a few computed answers.

For a general integrand, the composite midpoint quadrature rule and composite trapezoidal rule give second-order quadrature error bounds when $f$ has two continuous derivatives on $[a,b]$ and a known bound on $|f^{\prime \prime }|$. These hypotheses describe behavior between samples. A formal error formula containing an unknown derivative is not a numerical accuracy certificate or a justified sample count. Finite samples do not certify quadrature accuracy makes the missing information explicit. Errors in the evaluations and roundoff error must also be considered; noise sensitivity of a quadrature rule treats the former.

More elaborate choices exploit other information: Gaussian quadrature obtains high polynomial exactness from carefully chosen quadrature nodes, and Simpson's rule uses a quadratic interpolant. Adaptive numerical integration compares approximations and concentrates samples where they disagree, often saving evaluations. Such a difference is an error estimate, not a certified bound without further assumptions: both approximations can miss the same feature.^[ref]

A quadrature weight multiplies a sampled integrand value in a quadrature rule. For ordinary numerical integration, its units match those of the integration variable. Positive quadrature weights that sum to $b-a$ make $Q(f)/(b-a)$ a weighted average of the samples; signed quadrature weights can magnify sample perturbations, as quantified by noise sensitivity of a quadrature rule.