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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.

Simpson's rule approximates a definite integral by integrating the quadratic polynomial obtained by polynomial interpolation at two endpoints and their midpoint. For an interval $[a,b]$ with $a<b$, write $m=(a+b)/2$ and $h=(b-a)/2$. The three quadrature nodes are $a,m,b$, and the quadrature rule isThus its quadrature weights are $h/3,4h/3,h/3$. Their sum is $2h=b-a$, as required for exactness on constant functions. This is the three-node Newton-Cotes closed quadrature; its degree of precision of a quadrature rule is three, even though the interpolating polynomial has polynomial degree at most two.^[ref]

For an original worked example, take the cubic polynomial $f(x)=x^3-2x+4$ on the interval $[-1,2]$. Here $m=1/2$ and $h=3/2$, so the function evaluations giveSimpson's rule givesAn antiderivative supplies an independent exact definite integral:The quadratic polynomial through the same quadrature nodes is $p_2(x)=\frac{3}{2}{x}^2-\frac{1}{2}x+3$. It differs from $f$ between the quadrature nodes, but its definite integral agrees exactly. The picture shows why: the signed polynomial interpolation error has two lobes with opposite integrals.

Three function evaluations therefore capture more information about an integral than about the whole function. When an antiderivative is unavailable, composite Simpson's rule repeats this inexpensive local construction over a partition of an interval. For a sufficiently smooth function, halving the node spacing reduces its truncation error by approximately a factor of sixteen.