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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.

Take $n$ to be a positive even integer, as required by composite Simpson's rule. There are $n/2$ two-strip panels, each of width $2/n$. Applying the supplied panel bound with the global fourth-derivative minimum $0$ and maximum $1$, and summing, givesHere $E_n$ is the nonnegative absolute error. In this example Simpson's rule overestimates the integral, so $E_n=S_n-I$. Each panel has that same sign because its fourth derivative is nonnegative; cancellation is not being assumed.^[ref]

Choose equation (1), evaluated by refined composite Simpson's rule. It gives $y$ directly as a one-dimensional definite integral with fixed endpoints and a smooth integrand. It therefore avoids root finding altogether. Before evaluating, use the trigonometric identityThis equivalent integrand is nonnegative throughout $[0,\pi /2]$, avoiding cancellation between the original two terms. Equation (2) instead subtracts two potentially large exponential functions, while equation (3) would ordinarily put a numerical integration inside a root finding loop.

For even $n$, let $h=(\pi /2)/n$, $\varphi (x)=2\sin x\cos ((\pi /4)\cos x)$, and computeThen double $n$. The integrand has bounded derivatives of every fixed order on the integration interval; in the fourth-order error regime, the estimated remaining truncation error in $Y_{2n}$ is $|Y_{2n}-Y_n|/15$. The factor $15$ follows by eliminating the leading term $C{h}^4$ between the two mesh widths.^[ref]

Stop when this estimate is comparable to ${\epsilon }_{\mathrm{mach}}\max (1,|Y_{2n}|)$ and further refinement shows stagnation at the roundoff error scale. Check that the differences previously decreased consistently with fourth-order convergence; differences alone are not a rigorous error certificate if cancellation or roundoff error dominates. Compensated summation helps keep the summation error near the precision limit. Refining indefinitely after this point increases work without reliably adding meaningful digits. This answers the requested choice and termination strategy without requiring an explicit value of $y$.

Let $h=(b-a)/n$ and let $M_4={\max }_{x\in [a,b]}|f^{(4)}(x)|$. A composite Simpson's rule panel covers two strips, so it has width $2h$ and absolute error at mostThere are $n/2$ panels. By the triangle inequality, the absolute error of their sum is at most the sum of their absolute errors. HenceThis assumes even $n$ and, for the usual Simpson remainder theorem, $f\in C^4[a,b]$. In particular its fourth derivative is bounded on the interval. It is the error bound for composite Simpson's rule, expressed in strip count rather than panel count.^[ref]

Simpson baseline integration for pulse-energy correction subtracts a three-point estimate of integrated background from a measured gate integral. For a baseline $b(t)$ on $[0,T]$, Simpson's rule givesIt is exact for a quadratic or cubic baseline, provided the three values represent background rather than pulse-contaminated samples. For a four-times differentiable baseline, the remainder is ${\int }_0^{T}bdt-B_S=-T^5{b}^{(4)}(\xi )/2880$ for some $0<\xi <T$.^[ref]

For an original chosen quadratic background, take $T=6\mu \mathrm{s}$ with beginning, midpoint and end values $2$, $3$ and $2\mathrm{mW}$. Its integrated baseline is $16\mathrm{nJ}$, whereas endpoint averaging would give $12\mathrm{nJ}$. A raw gate integral of $116\mathrm{nJ}$ therefore gives $100\mathrm{nJ}$ after the three-point correction. Overestimating only the midpoint by $0.50\mathrm{mW}$ would subtract an extra $2\mathrm{nJ}$.

The uncertainty of a three-point baseline integral is separate from its quadrature error. With baseline-sample covariance matrix $C_b$ and $\mathbf{w}=(T/6)(1,4,1{)}^T$, its variance is ${\mathbf{w}}^T{C}_b\mathbf{w}$. Independent samples of common standard uncertainty ${\sigma }_b$ give $u(B_S)=T{\sigma }_b/\sqrt{2}$; a perfectly shared additive baseline error of the same uncertainty gives $T{\sigma }_b$ instead. Covariance terms cannot be discarded.^[ref]

A matched blank acquisition or independently measured background can supply the midpoint. Historical NIST pulse-energy apparatus used before/after baselines for background and thermal-drift compensation.^[ref] The three-point example is an original extension, not a claim about that instrument's processing. Higher polynomial exactness does not compensate for an incorrectly identified baseline.

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.