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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.

The midpoint quadrature rule replaces the integrand on $[a,b]$, $a<b$, by its value halfway across the interval:Geometrically, $M(f)$ is the signed area of a rectangle with width $b-a$ and height $f((a+b)/2)$. Its single quadrature node is $(a+b)/2$ and its quadrature weight is $b-a$. Sampling in the interior is useful when endpoint evaluation is inconvenient, although an endpoint singularity can invalidate the quadrature error estimates below.

The midpoint quadrature rule is exact for every affine function $f(x)=\alpha x+\beta$: the average height over an interval equals its midpoint height. For example, on $[0,2]$, $f(x)=2x+1$ gives $I(f)=6$ and $M(f)=2f(1)=6$. Positive and negative area discrepancies cancel for a sloping straight line. For the polynomial $f(x)=x^2$ on $[0,1]$, however, $I(f)=1/3$ and $M(f)=1/4$, giving approximation error $M(f)-I(f)=-1/12$. Thus its degree of precision of a quadrature rule is exactly one.^[ref]

For the polynomial $f(x)=x^2$ on $[0,1]$, four panels give composite midpoint quadrature rule value $21/64$ and composite trapezoidal rule value $11/32$, bracketing the exact integral $1/3$. Individual midpoint rectangles can cross the curve; it is their total signed area that underestimates the integral.

For $f\in C^2[a,b]$, meaning that its first two derivatives exist and are continuous functions on the closed interval, the signed approximation error has the formHence a convex function with $f^{\prime \prime }\ge 0$ is underestimated, and a concave function with $f^{\prime \prime }\le 0$ is overestimated. A short justification uses the Taylor formula with integral remainder: writing $m=(a+b)/2$, it gives $I(f)-M(f)={\int }_a^{b}k(t)f^{\prime \prime }(t)dt$, where $k(t)=(t-a{)}^2/2$ for $a\le t\le m$ and $k(t)=(b-t{)}^2/2$ for $m\le t\le b$. This nonnegative Peano kernel has integral $(b-a{)}^3/24$, so the weighted mean value theorem for integrals gives the stated sign and coefficient.

For a real-valued linear error functional annihilating $P_n$, the Peano kernel iswhere $x$ is the argument on which the functional acts. Under the standard boundedness and extension hypotheses, the Peano kernel theorem statesFor example, a bounded functional on continuous functions, represented by a finite signed measure and extended to the truncated powers, has this representation. Quadrature error functionals have precisely this structure.

The source's bare linearity assumption on $C^{n+1}$ is not by itself a complete theorem hypothesis. Truncated powers need not belong to that domain, and interchange of the functional with an integral needs justification. For instance $X[u]=u^{(n+1)}(t_0)$ annihilates $P_n$ but would require a Dirac point mass rather than an ordinary integrable kernel. The statement above supplies the usual functional class needed for the requested theorem and applies directly to part (c).

Let $L(f)=\mathcal{T}[f]-f^{\prime }(1/3)$. Direct substitution shows $L(1)=L(x)=L(x^2)=0$, so the functional annihilates all quadratic polynomials. Taylor's integral-remainder formula isApplying $L$ and interchanging the finite evaluations and derivative with the continuous integral gives the Peano kernel representation $L(f)={\int }_0^{1}K(t)f^{\prime \prime \prime }(t)dt$, whereAll three expressions are nonnegative; the middle quadratic has positive leading coefficient and discriminant $-1/9$. Therefore the best possible norm bound has constant ${\int }_0^{1}K(t)dt$. Evaluate that integral particularly simply by using $f(x)=x^3/6$, for which $f^{\prime \prime \prime }=1$:It follows thatThe same cubic attains equality, so the constant is sharp, not merely sufficient. This is the sharp Peano bound for an interior three-point first derivative.

Let $L(f)=f^{\prime }(0)-\mu (f)$. Direct evaluation on $1,x,x^2$ gives $L=0$, so the Peano kernel theorem givesThis representation also follows by applying $L$ to the Taylor expansion of $f$ with integral remainder. The derivative at zero of the truncated quadratic is zero for $t\ge 0$, and its sampled values giveThe Peano kernel is nonnegative, andHence $|L(f)|\le \frac{1}{3}\Vert f^{(3)}{\Vert }_{\infty }$. For $f(x)=x^3/6$, the third derivative is $1$, while $f^{\prime }(0)=0$ and $\mu (f)=-1/3$. Thus equality holds, proving the least constant is $c=1/3$, attained by $f(x)=x^3/6$.

Start with Taylor's theorem with integral remainder, written on the whole interval using truncated powers:For the Peano kernel theorem, require that $L$ be a linear error functional, annihilate all polynomials of degree at most $k$, and permit interchange with this integral. These conditions hold, in particular, for finite linear combinations of point evaluations of $f$ and its derivatives through order $k$, with the truncated-power derivatives interpreted almost everywhere in $\theta$; dominated integration justifies the interchange. Another sufficient setting is a bounded functional on a differentiability space where the truncated-power integral is convergent in its norm. Polynomial exactness alone, without linearity and an integration-interchange justification, is not sufficient.

Applying $L$ annihilates the Taylor polynomial and givesThis is the unnormalized Peano kernel convention used by the question; including $1/k!$ within the kernel instead removes the factor outside the integral.

For the stated derivative error, constants and linear and quadratic polynomials are annihilated, so $k=2$. Acting on $(x-\theta {)}_{+}^2$ givesThus the sharp central-secant derivative error satisfiesHence $c=1/6$ works. It is the minimum: for $f(x)=x^3$, $L(f)=3-4=-1$ and $\Vert f^{\prime \prime \prime }{\Vert }_{\infty }=6$, so no smaller constant can satisfy the inequality. ThereforeThe sign-definite kernel explains why a constant third derivative attains the bound exactly.

Assume $L$ is linear, annihilates all polynomials of degree at most $k$, is defined on the truncated-power kernels and permits interchange with the remainder integral. For $k\ge 1$, a convenient sufficient condition is that $L$ be a bounded linear functional on $C[a,b]$; more general approximation functionals need the corresponding continuity/interchange property on their domain. Taylor's integral remainder givesApplying $L$, the polynomial terms vanish, and the Peano kernel theorem givesHere $t_{+}=\max (t,0)$. This is the printed raw-kernel convention; the normalized Peano kernel includes $1/k!$ inside its definition instead, and then has no factorial outside the integral. The interchange is justified by boundedness applied to the limit of Riemann sums of the continuous kernel-valued integrand. For a kernel with $k=0$, its discontinuity requires a function space/interchange assumption accommodating the step functions as well.

For the specified Simpson's rule error, direct evaluation on $1,x,x^2,x^3$ gives zero, so $k=3$. For $0\le \theta \le 2$,Simplifying separately on the two halves givesBoth pieces are strictly negative in the open interval; they agree at the midpoint. Their integral is $-1/15$, as direct polynomial integration shows. ThusThis is optimal, not merely an upper bound. For $f(x)=x^4/24$, the fourth derivative is identically one and $L(f)=-1/90$, so the minimum constant is $\rho =1/90$.

For the Peano kernel theorem, the integral Taylor remainder isHere $u_{+}=\max (u,0)$. The error functional must be linear, vanish on every polynomial of degree at most $k$, and permit its action to be interchanged with this remainder integral. The last regularity condition holds for the finite combinations of point evaluations and derivative evaluations used below, by applying the corresponding differentiated remainder formula. Under these conditions the polynomial terms vanish, yielding

For the three-point asymmetric second-derivative formula, choose the error convention $\lambda (f)=f^{\prime \prime }(2)-L(f)$. Exactness on $1,x,x^2$ giveshenceApplying $\lambda$ to $(x-\theta {)}_{+}^2$, for almost every $\theta \in [0,3]$, gives the Peano kernelThe value at $\theta =2$ is immaterial to the integral. The sketch rises from zero through $2/3$ at one to the left limit $5/3$ at two, jumps down to the right limit $-1/3$, and then rises to zero at three.

The sharp error constants for a Peano kernel are the corresponding integral-operator norms. HereAs $k!=2$,Indeed, $|\lambda (f)|\le (\Vert K{\Vert }_{\infty }/2)\Vert f^{(3)}{\Vert }_1$ and $|\lambda (f)|\le (\Vert K{\Vert }_1/2)\Vert f^{(3)}{\Vert }_{\infty }$. They are best possible even for $f\in C^3$: for $r$, choose continuous nonnegative third derivatives concentrated just left of two with unit integral; for $s$, approximate the sign of $K$ by continuous functions of supremum norm one. Integrating any such third derivative three times supplies an admissible $f$, and the resulting ratios approach the stated constants.

For $f(x)=x^3$, $L(f)=-1+27/3=8$, while $f^{\prime \prime }(2)=12$, so $\lambda (f)=4$. Also $\Vert f^{(3)}{\Vert }_1=18$ and $\Vert f^{(3)}{\Vert }_{\infty }=6$. The two bounds explicitly read

The Peano kernel theorem says that if a continuous linear error functional $L$ annihilates all polynomials of degree less than $r$ and permits the required interchange with integration, then for $f\in C^r[a,b]$,It follows by applying $L$ to the integral-remainder Taylor formula with integral remainder. Here choose $r=2$ andThis functional annihilates constants and linear functions (indeed quadratics too), and its Peano kernel isThus ${\int }_0^3|K|d\theta =1/2$ and $\max |K|=1/3$. The integral representation givesFor the first inequality, continuous functions $f^{\prime \prime }$ with $\Vert f^{\prime \prime }{\Vert }_{\infty }\le 1$ can approximate $\mathrm{sgn}K$ away from its zeros, making $L(f)$ arbitrarily close to $\int |K|$. Such derivatives integrate twice to admissible $C^2$ functions, proving no smaller $\alpha$ works. For the second, use continuous $f^{\prime \prime }$ of $L^1$ norm one concentrated near $\theta =1$ or $2$, where $|K|=1/3$, with the sign of $K$ there. This makes $|L(f)|$ approach $1/3$, proving optimality of $\beta$.

For $f=x^3$, $L(f)=2$, $\max |f^{\prime \prime }|=18$ and $\Vert f^{\prime \prime }{\Vert }_1=27$. Both right-hand bounds equal nine, so both inequalities hold.

Let $x_1,\dots ,x_k$ be the distinct interior roots of $p_n$ where its sign changes, and put $q(x)={\prod }_{j=1}^k(x-x_j)$. If $k<n$, orthogonality gives ${\int }_a^{b}w{p}_{n}q=0$. But $p_{n}q$ has constant sign on $(a,b)$, apart from its isolated zeros, because every sign change has been canceled. It is not identically zero, and the positive weight makes its integral nonzero, a contradiction. Thus $k\ge n$. Degree $n$ forces exactly $n$ distinct simple roots, all inside $(a,b)$.

For distinct nodes, let ${\ell }_i(x)={\prod }_{j\ne i}(x-c_j)/(c_i-c_j)$. Lagrange interpolation polynomial gives $f={\sum }_{i}f(c_i){\ell }_i$ for every polynomial of degree at most $n-1$. Integrating this identity proves exactness with $b_i={\int }_a^{b}w{\ell }_i$.

For the Gaussian quadrature nodes, divide any $f\in {\mathbb{P}}_{2n-1}$ as $f=q{p}_n+r$, with $q,r\in {\mathbb{P}}_{n-1}$. The integral of $q{p}_n$ vanishes by orthogonality, and its nodal values vanish because $p_n(c_i)=0$. The remaining polynomial $r$ is integrated exactly by the preceding interpolation result. HenceIts weights are positive: exactness applied to ${\ell }_i^2$ gives $b_i={\int }_a^{b}w{\ell }_i^2>0$.

Define the error functional $L(f)={\int }_a^{b}w(x)f(x)dx-{\sum }_i{b}_{i}f(c_i)$. It annihilates polynomials of degree at most $2n-1$. For $f\in C^{2n}[a,b]$, the Peano kernel theorem giveswhere a formal Peano kernel with the normalization made explicit isNo evaluation of $K$ is needed. The supplied inner product presupposes an integrable positive weight on the finite interval, so the integral and point-evaluation error functional is bounded on continuous functions.

Repeated integration by parts inproduces the first $k+1$ terms of the Taylor polynomial and leaves $f(x)$, proving the integral remainder in Taylor theoremApply the linear functional ${\lambda }_k$ to this identity. It annihilates the Taylor polynomial, and commuting it with integration gives the Peano kernel theorem

For the midpoint quadrature rule on $[0,1]$, $e[1]=e[x]=0$, so $k=1$. Its Peano kernel isFor $f(x)=x^2$, the formula givesagreeing with ${\int }_0^1{x}^{2}dx-(1/2{)}^2=1/3-1/4$.

For $h>0$, the third forward difference $L_{h}u=-u(0)+3u(h)-3u(2h)+u(3h)$ annihilates polynomials of degree at most two. Its Peano kernel is $K_h(t)=\frac{1}{2}[3(h-t{)}_{+}^2-3(2h-t{)}_{+}^2+(3h-t{)}_{+}^2]$ for $0\le t\le 3h$, extended by zero outside that interval. It is $h^2$ times the quadratic cardinal B-spline evaluated at $t/h$, is positive on $(0,3h)$, and has integral $h^3$. The Peano kernel theorem therefore represents $L_{h}u$ as a positive weighted average of the third derivative: $L_{h}u=h^3{u}^{(3)}(\xi )$ for some interior $\xi$ when that derivative is continuous. The kernel is the smooth error weight hidden inside a discrete difference.

For $f\in C^3[x-h,x+h]$, the Peano kernel for the error of the central secant slope has one sign. Integrating its absolute value gives the displayed constant $h^2/6$, and a cubic polynomial attains it. At $x=1$, $h=1$, the unnormalized order-two kernel is $-{\theta }^2/2$ on $[0,1]$ and $-(2-\theta {)}^2/2$ on $[1,2]$; the integral representation includes the factor $1/2!$.

For $L(f)=-5f(0)/3+4f(1/2)/3+f(1)/3-f^{\prime }(1/3)$ on $C^3[0,1]$, quadratic polynomials are annihilated. Taylor's formula with integral remainder gives $L(f)={\int }_0^{1}K(t)f^{\prime \prime \prime }(t)dt$ with $K(t)=2(1/2-t{)}_{+}^2/3+(1-t{)}^2/6-(1/3-t{)}_{+}$. This Peano kernel is nonnegative throughout the interval and integrates to $1/36$, giving the sharp bound; $f(x)=x^3/6$ attains equality. The derivative is at an interior point, so this is distinct from an endpoint differentiation formula.

For $L(f)=f^{\prime }(0)+\frac{3}{2}f(0)-2f(1)+\frac{1}{2}f(2)$ on $C^3[0,2]$, the Peano kernel isThe Taylor expansion with integral remainder gives $L(f)={\int }_0^{2}K(t)f^{(3)}(t)dt$, because $L$ annihilates polynomials of degree at most two. Since $K\ge 0$ and ${\int }_0^{2}K=1/3$, the sharp bound is $|L(f)|\le \frac{1}{3}\Vert f^{(3)}{\Vert }_{\infty }$. Equality holds for $f(x)=x^3/6$.

Matching the monomials $1,x,x^2$ gives $f^{\prime \prime }(2)\approx (2/3)f(0)-f(1)+(1/3)f(3)$. With error defined as the true derivative minus this approximation, its Peano kernel is $K(\theta )=21_{\{\theta <2\}}+(1-\theta {)}_{+}^2-(3-\theta {)}^2/3$ on $[0,3]$. The derivative evaluation causes a jump at two. Applying the Peano kernel theorem gives $\lambda (f)=\frac{1}{2}{\int }_0^{3}K{f}^{(3)}$.

As a quadrature rule, the trapezoidal rule replaces an integrand by the linear interpolation joining its two endpoint values:For nonnegative endpoint values, this is the area of a trapezoid; in general it is a signed area. The quadrature nodes are $a,b$, each with quadrature weight $(b-a)/2$. Unlike the midpoint quadrature rule, it uses endpoint samples, which can be reused between adjacent panels.

Every affine function $f(x)=\alpha x+\beta$ agrees with its linear interpolation, so the trapezoidal rule integrates it exactly. On $[0,2]$, $f(x)=2x+1$ gives $T(f)=(2/2)(1+5)=6=I(f)$. On $[0,1]$, the polynomial $f(x)=x^2$ instead gives $T(f)=1/2$, exceeding $I(f)=1/3$ by $1/6$. Its degree of precision of a quadrature rule is exactly one.

If $f\in C^2[a,b]$, its signed approximation error satisfiesA convex function lies below its endpoint chord, so the trapezoidal rule overestimates its integral; a concave function gives the reverse inequality. The error formula assumes that the second derivative exists and is a continuous function throughout the closed interval.^[ref] A short derivation is the identityobtained by two integrations by parts. The nonnegative Peano kernel has integral $(b-a{)}^3/12$, and the weighted mean value theorem for integrals gives the formula.

The trapezoidal rule for an ordinary differential equation averages vector fields at the old and new states and is A-stable.

This time-stepping use applies the same trapezoidal rule to the integral of the derivative in an ordinary differential equation. That connection explains its place under implicit Runge-Kutta methods; its quadrature behavior and the nonlinear issue in trapezoidal rule fails B-stability concern different uses of the same endpoint averaging.