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On a partition of an interval $a=x_0<x_1<\cdots <x_N=b$, apply the trapezoidal rule separately to each panel and add:For uniform spacing $h=(b-a)/N$, this composite trapezoidal rule becomesEach interior sample occurs in two panels; their half-quadrature weights combine to $h$. The endpoint quadrature weights are $h/2$.

A linear spline is a continuous function whose restriction to every interval $[x_{i-1},x_i]$ is a polynomial of polynomial degree at most one. With $h_i=x_i-x_{i-1}$ and ${\ell }_i=\ell (x_i)$, linear interpolation givesIntegrating the two linear polynomials and adding the contributions givesThis is the composite trapezoidal rule applied to the nodal values, and it is exact for the linear spline itself, including nonuniform spline knots.

A cubic spline on this partition of an interval is a function $S\in C^2[x_0,x_n]$ that restricts to a polynomial of polynomial degree at most three on every interval $[x_{i-1},x_i]$. Thus its second derivative is continuous and piecewise of polynomial degree at most one: $S^{\prime \prime }$ is a linear spline on the same spline knots. Its third derivative is constant on each open interval and can jump across the spline knots; its fourth derivative vanishes inside each open interval.

A quadrature error is the difference between an integral and its finite weighted approximation. Its bound depends on both the sampled function and how the weights are formed. Suppose $|f(x)-f(y)|\le C|x-y{|}^{\alpha }$ on $[a,b]$, with $0<\alpha \le 1$. For a partition of an interval of mesh of a partition $\Delta$ and left endpoints $x_{i-1}$,Indeed, integrate the Hölder continuity bound on each cell and add. For an integrable weight $w$, the same proof gives $C\Vert w{\Vert }_1{\Delta }^{\alpha }$ if each discrete weight is the exact cell integral ${\int }_{x_{i-1}}^{x_i}w(x)dx$. Sampling $w$ too introduces another error.

A different endpoint estimate concerns a normalized window $w=1_{[u,v]}/h$, where $h=v-u>0$. Approximate this indicator function by its left-endpoint value on each cell. Only cells meeting $u$ or $v$ can disagree on a set of positive length; their total length is at most $2\Delta$. Therefore its L1 norm error is at most $2\Delta /h$. This is the source of a $\Delta /h$ bound; it is not the Hölder continuity estimate for a smooth integrand.

A quadrature rule approximates an integral by a finite weighted sum of sampled function values:The quadrature nodes $x_i$ specify where to sample, and the quadrature weights $w_i$ specify how much each sample contributes. The density $\rho (x)$ specifies the target integral, whereas the discrete quadrature weights belong to the approximation. Ordinary numerical integration has $\rho =1$. Gaussian quadrature and the midpoint quadrature rule choose interior quadrature nodes; the trapezoidal rule includes endpoints.^[ref]

For ordinary numerical integration, exactness on the constant function $1$ requires ${\sum }_i{w}_i=b-a$; exactness on $x$ additionally requires ${\sum }_i{w}_i{x}_i=(b^2-a^2)/2$. These two checks make the quadrature rule exact for every affine function. For example, both the midpoint quadrature rule and trapezoidal rule give $6$ for ${\int }_0^2(2x+1)dx$. Testing progressively higher polynomial degrees defines the degree of precision of a quadrature rule. Exactness describes a class of integrands, rather than promising exact results for every function.

Composite midpoint quadrature rules and composite trapezoidal rules apply a simple quadrature rule on every panel of a partition of an interval. This buys accuracy by shrinking the panels while keeping the elementary formula fixed. For continuous functions sampled exactly, their quadrature errors tend to zero as the largest panel width tends to zero. The familiar second-order bounds require stronger assumptions about derivatives. The signed approximation error used here is $Q(f)-I_{\rho }(f)$; some sources define quadrature error with the opposite sign, so the convention must accompany any formula.

For continuous local martingales $M,N$ and a sequence in which each term is a partition of an interval whose mesh tends to zero, the sumsconverge in the sense of uniform convergence on compacts in probability to a continuous increasing process. To identify the limit, put $C=[M]+[N]$ and choose Radon-Nikodym derivativesIf $(U,V)$ is a centered bivariate normal distribution with covariance matrix $\left(\begin{array}{cc}a& c\\ c& b\end{array}\right)$, thenLocalizing, representing the pair as stochastic integrals against a two-dimensional Brownian motion, and approximating the integrands by bounded predictable step processes proves the convergence. The step-process case follows from the weak law of large numbers for independent Gaussian increments; the Burkholder-Davis-Gundy inequality controls the approximation error. Since $|\mathbb{E}[UV]|\le \mathbb{E}|UV|\le \sqrt{\mathbb{E}{U}^2\mathbb{E}{V}^2}$,

For a bounded real function on a closed finite interval, the Riemann integral exists when the supremum of its lower Darboux sums equals the infimum of its upper Darboux sums. Their common value is ${\int }_a^{b}f(x)dx$. Equivalently, for every $ϵ>0$ there is a partition of an interval whose upper-minus-lower sum is less than $ϵ$.^[ref]

For $f(x)=x$ on $[0,1]$, uniform partitions into $n$ pieces give lower and upper sums $1/2-1/(2n)$ and $1/2+1/(2n)$, proving that the integral is $1/2$. The Dirichlet function has lower sums zero and upper sums one on this interval, so it is not Riemann integrable. The definition explains why boundedness alone is insufficient.

Riemann integration defines an integral by requiring that every sufficiently fine tagged Riemann sum approximate the same number. For a bounded function $f:[a,b]\to \mathbb{R}$, a partition of an interval $a=x_0<\cdots <x_n=b$ and tags ${\xi }_i\in [x_{i-1},x_i]$ giveThe mesh of a partition must tend to zero, and the answer must be independent of the tags and the chosen partitions of an interval. The equivalent Darboux sum formulation traps every Riemann sum between a lower Darboux sum and an upper Darboux sum. Their common limiting value, when it exists, is the Riemann integral.^[ref]

For $f(x)=x$ on $[0,1]$, equal-width partitions of an interval give lower and upper values $1/2\mp 1/(2n)$, so the Riemann integral is $1/2$. For the Dirichlet function, every nondegenerate subinterval has infimum zero and supremum one: lower and upper values remain zero and one. Refining the partition of an interval cannot resolve that gap. Thus boundedness alone does not ensure Riemann integrability; the oscillation of a function on an interval must become small in the weighted aggregate.

For a bounded function on a fixed closed interval, Riemann integrability with Riemann integral $I$ is equivalent to convergence of all tagged Riemann sums to $I$ as the mesh of a partition tends to zero. Precisely, for every $\epsilon >0$ there is $\delta >0$ such that every partition of an interval of mesh less than $\delta$, with any tags in the corresponding closed subintervals, has its Riemann sum within $\epsilon$ of $I$.

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.

On a real interval, a step function is constant on the open cells of a finite partition of an interval, with endpoint values specified separately. Equivalently, up to those finitely many point values it is a finite linear combination of interval indicator functions. For example, a function equal to two on $[0,1/4)$, one on $[1/4,3/4)$ and zero on $[3/4,1]$ is a step function. Its integral is the sum of rectangle heights times widths.

A simple function is more general in measure theory: its finitely many value sets can be arbitrary measurable sets. The Dirichlet function $1_{\mathbb{Q}}$ on $[0,1]$ is simple but is not an interval step function, because every nonempty interval contains both rational and irrational points. Merely being constant on a finite or countable measurable partition therefore does not characterize the usual interval-based step functions.

Countably many jumps can be considered under explicitly stated extensions of the term. The finite-partition convention is the one used for elementary Riemann approximations; the measurable finite-range convention belongs to simple functions in Lebesgue integration.

On a partition of an interval symmetric about zero, the composite trapezoidal rule gives opposite quadrature nodes equal quadrature weights. Samples of an odd function cancel pairwise, and a zero quadrature node contributes zero. For a continuous odd function, its exact integral over $[-L,L]$ is also zero. Adding any affine function preserves exactness by linearity of the integral and of the quadrature rule. For example, $1+\sin x$ is nonpolynomial but both the exact integral and every uniform-grid composite trapezoidal rule value on $[-1,1]$ are two.

The total variation of a function $f:[a,b]\to \mathbb{R}$ measures its accumulated rises and falls, rather than just its change between the endpoints. For each partition of an interval $\Pi =(a=t_0<\cdots <t_n=b)$, add the absolute values of its sampled increments. Then take the supremum over all such partitions of an interval:The function has bounded variation when $V_a^b(f)<\infty$. This definition uses its value at every point of the interval, including both endpoints.^[ref]

For a nondecreasing function, all increments are nonnegative and the sum telescopes to $f(b)-f(a)$. More generally, the triangle inequality gives $|f(b)-f(a)|\le V_a^b(f)$. Equality can fail dramatically when the function changes direction: an object can travel a long distance and finish near its starting position. Here we measure movement of the function's value; arc length of its function graph also includes horizontal movement.

For a path of finite variation, its total-variation process iswhere the supremum is over every partition of an interval of $[0,t]$. It is an increasing process, and the signed measure $dA$ satisfies $|dA|=dV(A)$.

If Riemann integrable functions $f_n:[a,b]\to \mathbb{R}$ converge by uniform convergence to $f$, then $f$ is Riemann integrable andThis concerns one fixed compact interval, with finite length.^[ref]

Here is a proof outline that also establishes Riemann integrability of the uniform limit. Assume $a<b$, and put $\ell =b-a$ and $E_N=\Vert f_N-f{\Vert }_{\infty }$. One bounded $f_N$ and finite $E_N$ show that $f$ is a bounded function. For any partition of an interval $P$, its upper Darboux sum and lower Darboux sum satisfyGiven $\eta >0$, choose $N$ with $E_N<\eta /(4\ell )$, then a partition of an interval with $U(f_N,P)-L(f_N,P)<\eta /2$. The displayed gap is below $\eta$, proving Riemann integrability. Finally, $|f_n-f|\le E_n$ and monotonicity of the Riemann integral give the stated error estimate. When $a=b$, every integral is zero.

For the polynomial functions $f_n(x)=x^n$ on $[0,0.8]$, direct integration and the uniform bound giveAt $n=21$, the exact integral is approximately $0.0003353953$, while the general error bound is approximately $0.0073786976$. The bound certifies the result without an exact antiderivative. On $[0,1]$, these same integrals still tend to zero, although there is no uniform convergence: sufficient conditions need not be necessary. On an unbounded interval, even uniform convergence can fail to control total area, as uniformly vanishing functions can retain a nonzero integral shows.

Uniform convergence alone also fails to preserve derivatives: $f_n(x)=\sin (n^{2}x)/n$ converges uniformly to zero on $[-1,1]$, while $f_n^{\prime }(0)=n$. The small amplitude does not control the slopes of the sine oscillations.

For each positive integer $n$, put $f_n(t)=\sin (2\pi nt)/n$ on $[0,1]$. Its supremum norm is exactly $1/n$, so this sequence of functions converges to zero by uniform convergence. Nevertheless,Each of its $n$ complete sine oscillations contributes $4/n$: rise from zero to $1/n$, fall through zero to $-1/n$, then return to zero. Taking a partition of an interval through all extrema attains $4$. Every further partition refinement merely subdivides monotone function segments, so the same reasoning as in partition example of total variation proves the upper bound too.

For a continuously differentiable function on a closed bounded interval, total variation also satisfiesApplied here, the derivative is $2\pi \cos (2\pi nt)$, and integrating its absolute value over $n$ periods gives $4$. The differentiability hypothesis matters: the partition definition applies much more widely than this derivative formula.^[ref] Small pointwise size and small accumulated movement are different approximation goals.

A continuous function on a compact interval is uniformly continuous by the Heine-Cantor theorem. Choose a partition of an interval whose widths lie below the input tolerance for a desired output error $\epsilon$. On each half-open partition subinterval, use the value of the function at its left endpoint, and assign the rightmost endpoint its own value. This defines a step function whose error is strictly below $\epsilon$ at every input. Finitely assigned partition-point values are allowed and do not alter the Riemann integral.

For a bounded function $f$ and a partition of an interval $P$, the upper Darboux sum multiplies the supremum of $f$ on each subinterval by its length and adds the results.

For observations of an Itô integral on a partition of an interval, the statistic ${\sum }_{i}g(t_{i-1})(X_{t_i}-X_{t_{i-1}}{)}^2$ estimates $\int g(t)\sigma (t{)}^{2}dt$. With a deterministic noise coefficient the squared increments are independent, and the fluctuation has variance $2{\sum }_{i}g(t_{i-1}{)}^2({\int }_{t_{i-1}}^{t_i}{\sigma }^2{)}^2$.