Riemann sum

Types of Riemann sums

Specific choices of ${\displaystyle x_i^{\ast }}$ give different types of Riemann sums:

• If ${\displaystyle x_i^{\ast }=x_{i-1}}$ for all i, the method is the left rule and gives a left Riemann sum.
• If ${\displaystyle x_i^{\ast }=x_i}$ for all i, the method is the right rule and gives a right Riemann sum.
• If ${\displaystyle x_i^{\ast }=(x_i+x_{i-1})/2}$ for all i, the method is the midpoint rule and gives a middle Riemann sum.
• If ${\displaystyle f(x_i^{\ast })=supf([x_{i-1},x_i])}$ (that is, the supremum of $f$ over ${\displaystyle [x_{i-1},x_i]}$), the method is the upper rule and gives an upper Riemann sum or upper Darboux sum.
• If ${\displaystyle f(x_i^{\ast })=inff([x_{i-1},x_i])}$ (that is, the infimum of $f$ over ${\displaystyle [x_{i-1},x_i]}$), the method is the lower rule and gives a lower Riemann sum or lower Darboux sum.

All these Riemann summation methods are among the most basic ways to accomplish numerical integration. Loosely speaking, a function is Riemann integrable if all Riemann sums converge as the partition "gets finer and finer".

While not derived as a Riemann sum, taking the average of the left and right Riemann sums is the trapezoidal rule and gives a trapezoidal sum. It is one of the simplest of a very general way of approximating integrals using weighted averages. This is followed in complexity by Simpson's rule and Newton–Cotes formulas.

Any Riemann sum on a given partition (that is, for any choice of ${\displaystyle x_i^{\ast }}$ between ${\displaystyle x_{i-1}}$ and ${\displaystyle x_i}$) is contained between the lower and upper Darboux sums. This forms the basis of the Darboux integral, which is ultimately equivalent to the Riemann integral.

Riemann summation methods

The four Riemann summation methods are usually best approached with subintervals of equal size. The interval [a, b] is therefore divided into ${\displaystyle n}$ subintervals, each of length $${\displaystyle \mathrm{\Delta }x=\frac{b-a}{n}.}$$

The points in the partition will then be $${\displaystyle a,a+\mathrm{\Delta }x,a+2\mathrm{\Delta }x,\dots ,a+(n-2)\mathrm{\Delta }x,a+(n-1)\mathrm{\Delta }x,b.}$$

Left rule

For the left rule, the function is approximated by its values at the left endpoints of the subintervals. This gives multiple rectangles with base Δx and height f(a + iΔx). Doing this for i = 0, 1, ..., n − 1, and summing the resulting areas gives $${\displaystyle S_{\mathrm{l}\mathrm{e}\mathrm{f}\mathrm{t}}=\mathrm{\Delta }x\left[f(a)+f(a+\mathrm{\Delta }x)+f(a+2\mathrm{\Delta }x)+\cdots +f(b-\mathrm{\Delta }x)\right].}$$

The left Riemann sum amounts to an overestimation if f is monotonically decreasing on this interval, and an underestimation if it is monotonically increasing. The error of this formula will be $${\displaystyle \left|{\int }_a^{b}f(x)dx-S_{\mathrm{l}\mathrm{e}\mathrm{f}\mathrm{t}}\right|\le \frac{M_1(b-a{)}^2}{2n},}$$ where ${\displaystyle M_1}$ is the maximum value of the absolute value of ${\displaystyle f^{\mathrm{\prime }}(x)}$ over the interval.

Right rule

For the right rule, the function is approximated by its values at the right endpoints of the subintervals. This gives multiple rectangles with base Δx and height f(a + iΔx). Doing this for i = 1, ..., n, and summing the resulting areas gives $${\displaystyle S_{\mathrm{r}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}}=\mathrm{\Delta }x\left[f(a+\mathrm{\Delta }x)+f(a+2\mathrm{\Delta }x)+\cdots +f(b)\right].}$$

The right Riemann sum amounts to an underestimation if f is monotonically decreasing, and an overestimation if it is monotonically increasing. The error of this formula will be $${\displaystyle \left|{\int }_a^{b}f(x)dx-S_{\mathrm{r}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}}\right|\le \frac{M_1(b-a{)}^2}{2n},}$$ where ${\displaystyle M_1}$ is the maximum value of the absolute value of ${\displaystyle f^{\mathrm{\prime }}(x)}$ over the interval.

Midpoint rule

For the midpoint rule, the function is approximated by its values at the midpoints of the subintervals. This gives f(a + Δx/2) for the first subinterval, f(a + 3Δx/2) for the next one, and so on until f(b − Δx/2). Summing the resulting areas gives $${\displaystyle S_{\mathrm{m}\mathrm{i}\mathrm{d}}=\mathrm{\Delta }x\left[f\left(a+\frac{\mathrm{\Delta }x}{2}\right)+f\left(a+\frac{3\mathrm{\Delta }x}{2}\right)+\cdots +f\left(b-\frac{\mathrm{\Delta }x}{2}\right)\right].}$$

The error of this formula will be $${\displaystyle \left|{\int }_a^{b}f(x)dx-S_{\mathrm{m}\mathrm{i}\mathrm{d}}\right|\le \frac{M_2(b-a{)}^3}{24n^2},}$$ where ${\displaystyle M_2}$ is the maximum value of the absolute value of ${\displaystyle f^{\mathrm{\prime }\mathrm{\prime }}(x)}$ over the interval. This error is half of that of the trapezoidal sum; as such the middle Riemann sum is the most accurate approach to the Riemann sum.

Trapezoidal rule

For the trapezoidal rule, the function is approximated by the average of its values at the left and right endpoints of the subintervals. Using the area formula ${\displaystyle \frac{1}{2}h(b_1+b_2)}$ for a trapezium with parallel sides b₁ and b₂, and height h, and summing the resulting areas gives $${\displaystyle S_{\mathrm{t}\mathrm{r}\mathrm{a}\mathrm{p}}=\frac{1}{2}\mathrm{\Delta }x\left[f(a)+2f(a+\mathrm{\Delta }x)+2f(a+2\mathrm{\Delta }x)+\cdots +f(b)\right].}$$

The error of this formula will be $${\displaystyle \left|{\int }_a^{b}f(x)dx-S_{\mathrm{t}\mathrm{r}\mathrm{a}\mathrm{p}}\right|\le \frac{M_2(b-a{)}^3}{12n^2},}$$ where ${\displaystyle M_2}$ is the maximum value of the absolute value of ${\displaystyle f^{″}(x)}$.

The approximation obtained with the trapezoidal sum for a function is the same as the average of the left hand and right hand sums of that function.

Connection with integration

For a one-dimensional Riemann sum over domain ${\displaystyle [a,b]}$, as the maximum size of a subinterval shrinks to zero (that is the limit of the norm of the subintervals goes to zero), some functions will have all Riemann sums converge to the same value. This limiting value, if it exists, is defined as the definite Riemann integral of the function over the domain, $${\displaystyle {\int }_a^{b}f(x)dx=\underset{\Vert \mathrm{\Delta }x\Vert \to 0}{lim}\sum _{i=1}^{n}f(x_i^{\ast })\mathrm{\Delta }{x}_i.}$$

For a finite-sized domain, if the maximum size of a subinterval shrinks to zero, this implies the number of subinterval goes to infinity. For finite partitions, Riemann sums are always approximations to the limiting value and this approximation gets better as the partition gets finer. The following animations help demonstrate how increasing the number of subintervals (while lowering the maximum subinterval size) better approximates the "area" under the curve:

Since the red function here is assumed to be a smooth function, all three Riemann sums will converge to the same value as the number of subintervals goes to infinity.

Example

Taking an example, the area under the curve y = x² over [0, 2] can be procedurally computed using Riemann's method.

The interval [0, 2] is firstly divided into n subintervals, each of which is given a width of ${\displaystyle \frac{2}{n}}$; these are the widths of the Riemann rectangles (hereafter "boxes"). Because the right Riemann sum is to be used, the sequence of x coordinates for the boxes will be ${\displaystyle x_1,x_2,\dots ,x_n}$. Therefore, the sequence of the heights of the boxes will be ${\displaystyle x_1^2,x_2^2,\dots ,x_n^2}$. It is an important fact that ${\displaystyle x_i=\frac{2i}{n}}$, and ${\displaystyle x_n=2}$.

The area of each box will be ${\displaystyle \frac{2}{n}\times x_i^2}$ and therefore the nth right Riemann sum will be: $${\displaystyle \begin{array}{cc}S& =\frac{2}{n}{\left(\frac{2}{n}\right)}^2+\cdots +\frac{2}{n}{\left(\frac{2i}{n}\right)}^2+\cdots +\frac{2}{n}{\left(\frac{2n}{n}\right)}^2\\ & =\frac{8}{n^3}\left(1+\cdots +i^2+\cdots +n^2\right)\\ & =\frac{8}{n^3}\left(\frac{n(n+1)(2n+1)}{6}\right)\\ & =\frac{8}{n^3}\left(\frac{2n^3+3n^2+n}{6}\right)\\ & =\frac{8}{3}+\frac{4}{n}+\frac{4}{3n^2}.\end{array}}$$

If the limit is viewed as n → ∞, it can be concluded that the approximation approaches the actual value of the area under the curve as the number of boxes increases. Hence: $${\displaystyle \underset{n\to \mathrm{\infty }}{lim}S=\underset{n\to \mathrm{\infty }}{lim}\left(\frac{8}{3}+\frac{4}{n}+\frac{4}{3n^2}\right)=\frac{8}{3}.}$$

This method agrees with the definite integral as calculated in more mechanical ways: $${\displaystyle {\int }_0^2{x}^{2}dx=\frac{8}{3}.}$$

Because the function is continuous and monotonically increasing over the interval, a right Riemann sum overestimates the integral by the largest amount (while a left Riemann sum would underestimate the integral by the largest amount). This fact, which is intuitively clear from the diagrams, shows how the nature of the function determines how accurate the integral is estimated. While simple, right and left Riemann sums are often less accurate than more advanced techniques of estimating an integral such as the Trapezoidal rule or Simpson's rule.

The example function has an easy-to-find anti-derivative so estimating the integral by Riemann sums is mostly an academic exercise; however it must be remembered that not all functions have anti-derivatives so estimating their integrals by summation is practically important.

Higher dimensions

The basic idea behind a Riemann sum is to "break-up" the domain via a partition into pieces, multiply the "size" of each piece by some value the function takes on that piece, and sum all these products. This can be generalized to allow Riemann sums for functions over domains of more than one dimension.

While intuitively, the process of partitioning the domain is easy to grasp, the technical details of how the domain may be partitioned get much more complicated than the one dimensional case and involves aspects of the geometrical shape of the domain.

Two dimensions

In two dimensions, the domain ${\displaystyle A}$ may be divided into a number of two-dimensional cells ${\displaystyle A_i}$ such that $A=\bigcup _i{A}_i$. Each cell then can be interpreted as having an "area" denoted by ${\displaystyle \mathrm{\Delta }{A}_i}$. The two-dimensional Riemann sum is $${\displaystyle S=\sum _{i=1}^{n}f(x_i^{\ast },y_i^{\ast })\mathrm{\Delta }{A}_i,}$$ where ${\displaystyle (x_i^{\ast },y_i^{\ast })\in A_i}$.

Three dimensions

In three dimensions, the domain ${\displaystyle V}$ is partitioned into a number of three-dimensional cells ${\displaystyle V_i}$ such that $V=\bigcup _i{V}_i$. Each cell then can be interpreted as having a "volume" denoted by ${\displaystyle \mathrm{\Delta }{V}_i}$. The three-dimensional Riemann sum is $${\displaystyle S=\sum _{i=1}^{n}f(x_i^{\ast },y_i^{\ast },z_i^{\ast })\mathrm{\Delta }{V}_i,}$$ where ${\displaystyle (x_i^{\ast },y_i^{\ast },z_i^{\ast })\in V_i}$.

Arbitrary number of dimensions

Higher dimensional Riemann sums follow a similar pattern. An n-dimensional Riemann sum is $${\displaystyle S=\sum _{i}f(P_i^{\ast })\mathrm{\Delta }{V}_i,}$$ where ${\displaystyle P_i^{\ast }\in V_i}$, that is, it is a point in the n-dimensional cell ${\displaystyle V_i}$ with n-dimensional volume ${\displaystyle \mathrm{\Delta }{V}_i}$.

Generalization

In high generality, Riemann sums can be written $${\displaystyle S=\sum _{i}f(P_i^{\ast })\mu (V_i),}$$ where ${\displaystyle P_i^{\ast }}$ stands for any arbitrary point contained in the set ${\displaystyle V_i}$ and ${\displaystyle \mu }$ is a measure on the underlying set. Roughly speaking, a measure is a function that gives a "size" of a set, in this case the size of the set ${\displaystyle V_i}$; in one dimension this can often be interpreted as a length, in two dimensions as an area, in three dimensions as a volume, and so on.