Riemann sum

In mathematics, a Riemann sum is a summation of a large number of small partitions of a region. It may be used to define the integration operation. The method was named after German mathematician Bernhard Riemann.

Definition

Let f : D → R be a function defined on a subset, D, of the real line, R. Let I = [a, b] be a closed interval contained in D, and let

P=\left\{[x_{0},x_{1}),[x_{1},x_{2}),\dots ,[x_{n-1},x_{n}]\right\},

be a partition of I, where

a=x_{0}<x_{1}<x_{2}<\cdots <x_{n}=b.

The Riemann sum of f over I with partition P is defined as

S=\sum _{i=1}^{n}f(x_{i}^{*})(x_{i}-x_{i-1}),\quad x_{i-1}\leq x_{i}^{*}\leq x_{i}.

The choice of $x_{i}^{*}$ in the interval $[x_{i-1},x_{i}]$ is arbitrary.

Example: Specific choices of $x_{i}^{*}$ give us different types of Riemann sums:

If $x_{i}^{*}=x_{i-1}$ for all i, then S is called a left Riemann sum.
If $x_{i}^{*}=x_{i}$ for all i, then S is called a right Riemann sum.
If $x_{i}^{*}={\tfrac {1}{2}}(x_{i}+x_{i-1})$ for all i, then S is called a middle Riemann sum.
The average of the left and right Riemann sum is the trapezoidal sum.
If it is given that

S=\sum _{i=1}^{n}v_{i}(x_{i}-x_{i-1}),

where

v_{i}

is the supremum of f over

[x_{i-1},x_{i}]

, then S is defined to be an upper Riemann sum.

Similarly, if $v_{i}$ is the infimum of f over $[x_{i-1},x_{i}]$ , then S is a lower Riemann sum.

Riemann sum methods of x³ over [0,2] using 4 subdivisions

Left

Right

Middle

Trapezoidal

Simpson's Method

Any Riemann sum on a given partition (that is, for any choice of $x_{i}^{*}$ between $x_{i-1}$ and $x_{i}$ ) is contained between the lower and the upper Riemann sums. A function is defined to be Riemann integrable if the lower and upper Riemann sums get ever closer as the partition gets finer and finer. This fact can also be used for numerical integration.

Methods

The four methods of Riemann summation are usually best approached with partitions of equal size. The interval [a, b] is therefore divided into n subintervals, each of length

\Delta x={\frac {b-a}{n}}.

The points in the partition will then be

a,a+\Delta x,a+2\Delta x,\ldots ,a+(n-2)\Delta x,a+(n-1)\Delta x,b.

Right sum

f is here approximated by the value at the right endpoint. This gives multiple rectangles with base Δx and with the first partition having the height f(a + Δx''). The right Riemann sum amounts to an underestimation if f is monotonically decreasing, and an overestimation if it is monotonically increasing.

Middle sum

Approximating f at the midpoint of intervals gives f(a + Q/2) for the first interval, for the next one f(a + 3Q/2), and so on until f(b − Q/2). Summing up the areas gives

Q\left[f(a+{\tfrac {Q}{2}})+f(a+{\tfrac {3Q}{2}})+\cdots +f(b-{\tfrac {Q}{2}})\right].

The error of this formula will be

\left\vert \int _{a}^{b}f(x)\,dx-A_{\mathrm {mid} }\right\vert \leq {\frac {M_{2}(b-a)^{3}}{24n^{2}}},

where $M_{2}$ is the maximum value of the absolute value of $f^{\prime \prime }(x)$ on the interval.

Trapezoidal rule

In this case, the values of the function f on an interval are approximated by the average of the values at the left and right endpoints. In the same manner as above, a simple calculation using the area formula

A={\tfrac {1}{2}}h(b_{1}+b_{2})

for a trapezium with parallel sides b₁, b₂ and height h produces

{\tfrac {1}{2}}Q\left[f(a)+2f(a+Q)+2f(a+2Q)+2f(a+3Q)+\cdots +f(b)\right].

The error of this formula will be

\left\vert \int _{a}^{b}f(x)\,dx-A_{\mathrm {trap} }\right\vert \leq {\frac {M_{2}(b-a)^{3}}{12n^{2}}},

where $M_{2}$ is the maximum value of the absolute value of $f^{\prime \prime }(x).$

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

Example

A visual representation of the area under the curve y = x² for the interval from 0 to 2

Taking an example, the area under the curve of y = x² between 0 and 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 ${\tfrac {2}{n}}$ ; these are the widths of the Riemann rectangles. Because the right Riemann sum is to be used, the sequence of x coordinates for the boxes will be $x_{1},x_{2},\ldots ,x_{n}$ . Therefore, the sequence of the heights of the boxes will be $x_{1}^{2},x_{2}^{2},\ldots ,x_{n}^{2}$ . It is an important fact that $x_{i}={\tfrac {2i}{n}}$ , and $x_{n}=2$ .

The area of each box will be ${\tfrac {2}{n}}\times x_{i}^{2}$ and therefore the nth right Riemann sum will be:

{\begin{aligned}S&={\frac {2}{n}}\times \left({\frac {2}{n}}\right)^{2}+\cdots +{\frac {2}{n}}\times \left({\frac {2i}{n}}\right)^{2}+\cdots +{\frac {2}{n}}\times \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{aligned}}

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: