The Riemann sum is the name given to the approximate calculation of a definite integral, by means of a discrete sum with a finite number of terms. A common application is the area approximation of functions on a graph.

It was the German mathematician Georg Friedrich Bernhard Riemann (1826-1866) who first offered a rigorous definition of the integral of a function on a given interval. He made it known in an article published in 1854.

The Riemann sum is defined over a function y = f(x), with x belonging to the closed interval [a, b]. A partition P of n elements is carried out over this interval:

P = {x0= a, x1, x2, …, xn= b}

This means that the interval is divided as follows:

Here twhat is between xk-1 yxwhat:

xk-1 ≤ youwhat ≤ xwhat

Figure 1 graphically shows the Riemann sum of the function f in the interval [x0, x4] on a partition of four subintervals, the rectangles of gray color.

The sum represents the total area of the rectangles and the result of this sum is numerically close to the area under the curve f, between the abscissas x=x0 yx=x4.

Of course, the approximation to the area under the curve is greatly improved as the number no of partitions is greater. In this way the sum converges to the area under the curve, when the number no of partitions tends to infinity.

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**Formulas and properties**

The Riemann sum of the function f(x) over the partition:

P = {x0= a, x1, x2, …, xn= b}

Defined on the interval [a, b]is given by:

S(P, f) = ∑k=1no f(tk)) (xk –xk-1)

where tk is a value in the interval [xk, xk-1]. In the Riemann sum, regular intervals of width Δx = (b – a)/n are usually used, where a and b are the minimum and maximum values of the abscissa, while n is the number of subdivisions.

In that case the right Riemann sum is:

Sd(f,n)= [f(a+Δx) +f(a+2Δx)+ …+f(a+(n-1)Δx)+f(b)]*Δx

while the *left Riemann sum* is expressed as:

SYo(f,n)= [f(a) +f(a+Δx)+ …+f(a+(n-1)Δx)]*Δx

Finally the central Riemann sum is:

Sc(f,n)= [f(a+ Δx/2) +f(a+ 3Δx/2)+ …+f(b- Δx/2 )]*Δx

D.depending on where point t is locatedwhat in the interval [xk, xk-1] the Riemann sum can overestimate or underestimate the exact value of the area under the curve of the function y = f(x). That is, the rectangles can protrude from the curve or be a little below it.

**The area under the curve**

The main property of the Riemann sum, and from which its importance comes, is that if the number of subdivisions tends to infinity, the result of the sum converges to the definite integral of the function:

The above expression corresponds to the definition of the Riemann integral and applies whenever the function f is continuous and smooth. For more particular functions there are other definitions of the integral (Stieltjes integral and Lebesgue integral).

**solved exercises**

**– Exercise 1**

Calculate the value of the definite integral between a = -2 to b = +2 of the function:

f(x) = x2

Make use of a Riemann sum. To do this, first find the sum for n regular partitions of the interval [a, b] and then take the mathematical limit for the case that the number of partitions tends to infinity.

**Solution**

These are the steps to follow:

-Firstly, the interval of the partitions is defined as:

Δx = (b – a)/n.

-Then the Riemann sum for the right corresponding to the function f(x) looks like this:

-Now substitute a=-2 and b=+2, so that the interval or step Δx = 4/n. That is to say that the Riemann sum for the function f(x) = x2 is:

-Next, the square binomial is developed:

*[-2 +(4i/n)]2 = 4 – (16 i /n) + (4/n)2 i2 *

-And then carefully substitute in the summation:

-The next step is to separate the sums and take out the constant quantities as a common factor of each sum. It is necessary to take into account that the index is i, therefore the numbers and the terms with *no* are considered constant:

-Each sum is evaluated, since for each of them there are appropriate expressions. For example, the first of the sums gives n:

Lto second is:

And the third result:

-Substituting the results of the sums in the Riemann sum, we finally obtain:

*S(f, n) = 16 – 64(n+1)/2n + 64(n+1)(2n+1)/6n2*

-Finally we have that the integral that we want to calculate is:

*= slimn➝∞ [16 – 64(n+1)/2n + 64(n+1)(2n+1)/6n2]=*

*= 16 -(64/2) + (64/3) = 16/3 = 5,333*

The reader can verify that this is the exact result, which can be obtained by solving the indefinite integral and evaluating the limits of integration by Barrow’s rule.

**– Exercise 2**

Approximately determine the area under the function:

f(x) = (1/√(2π)) e(-x2/2)

Between x=-1 and x=+1, using a central Riemann sum with 10 partitions. Compare with the exact result and estimate the percentage difference.

**Solution**

The step or increment between two successive discrete values is:

Δx = (1 – (-1)/10 = 0.2

So that the partition P on which the rectangles are defined looks like this:

P = {-1.0; -0.8; -0.6; -0.4; -0.2; 0.0; 0.2; 0.4; 0.6; 0.8; 1.0}

But since what is wanted is the central sum, the function f(x) will be evaluated at the midpoints of the subintervals, that is, on the set:

T = {-0.9; -0.7; -0.5; -0.3; -0.1; 0.1; 0.3; 0.5; 0.7; 0.9}.

The (central) Riemann sum looks like this:

S=f(-0.9)*0.2 + f(-0.7)*0.2+f(-0.5)*0.2+…+f(0.7)*0.2 +f(0.9)*0.2

Since the function f is symmetric, it is possible to reduce the sum to only 5 terms and the result is multiplied by two:

S = 2*0.2*{f(0.1)+ f(0.3)+ f(0.5)+ f(0.7)+ f(0.9)}

S = 2*0.2*{0.397+ 0.381+ 0.352+ 0.312+ 0.266}=0.683

The function given in this example is none other than the well-known Gaussian bell (normalized, with mean equal to zero and standard deviation one). It is known that the area under the curve in the interval [-1,1] for this function is 0.6827.

This means that the approximate solution with just 10 terms matches the exact solution to three decimal places. The percentage error between the approximate and the exact integral is 0.07%.

**References**

Casteleiro, JM, & Gómez-Álvarez, RP (2002). Integral calculus (Illustrated ed.). Madrid: ESIC Editorial.

Unican. History of the concept of integral. Retrieved from: repositorio.unican.es

UIS. Riemann sums. Retrieved from: matematicas.uis.edu.co

Wikipedia. Riemann sum. Recovered from: en.wikipedia.com

Wikipedia. Riemann integration. Recovered from: en.wikipedia.com