21 

 

Mathematics Education Journals 

Vol. 6 No. 1 February 2022 

 

 

ISSN : 2579-5724   

ISSN : 2579-5260 (Online) 

http://ejournal.umm.ac.id/index.php/MEJ 

 

 

 

     Statistical Simmulation of  Definite Integral Based on Uppersum and 

Lowersum Random Partitions using Geogebra 
   

Widiya Astuti Alam Sur  

 
Politeknik Negeri Tanah Laut 

Email: widiyasur@politala.ac.id  

Abstract 

This paper aims to demonstrate the ability of Geogebra application in 

presenting statistical model simulations. Statistical simulation on Geogebra 

is based on the value of the random partition of the upper sums and lower 

sums on definite integral concepts. Random partitioning of the upper sums 

and lower sums in determining a definite integral value is used to determine 

the statistical distribution of the resulting random variables. Because the 

partition value data is obtained randomly, statistical tests can be carried out 

to determine the type of distribution on the random data obtained. Based on 

the Kolmogorov Smirnov test, the data for the subinterval partition random 

variables at the upper sums and lower sums using Goegebra follow a specific 

statistical distribution. It was found that the statistical simulations of random 

partitions of the upper sums and lower sums are based on definite integral 

concepts following the Burr 4, Log-Pearson 3, and Pearson 6 distribution 

parameters. The results of this statistical test simulation show that apart from 

being an application for visualizing mathematical concepts geometrically, 

Geogebra can also analyze mathematical concepts statistically. 

Keywords: uppersum; lowersum; integral; Geogebra 
 

 

INTRODUCTION 
 

The need to visualize mathematics concepts by forming the images (either 

manually with pencil and paper or with technology) or using such images has 

effectively discovered the understanding of mathematics (Caligaris, 2014). To 

achieve deep understanding, the students have to learn how can represent their 

ideas. The teacher's creativity is needed to choose and use the appropriate learning 

media to bridge the student's mathematical ability. Milovanović (2011) showed that 

using multimedia in math classes about definite integral is highly interesting for 

students and showed better theoretical, practical, and visual knowledge because it 

can give visualization possibilities, animations, and illustrations. 

Along with the development of technology, many choices of mathematics 

learning media can be used. The software that can be obtained from the internet 

freely is Geogebra. Geogebra is a dynamic mathematical software that can be used 

for all the education levels from the primary to the university (Hohenwarter, 2007). 

GeoGebra is already used by more than 100 million students around the 

world. GeoGebra can minimize the difficulties of students who get Calculus 

subjects, especially students majoring in Natural Sciences and Engineering. Arini 

(2019) stated that the advantages of using GeoGebra are helping convey the 

mailto:widiyasur@politala.ac.id


 
 

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Mathematics Education Journals 

Vol. 6 No. 1 February 2022 

 

 

ISSN : 2579-5724   

ISSN : 2579-5260 (Online) 

http://ejournal.umm.ac.id/index.php/MEJ 

 

 

 

Calculus concept material to be more interesting. The material, especially concepts 

of functions, limits, derivatives, and integrals, provides a more realistic image, 

especially for more complex calculus material, and provides a faster and more 

accurate solution. 

Caligaris (2014) stated that understanding the definite integral definitions 

and theorems in mathematics takes symbolic representation or graphics to describe. 

With Geogebra, an interactive application can illustrate the mathematics materials 

that require visualization, especially on Calculus material. Furthermore, Serhan 

(2015) stated that most students only knew the procedure and calculation steps to 

solve the calculus problem, especially the indefinite integral problem. In fact, to 

understand the concept of definite integral, many things that students can explore 

and get a piece of additional knowledge, both its relationship with calculus itself or 

its relationship with the other science, for example, statistics.  

Nur’aini (2017) used Geogebra to draw and calculate the geometries 

mathematically as a catalyst to make mathematics learning to be more realistic. The 

implementation of Geogebra can be used to visualize and determine: the application 

of the Pythagoras Theorem, the angle of the clock, and geometric transformation. 

Sari (2016) did the research about Geogebra-assisted learning media (module) that 

was developed received an assessment for the interesting category and was worthy 

of being used as a learning media for derivative. 

Integrating educational technology in the teaching and learning of definite 

integral creates a conceptually rich learning environment. Kado & Dem (2020) 
found that GeoGebra software can enhance and significantly improve students’ 

conceptual understanding of definite integral. The computer-assisted instruction 

method using GeoGebra was found to contribute to teaching the definite integral 

topic positively. Because all the study about using GeoGebra to visualize the 

materials of mathematics gives the preliminary study about visualizing the definite 

integral concept, this study gives the exploration in using Geogebra to construct the 

definite integral concept.  

Two things that become the basic idea to define and construct the definite 

integral are the case to calculate the traveled distance of a moving object with the 

velocity and the case to calculate the area. Calculating the traveled distance of an 

object that moves at certain time intervals will provide how to calculate the area 

under the curve. Caligaris (2015) used GeoGebra to illustrate the integral function 

depends on the upper limit o integration and constructed the value changes to 

improve the presentation of content taught, allowing dynamic visualization. The 

paper concludes that incorporating the Geogebra Applets is a much more effective 

teaching methodology than traditional one to facilitate the learning of the 

fundamental concepts of Calculus. 

Defining the definite integral concept is based on the partition of the interval 

as the lower sum and upper sum with the same and random subinterval length. 

Calculation of the area under the curve becomes the basic idea for understanding 

the definite integral materials. The results of the exploration and material design of 

the definite integral are expected to form new knowledge about concepts and 



 
 

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Mathematics Education Journals 

Vol. 6 No. 1 February 2022 

 

 

ISSN : 2579-5724   

ISSN : 2579-5260 (Online) 

http://ejournal.umm.ac.id/index.php/MEJ 

 

 

 

application of the definite integral concept material by visualizing it on Geogebra 

(Sur, 2020).  

GeoGebra can design the same and random subinterval to present the 

definite integral dynamically. The length of the same or random subinterval of the 

integral concept was the basic idea in this research. Since GeoGebra can create the 

partitions randomly with an unlimited number of partitions, the authors were 

interested in knowing whether the random partitions in Geogebra statistically have 

a particular distribution. 

This study illustrates and presents definite integral concepts with analytical 

and statistical approaches by using GeoGebra. Exploration of the concepts is related 

to the comparation of the lower sum and upper sum of the integran function based 

on the subinterval length. Exploration is presented to construct and discover new 

things in understanding the definite integral concepts, particularly about the 

statistical model of an upper sum and lower sum partitions of the definite integral 

concept. 

 

RESEARCH METHOD 

This research appropriates the definite integral concept materials to make 

the simulation, illustration, and then analyze the comparison between upper sum 

and lower sum geometrically by using Geogebra 5.0 version. After the simulation, 

illustration, and analysis, we perform the statistical test of upper sum and lower sum 

random partition associated with the definite integral concepts to find out the 
statistical distribution of its generated random variable. To obtain the numerical 

statistic of random partitions, we used with Kolmogorov Smirnov test. Based on 

the test results, we will determine if the random partition data follows the specified 

distribution or does not follow the specified distribution.  This anaysis uses the 

following hypothesis: 

 

𝐻0: Data follows the specified distribution 
𝐻1: Data do not follow the specified ditribution 

If data follows the specified distribution, then the distribution of random 

variable data in every subinterval partition can be determined using EasyFit version 

5.5. 

 
RESULTS AND DISCUSSION 

Elementary Concepts About Uppersum and Lowersum by Using  Geogebra 

In case to define the definite integral concepts, let the definition of Uppersum and 

Lowersum of a function. Given P is a finite ordered points between a dan b, 𝑃 =
{𝑥0, 𝑥1, 𝑥2, … , 𝑥𝑛−1, 𝑥𝑛}, with 𝑎 = 𝑥0 < 𝑥1 < 𝑥2, … , < 𝑥𝑛−1 < 𝑥𝑛 = 𝑏. The set 𝑃 
is said a partition of interval [𝑎, 𝑏], that devides [𝑎, 𝑏] into 𝑛 subinterval,with 𝑖𝑡ℎ 
subinterval is [𝑥−1, 𝑥𝑖]. The length of 𝑖

𝑡ℎ from 𝑃 is ∆𝑥𝑖 = 𝑥𝑖 − 𝑥𝑖−1, for 1 ≤ 𝑖 ≤ 𝑛. 
(Salas, 2007) 

 

Based on theorem Adams (2010) about minimum and maksimum value of a 

function, we know  if 𝑓: [𝑎, 𝑏] → ℝ is continu on [𝑎, 𝑏], then in every subinterval 



 
 

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ISSN : 2579-5724   

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[𝑥𝑖−1, 𝑥𝑖], there exist 𝑙𝑖, 𝑢𝑖 ∈ [𝑥𝑖−1, 𝑥𝑖], such that 𝑓 minimum at 𝑓(𝑙𝑖) and maximum 
at 𝑓(𝑢𝑖) 
 

Defining the definite integral based on the partision of the interval as the lower and 

the upper limit with the same and random subinteval length, can be explored by 

using geogebra. Consider the region A is an area bounded by the graph of a 

continuous function 𝑦 = 𝑓(𝑥),the 𝑥-axis and between vertikal line 𝑥 = 𝑎 , and 𝑥 =
𝑏. Region 𝐴 can be estimated by dividing region 𝐴 into 𝑛 subregion 𝐴1, 𝐴2, … , 𝐴𝑛. 
By the worksheet of Geogebra, we can draw the following ilustration 

 

 

Fig. 1. Ilustration of dividing region A into n subregions  

Interval [a,b] is divided into a finite number of subintervals [𝑎 =

𝑥0, 𝑥1]; [𝑥1, 𝑥2]; … [𝑥𝑛−1, 𝑥𝑛 =  𝑏]. Because 𝑓 continu on interval [a, b], then f  

continu on every subintervals  [𝑥𝑖−1, 𝑥𝑖], with 𝑖 = 1,2, … , 𝑛 . It means f  maksimum 

and minimum at every points on that subinterval. Because of that, there exist the 

numbers 𝑙𝑖 and 𝑢𝑖 at [𝑥𝑖−1, 𝑥𝑖], such that 𝑓(𝑙𝑖) ≤   𝑓(𝑥)  ≤  𝑓(𝑢𝑖) with 𝑥𝑖−1 ≤ 𝑥 ≤

𝑥𝑖 (Bartle, 2000). 

Let’s denote 𝑀𝑖 =  𝑓(𝑢𝑖)  the maximum value of f on [𝑥𝑖−1, 𝑥𝑖],  and 𝑚𝑖 =
𝑓(𝑙𝑖) the minimum valueof 𝑓 on[𝑥𝑖−1, 𝑥𝑖]. If  we take any 𝑖

𝑡ℎsubitervals on [𝑎, 𝑏], 
then we can consider the rectangles 𝑟𝑖 and 𝑅𝑖, with 𝑟𝑖 ⊆ 𝐴𝑖 ⊆ 𝑅𝑖. It means area of 
𝑟𝑖 ≤ area of 𝐴𝑖 ≤ area of 𝑅𝑖. Since the area of regctamgle is the length times the 
width, then 𝑚𝑖∆𝑥𝑖 ≤ area of 𝐴𝑖 ≤ 𝑀𝑖∆𝑥𝑖, with ∆𝑥𝑖 = (𝑥𝑖 − 𝑥𝑖−1). It holds for 𝑖 =
1,2, … , 𝑛. The sum of the minimum value approaches 𝑚1∆𝑥1 + 𝑚2∆𝑥2 + ⋯ +
𝑚𝑛∆𝑥𝑛 ≤ area of 𝐴, and the sum of the maximum value approaches area of 𝐴 ≤
𝑀1∆𝑥1 + 𝑀2∆𝑥2 + ⋯ + 𝑀𝑛∆𝑥𝑛  (Stewart, 2010). 

The sum of minimum values on every subintervals is defined as Lowersum, and 

the sum of maximum values one every subintervals  is defined as Uppersum. By 

using Geogebra, we can prove that the more particies we make, the more we can 

reach one and only one number between the uppersum and the lowersum value. 

This such number will be an area of A, which is defined as definite integral. 

In case  to determine the uppersum and lowersumm of a function, the 

subinterval length of partition 𝑃 can be divided  not only into 𝑛  subinterval with 

the same length ∆𝑥𝑖 =
𝑏−𝑎

𝑛
, ∀𝑖 = 1,2, … , 𝑛 , but also with the difference or random 

subinterval length(Stewart, 2008). By using geogebra, we can make a simmulation 



 
 

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Vol. 6 No. 1 February 2022 

 

 

ISSN : 2579-5724   

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that is compare the integral function based on the uppersum and lowersum between 

the same subinterval length with random subinterval  length.  

 

Simmulation of Integrable Function Based on Lowersum and Uppersum 

Partition  

The bounded function of definite interval, can be defined as integrable function if 

the supremum of its lowersum is equal to infimum of its uppersum. By using 

Geogebra, we can make the simmulation that is  presenting for  all  ε>0, the bounded 

function f:[a,b]→R is integrable on [a,b] if the difference of its uppersum and 

lowersum less than ε. 

 

Integrable Function Based on Uppersum dan Lowersum Partitions with The Same 

Subinterval Length 

We created a simulation for a bounded function 𝑓: [1,2] → ℝ, with 𝑓(𝑥) =
1

𝑥
; ∀𝑥 ∈

[1,2]. The simulation by Geogebra is divided into 2 cases, i.e the same subintervals, 
and random subintervals. We compare the minimum partition obtained of uppersum 

and lowersumt, with 𝜀 =
1

100
,

1

200
, … ,

1

1000
.  Suppose Π[1,2] is the set of all 

partitions 𝑃 at [1,2] with Π[1,2] = {𝑃1, 𝑃2, … , 𝑃10}. 𝑈(𝑓, 𝑃) and 𝐿(𝑓, 𝑃) are 

uppersum and lowersum of 𝑓 =
1

𝑥
, 𝑥 ∈ [1,2] on  partition 𝑃. Then, for all the same 

subinterval length with ∆𝑥 =
1

𝑛
 we obtain : 

a. Given 𝜀 =
1

100
= 0,001 > 0, then the difference of uppersum and lowersum 

is 𝑈(𝑓, 𝑃1) − 𝐿(𝑓, 𝑃1)  =  0, 6981 − 0,6883 =  0, 0098 < 𝜀, with the 
number of minimum partition is n = 51. Partition 𝑃1 𝑜𝑛 [1, 2 ]when 𝑛 =
51 is written by 𝑃1 = {1, 1.02, 1.03, … ,1.98, 1.99, 2}. 

b. Given 𝜀 =
1

200
= 0,005 > 0, then the difference of uppersum and lowersum 

is 𝑈(𝑓, 𝑃2) − 𝐿(𝑓, 𝑃2)  =  0, 6956 − 0,6907 =  0, 0049 < 𝜀, with the 
number of minimum partition is n = 102. Partition 𝑃1 𝑜𝑛 [1, 2]when 𝑛 =
51 is written by 𝑃1 = {1, 1.02, 1.03, … ,1.98, 1.99, 2}. 

c. Given 𝜀 =
1

300
= 0,0033 > 0, then the difference of uppersum and lowersum 

is 𝑈(𝑓, 𝑃3) − 𝐿(𝑓, 𝑃3) =  0,6948 − 0, 6915 = 0, 0032 < 𝜀, with the number 
of minimum partition is n = 154. Partition 𝑃3 𝑜𝑛 [1, 2]when 𝑛 = 154 is written 
by 𝑃3 = {1, 1.01, 1.01, . .1.98, 1.99, 1.99, 2}. 

d. The same  process is continued till for partition 𝑃10, with 𝜀 =
1

1000
  

e. Given 𝜀 =
1

1000
= 0,001 > 0, then the difference of uppersum and lowersum 

is 𝑈(𝑓, 𝑃10) − 𝐿(𝑓, 𝑃10) =  0,6936 − 0, 6927 = 0, 0009 < 𝜀, with the 
number of minimum partition is n =527. Partition 𝑃10 on [1, 2]when 𝑛 =
527 is written by 𝑃10 = {1,1,1,1,1, 1.01, 1.01, … , 1.99,2,2,2, 2}. 

Simulation for the integrable function based on any value 𝜀 > 0 𝑜𝑛 𝑓(𝑥) =
1/𝑥 ; 𝑥 ∈ [1,2], with the same subinterval length is presented by Figure 2 (S, 2018) 

 



 
 

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ISSN : 2579-5724   

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Fig. 2. Simulation of integrable function for 𝜀 =
1

100
 with the same length 

subintervals 

 

Integrable Function Based on Uppersum dan Lowersum Partitions with The 

Random Subinterval Length 

 

With the same way, we can simulate the integrable function based on the 

lowersum and uppersums, for the random subinterval length. With the same 

function, suppose Π[1,2] is the set of all partitions 𝑃 at [1,2] with Π[1,2] =

{𝑃1, 𝑃2, … , 𝑃10}. 𝑈(𝑓, 𝑃) and 𝐿(𝑓, 𝑃) are uppersum and lowersum of 𝑓 =
1

𝑥
, 𝑥 ∈

[1,2] on  partition 𝑃. Then, for all the random subinterval length ∆𝑥, we obtain: 

a. Given 𝜀 =
1

100
= 0,01 > 0, then the difference of uppersum and lowersum 

is 𝑈(𝑓, 𝑃1) − 𝐿(𝑓, 𝑃1)  =  0, 6980 − 0,6884 =  0, 0096 < 𝜀, with the 
number of minimum partition is n = 94. 

b. Partition 𝑃1 𝑜𝑛 [1, 2]when  𝑛 = 94 is written by  𝑃1 =
{1, 1.01, 1.01,1.05, … ,1.94, 1.98, 1.99, 2} 

c. Given 𝜀 =
1

200
= 0,005 > 0, then the difference of uppersum and lowersum 

is 𝑈(𝑓, 𝑃2) − 𝐿(𝑓, 𝑃2)  =  0, 6955 − 0,6908 =  0, 0048 < 𝜀, with the 
number of minimum partition is n = 206.  

d. Partition 𝑃1 𝑜𝑛 [1, 2]when 𝑛 = 51 is written by 𝑃1 =
{1, 1.01, 1.03, … ,1.96, 1.98, 2, 2} 

e. Given 𝜀 =
1

300
= 0,0033 > 0, then the difference of uppersum and lowersum 

is 𝑈(𝑓, 𝑃3) − 𝐿(𝑓, 𝑃3) =  0,6947 − 0, 6915 = 0, 0032 < 𝜀, with the number 
of minimum partition is n = 306.  

f. Partition 𝑃3 𝑜𝑛 [1, 2]when 𝑛 = 306 is written by 𝑃3 = {1,1,1,1 1.01,
… ,1.98, 1.98, 1.98, 2}  

g. the same  process is continued till for partition 𝑃10, with 𝜀 =
1

1000
  

h. Given 𝜀 =
1

1000
= 0,001 > 0, then the difference of uppersum and lowersum 

is 𝑈(𝑓, 𝑃10) − 𝐿(𝑓, 𝑃10) =  0,6936 − 0, 6927 = 0, 0009 < 𝜀, with the 
number of minimum partition is n =1050.  



 
 

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Partition 𝑃10 on [1, 2]when 𝑛 = 1050 is written by 𝑃10 =
{1,1,1,1.01, … 1.99,2,2,2,2, 2}. 

 

Table 1.  The comparison of the partition between the same and random 

subintervals 

Π[1,2] 𝜀 > 0 
𝑈(𝑓, 𝑃) − 𝐿(𝑓, 𝑃)
< 𝜀 

Subinterval Length 

Same Random 

𝑃1 1/100 0,0098 n=51 n=94 
𝑃2 1/200 0,0049 n=102 n=206 
𝑃3 1/300 0,0032 n=154 n=306 
𝑃4 1/400 0,0024 n=205 n=435 
𝑃5 1/500 0,0019 n=257 n=571 
𝑃6 1/600 0,0017 n=304 n=635 
𝑃7 1/700 0,0013 n=371 n=735 
𝑃8 1/800 0,0012 n=415 n=872 
𝑃9 1/900 0,0011 n=477 n=982 
𝑃10 1/1000 0,0009 n=527 n=1050 

When we compared the partitions between random subintervals and the same 

subintervals length, it can be said that there are more partitions needed for the length 

of the random subinterval, to produce  the uppersum and lowersum convergen to a 

certain value, than the same subinterval length. The comparison of the partition 

between random and the same subintervals for the integrable function 𝑓(𝑥) =
1

𝑥
, ∀𝑥 ∈ [𝑎, 𝑏] is provided by Table 1. 

 

Fig. 3. Simulation of integrable function for 𝜀 =
1

100
 with the random length 

subintervals 

From the explanation about the same and random subintervals, it can be 

seen that the smaller 𝜀 value is taken, the larger 𝑛 partition it takes for the lowersum 
and uppersum to converge to a certain value. In other words, the difference between 

U(f, P) and L(f, P) gets smaller and closer to 𝜀, if n partition point gets bigger. That 
is how the theorem about the bounded function 𝑓: [𝑎, 𝑏] → ℝ is integrabel on[𝑎, 𝑏] 
iff for all 𝜀 > 0, there exist a partition 𝑃 ∈ Π[𝑎, 𝑏] such that 𝑈(𝑓, 𝑃) − 𝐿(𝑓, 𝑃) <
𝜀.(Bartle, 2000) 

 



 
 

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Statistical Test about Integral Value based on Random Subinterval Length  

Relating with the calculation of lower sum and upper sum based on random 

subinterval, we can present the statistical model to know the type of distribution on 

the poligons as result of a random subinterval partition. Because the length of 

subinterval gives the random value, we can make the number of partition with the 

difference value of partitions. As an example, on interval [1, 3], with number of 

partition is n = 10, we can note the  m sets of all partitions 𝑃 written by Π[1,3] =
{𝑃1, 𝑃2, 𝑃3, … , 𝑃𝑚} with: 

𝑃1 = {1, 1.23, 1.3, 1,35, 1.68, 2.01, 2.64, 2.71, 2.77, 2.97, 3} 
𝑃2 = {1, 1.11, 1.3, 1.55, 1.77, 1.97, 2.04, 2.43, 2.76, 2.84, 3} 
𝑃3 = {1, 1.33, 1.48, 1.83, 2, 2.22, 2.4, 2.6, 2.8, 3} 
𝑃4 = {1, 1.02, 1.18, 1.72, 1.74, 1.5, 2.05, 2.33, 2.71, 2.83, 3} 
𝑃5 = {1, 1.02, 1.27, 1.72, 1.85, 1.86, 2.05, 2.19, 2.54, 2.95, 3} 

⋮ 
𝑃𝑚 = {1 = 𝑥0, 𝑥1, 𝑥2, 𝑥3, 𝑥4, 𝑥5, 𝑥6, 𝑥7, 𝑥8, 𝑥9, 𝑥10 = 3} 

Based on that example, with random subinterval on[𝑎, 𝑏], we can make 𝑚 𝑠𝑒𝑡𝑠  
parition 𝑃 with 𝑛 partition points. It can be written as: 

𝑃1 = {𝑎 = 𝑥10, 𝑥11, 𝑥12, 𝑥13, 𝑥14, 𝑥15, … , 𝑥1𝑛 = 𝑏} 
𝑃2 = {𝑎 = 𝑥20, 𝑥21, 𝑥22, 𝑥23, 𝑥24, 𝑥25, … , 𝑥2𝑛 = 𝑏} 
𝑃3 = {𝑎 = 𝑥30, 𝑥31, 𝑥32, 𝑥33, 𝑥34, 𝑥35, … , 𝑥3𝑛 = 𝑏} 
𝑃4 = {𝑎 = 𝑥40, 𝑥41, 𝑥42, 𝑥43, 𝑥44, 𝑥45, … , 𝑥4𝑛 = 𝑏} 
𝑃5 = {𝑎 = 𝑥50, 𝑥51, 𝑥52, 𝑥53, 𝑥54, 𝑥55, … , 𝑥5𝑛 = 𝑏} 
                      ⋮ 
𝑃𝑚 = {𝑎 = 𝑥𝑚0, 𝑥𝑚1, 𝑥𝑚2, 𝑥𝑚3, 𝑥𝑚4, 𝑥𝑚5, … , 𝑥𝑚𝑛 = 𝑏} 

with 𝑥𝑚𝑛 is 𝑛
𝑡ℎ partition point at partition 𝑚 , with 𝑛 = 0,1,2, …, dan 𝑚 = 1,2, …. 

Then, we can obtain the area of the poligon as a lowersum of the function for every 

random subinterval. It can be presented as 𝐿𝑚×𝑛 matrix as bellow 

[
 
 
 
 
 
𝑙11
𝑙21
𝑙31
𝑙41
⋮

𝑙𝑚1

   

𝑙12
𝑙22
𝑙32
𝑙42
⋮

𝑙𝑚2

   

𝑙13
𝑙23
𝑙33
𝑙43
⋮

𝑙𝑚3

   

𝑙14
𝑙24
𝑙34
𝑙44
⋮

𝑙𝑚4

   

…
…
…
…
⋮
…

   

𝑙1𝑛
𝑙2𝑛
𝑙3𝑛
𝑙4𝑛
⋮

𝑙𝑚𝑛]
 
 
 
 
 

 

 

𝑙𝑖𝑗 is a lowersum at 𝑗
𝑡ℎ subinterval on 𝑖𝑡ℎ partition, with j = 1, 2,..., n, and i = 1, 

2,...,m. 

We can take a real valued function as an example. Let A⊆ ℝ, and 𝑔: 𝐴 →  ℝ, with  

𝑔(𝑥) =  2𝑥√1 + 𝑥2, ∀𝑥 ∈ 𝐴. By using geogebra, we form the random variable of 
𝐿 as a lower sum of 𝑓 on interval [1,3], with the number of patition  is 𝑛 = 50. If 
we made 5 types of random partition of subinterva length, then we presented the 

matrix as follow: 

𝐿𝑚×𝑛 =

[
 
 
 
 
0,00043
0,00110
0,30661
0,00052
0,00001

   

0, 00044
0,00118 0
0,37606
0,00053
0,00029

   

0,00025
0,00125
0,41889
0,00066
0,00031

   

0,00029
0, 00033
0,00061
0,00066
0,00031

   

…
…
…
…
…

   

0,16519
0,13901
0,2318
0,00027
0,00241]

 
 
 
 

             (1) 



 
 

29 

 

Mathematics Education Journals 

Vol. 6 No. 1 February 2022 

 

 

ISSN : 2579-5724   

ISSN : 2579-5260 (Online) 

http://ejournal.umm.ac.id/index.php/MEJ 

 

 

 

Based on the statitical analysis, with Kolmogorov Smirnov test by software EasiFit, 

we obtained the numerical statistic as follow at Table 2 

Table 2.  The comparison of the Statistical Test of Random Subinterval with Sign 

0,05 

Statisctical 

Test 
𝑃1 𝑃2 𝑃3 𝑃4 𝑃5 

D 0.02638 0,02356 0,02328 0,02653 0,2113 

P-Value 0,86794 0,938 0,94335 0,8636 0,97549 

Range 0,81023 0,41953 0,41881 0,35934 0,91562 

Mean 0,3001 0,03784 0,03819 0,03914 0,0399 

Variansi 0,00394 0,00212 0,00223 0,00243 0,00373 

StdDeviasi 0,06276 0,04607 0,04717 0,0493 0,06111 

Std. Error 0,00281 0,00206 0,00211 0,0022 0,00273 

 

Based on the results, we made the hypotesis tes on data of 𝐿𝑚×𝑛  as follows: 
 

𝐻0: Data follows the specifeid distribution 
𝐻1: Data do not follow the specified ditribution 

Statistical tes value of 𝐷 is represented by: 
Statistical test result of 𝑃1: 𝐷 = 0,02638 
Statistical test result of 𝑃2: 𝐷 = 0,02356 
Statistical test result of 𝑃3: 𝐷 = 0,02328 
Statistical test result of 𝑃4: 𝐷 = 0,02653 
Statistical test result of 𝑃5: 𝐷 = 0,02113 
Signifince Level : 𝛼 = 0,05  
Critical Value : 0,0607 

Critical Region : Reject 𝐻0 if 𝐷 > 0,0607 
Because all partitions gave statstical test 𝐷 < 0,0607, we can make a conclussion 
tht 𝐻0 is accepted. In other words, the random variable data of subinterval partitions 
on lowersum by using goegebra follwed the certain distribution. The distribution of 

all partition presented as: 

The first partition has a distribution Burr 4 parameter, with probability density 

function (pdf) is: 

𝑓(𝑥) =
𝛼𝑘 (

𝑥 − 𝛾
𝛽

)
𝛼−1

𝛽 (1 + (
𝑥 − 𝛾

𝛽
)
𝛼

)
𝑘+1

                 (2) 

with 𝑘 = 1,9582, 𝛼 = 1,1482, 𝛽 = 0,03125, and 𝛾 = 0,0000824 for 𝛾 ≤ 𝑥 ≤
+∞  
The second partition, has distribution 𝐵𝑢𝑟𝑟 4 𝑝𝑎𝑟𝑎𝑚𝑒𝑡𝑒𝑟, same as with the first 
partition ditribution. Its parameters are 𝑘 = 6,266, 𝛼 = 0,97426, 𝛽 = 0,20374, 
and 𝛾 = 0,00008 𝛾 ≤ 𝑥 ≤ +∞ The third partition has a distribution 𝐿𝑜𝑔 −
𝑃𝑒𝑟𝑠𝑜𝑛 3, with probability dnsity function (pdf) is: 
 

𝑓(𝑥) =
1

𝑥|𝛽|Γ(𝛼)
(
ln(𝑥) − 𝛾

𝛽
)

𝛼−1

exp (−
𝑙𝑛(𝑥) − 𝛾

𝛽
)         (3)  



 
 

30 

 

Mathematics Education Journals 

Vol. 6 No. 1 February 2022 

 

 

ISSN : 2579-5724   

ISSN : 2579-5260 (Online) 

http://ejournal.umm.ac.id/index.php/MEJ 

 

 

 

 

with 𝛼 = 9,574, 𝛽 = −0,41003, and 𝛾 = 0,01728 for 0 < 𝑥 ≤ 𝑒𝛾; 𝛽 < 0 and 
𝑒𝛾 ≤ 𝑥 < +∝; 𝛽 > 0  
The fourth partition has ditribution 𝑃𝑒𝑎𝑟𝑠𝑜𝑛 6 ,with probability density function 
(pdf) is: 

𝑓(𝑥) =
(
𝑥 − 𝛾

𝛽
)
𝛼−1

𝛽𝐵(𝛼1, 𝛼2)(1 + (
𝑥 − 𝛾

𝛽
)
𝛼1+𝛼2

                    (4) 

with 𝛼1 = 0,9371, 𝛼2 = 3,9549, 𝛽 = 0,12504 and 𝛾 = 0 for 𝛾 ≤ 𝑥 < +∞ 
The fifth partition has distribution 𝐿𝑜𝑔 − 𝑃𝑒𝑎𝑟𝑠𝑜𝑛 3 which is same with the 
distribution of third partition. Its paramaters are 𝛼 = 6,3222, 𝛽 = −0,56464, and 
𝛾 = −0,4218 for 0 < 𝑥 ≤ 𝑒𝛾; 𝛽 < 0 and 𝑒𝛾 ≤ 𝑥 < +∞; 𝛽 > 0. 

With the same process, we can determine the disitbution of uppersum based 

on random subinterval length of integral value of a real function.  We can make 

𝑚 𝑠𝑒𝑡𝑠  parition 𝑃 with 𝑛 partition points  
We can obtain the area of the poligon as an uppersum of the function for every 

random subinterval. It can be presented as 𝑈𝑚×𝑛 matrix as bellow 

𝑈𝑚×𝑛=

[
 
 
 
 
 
𝑢11
𝑢21
𝑢31
𝑢41
⋮

𝑢𝑚1

   

𝑢12
𝑢22
𝑢32
𝑢42
⋮

𝑢𝑚2

   

𝑢13
𝑢23
𝑢33
𝑢43
⋮

𝑢𝑚3

   

𝑢14
𝑢24
𝑢34
𝑢44
⋮

𝑢𝑚4

   

…
…
…
…
⋮
…

   

𝑢1𝑛
𝑢2𝑛
𝑢3𝑛
𝑢4𝑛
⋮

𝑢𝑚𝑛]
 
 
 
 
 

            (5) 

𝑢𝑖𝑗 is an uppersum at 𝑗
𝑡ℎ subinterval on 𝑖𝑡ℎ partition, with j = 1, 2,..., n, and i = 1, 

2,...,m. 

We can take a real valued function as an example. Let A⊆ ℝ, and 𝑔: 𝐴 →

 ℝ, with  𝑔(𝑥) =  2𝑥√1 + 𝑥2, ∀𝑥 ∈ 𝐴. By using geogebra, we form the random 
variable of U as an uppersum of 𝑓 on interval [1,3], with the number of partition  is 
𝑛 = 50. 

 

CONCLUSION 

Based on the simulation, we know that Geogebra can not only visualize the 

basics of definite integral concepts. Geogebra, with its combined ability to visualize 

the concept of the upper and lower sum of a function defined at certain intervals, 
can be used to explore the concept of definite integrals. One of the explorations is 

to do partitions of the same length and partitions of different lengths. Partitions of 

the same length give the number of partitions less than random partitions to 

determine the value of the integral approximation. Because of its ability to generate 

random partitions, Geogebra can also be used to explore statistical models of these 

random partitions. Based on the Kolmogorov Smirnov Normality test, we know 

that the upper sum and lower sum values of the functions that can be integrated are 

obtained with certain statistical distributions. Random partitions of upper and lower 

sum following a statistical distribution are Burr 4, Log-Pearson 3, and Pearson 6. 

 



 
 

31 

 

Mathematics Education Journals 

Vol. 6 No. 1 February 2022 

 

 

ISSN : 2579-5724   

ISSN : 2579-5260 (Online) 

http://ejournal.umm.ac.id/index.php/MEJ 

 

 

 

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