Different Solutions to a Mathematical Problem: A Case Study of Calculus 12

The International Journal Of Engineering And Science (IJES)
|| Volume || 5 || Issue || 8 || Pages || PP -40-44|| 2016 ||
ISSN (e): 2319 – 1813 ISSN (p): 2319 – 1805
www.theijes.com The IJES Page 40
Different Solutions to a Mathematical Problem: A Case Study of
Calculus 12
Nguyen Phu Loc1*,
Duong Huu Tong2,
Huynh Huu Dao Vu3
1,2
School of Education, Can Tho University, Vietnam
3
Tran Ngoc Hoang High School, Can Tho City
--------------------------------------------------------ABSTRACT-----------------------------------------------------------
An important duty of the mathematics teacher is to train and to develop thinking for students. To accomplish
this duty, teachers can organize creative activities for students through activities of solving problems. In
particular, there is an effective way to train students to think is that teachers can organize activities of solving
problems in many different ways. Based on this idea, we implement an experiment for students in grade 12 to
calculate integrals in various ways. The results of the study showed that students were active to find out
different solutions to the given problem.
Keywords: Calculus teaching, mathematics education, multi-solution to a problem, problem solving
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Date of Submission: 17 May 2016 Date of Accepted: 10 August 2016
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I. INTRODUCTION
Solving a problem in different ways has an important significance for the teaching of mathematics. Polya (1973,
6.61) insisted that encouraging students to get different results can help them obtain even the more elegant
solutions. To be able to solve the problem in different ways, students need to mobilize as much relevant
knowledge, then they continue to analyze, synthesize and evaluate the facts of the problem and finally find a
reasonable explanation. That process requires students to use the necessary thinking operations and logical
reasoning. Silver, Ghousseini, Gosen, Charalambous & Strawhun, (2005, p.228) argued that different solutions
can facilitate the connection of a problem with the various elements of the knowledge with which a student may
be familiar, thus leading to strengthen the networks of related ideas.
In reality, the current calculus 12 (Giải tích 12) textbook in Vietnam provides students with many problems
about calculating integrals (Hao, 2006). However, in our study, there is no any problem that requires students to
solve in many different ways. In this case study, we want to find out the answer to the following question:
The research question: For a mathematical problem with requirement of solving in different ways, are students
active to find out various solutions to the problem?
II. METHODOLOGY
2.1. Participants
The experiment was carried out in class 12A2 at the high school of Tran Ngoc Hoang, Can Tho City.
This class consisted of 31 students.
Time: started at 9 45 and ended 10:00, 01/29/2015.
2.2. Instrument and procedure
Students were requested to solve the following problem:
Calculate the following integral in different ways:
2
0
sin
4
I x d x

 
  
 
 .
Students worked individually (15 minutes). Students had to solve the problem printed on a paper. Every student's
behavior was shown on assignments. More specifically, students showed their ability in dealing with solving the
problem in various ways.
2.2.3. Pre-analyzing the problem
a. The class context of the experimental problem
The problem was organized for students after they had completed the primitive and integral topics. In particular,
the methods of finding primitive were mentioned.
b. The solutions expected to the problem

Preparation of Papers for the International Journal of Engineering and Science
Table 1 shows the possible solutions to the problem
Table 1: The solutions expected to the problem
Code Solution strategy Solution
S1 Applying the
definition of
primitive
'
2 2 2
0 0 0
sin co s co s
4 4 4
I x d x x d x x
  
        
             
      
 
2 2
cos cos 0
4 4 2 2
    
          
   
S2 Changing the
function under the
integral sign in the
supply side
2 2 2
0 0 0
sin co s sin
4 4 4
I x d x x d x x
  
       
          
     
 
3 2 2
sin sin 0
4 4 2 2
 
    
S3 Using the
relationship
between the odd
property of function
and its integral on
an interval.
Let
4
t x d x d t

     . Therefore, 0 ,
4 2 4
x t x t
  
      
4 4
4 4
sin sinI td t td t
 
 


   
  sinf t t is odd on ;
4 4
  
 
 
, then
4 4
4 4
sin sin 0I td t td t
 
 


    
S4 Using derivative
spellings 2 2 2
0 0 0
sin sin co s
4 4 4 4
I x d x x d x x
  
          
              
       
 
2 2
co s co s 0
4 4 2 2
    
         
   
S5 Changing the
function under the
integral sign in the
addition formula
 
2 2 2
0 0 0
2
sin sin cos cos sin cos sin
4 4 4 2
I x dx x x dx x x dx
  
     
        
   
  
 
2
0
2 2
sin co s sin co s sin 0 co s 0 0
2 2 2 2
x x

  
       
 
S6 Changing variables,
changing the limits
of the integral
Let
4
t x d x d t

     . Therefore, 0 ,
4 2 4
x t x t
  
      
4 4
4
4
4 4
2 2
sin sin cos cos cos 0
4 4 2 2
I tdt tdt t
 


 
 



 
             
 
 
S7 Changing variables,
but do not change
the limits of the
integral
Let
2
t x d t d x

     . Therefore, 0 , 0
2 2
x t x t
 
     
0 2
0
2
sin sin
4 4
I t d t t d t


    
       
   
 

2
0
3 2 2
co s co s co s 0
4 4 4 2 2
t

   
          
 
III. RESULTS AND DISCUSSION
Table 2: Statistics of the number of students with the right strategies, and the right solutions
S1 S2 S3 S4 S5
Number of
students
29 (93.5%) 14 (45.2%) 31 (100%) 18 (58.1%) 31 (100%)
Number of
students
doing
correctly
9 (31%) 12 (85.7%) 14 (45.2%) 17 (94.4%) 28 (90.3%)
The Table 2 indicated that the dominant solutions belonged to S5 (31/31 students, at 100% respectively). In
general, the participants were in the right direction, i.e., using integral’s linear property by converting the given
integral into addition (or subtraction) of the two integrals which might be used directly in the primitive table.
Moreover, some of them also added the formula  sin sin co s co s sina b a b a b   in their work. This could be
explained easily since these participants were familiar with applying the theoretical basis in solving problems.
Three students (representing 9.68 %) had inaccurate answers.
The solution of S3 was also as good as that of S5 (31/31 students, approximately 100%). However, only 14
students (representing 45.16 %) had the correct answers when choosing this strategy. Most errors were caused by
using wrong values after changing integral variables. Understanding from the concept of definite integrals on
 ; ,a b a b might lead learners to this mistake though they had been noted that: "In the case of a b or a b ,
there is a convention that      0;
a b a
a a b
f x d x f x d x f x d x     " , [ 2 , p.105 ] .
S1 solution was also chosen by many participants (29/31 students, accounted for 93.55%). In general, learners
had mastered primitive concept and consciously had a focus on this method. Though so, only 9 students
(representing 31.03%) got the correct answer. This might be apparently explained as in the current textbook
program, assignments can be solved through specifically formulas in the primitive table, i.e., changing variables
or solving through another unknown factor. Students do not probably have many opportunities to practice this
method when they do not practice finding different ways to get a problem result.
S4 solution was also taken by a number of learners (18/31 students, accounted for 58.06%). This was the highest
rate for those who had the right answer (17/18 students, accounted for 94.44%) when selecting and presenting
solutions by this method. Reflecting on the advantages (and disadvantages) of each solution in the survey
showed the superiority of differential method.
S2 solution was the least one to be chosen, because most students claimed that there was no difference between
two primitive functions: sin , and cos
4 4
x x
    
    
   
. These functions were available in the extended primitive
table. Meanwhile, for some students, finding primitive functions of cos
4
x
 
 
 
helped them to limit sign errors.
In other words,  sin 1 cos cos
4 4 4
x dx x C x C
        
              
      
 might lead to sign errors, while
cos sin
4 4
x dx x C
    
      
   
 rarely caused minus sign missing as in the previous function.
The errors that students might deal with when presenting solutions in different ways include these things.
Some of them wrote redundant symbols while transforming function under the integral sign (6 attendants). One
example of this error was student 19’s paper:
2
0
sin co s co s
4 2 4 4
x d x x x

         
           
      
 . In addition, a
few students missed the arc sign, or got the incorrect arc of trigonometric functions (4 students). Here was the

error of H31:
2 2 2 2
0 0 0 0
2 2 2 2
co s sin sin co s
2 2 2 2
xd x xd x x
   
     . Meanwhile, the right answer should be:
2 2 2 2
0 0 0 0
2 2 2 2
co s sin sin co s
2 2 2 2
xd x xd x x x
   
     . Besides, there are 6 attendants had the wrong sign. Here
was the error in H5’s task
2 2
0 0
sin sin
4 4 4
x d x x d x
 
       
        
     
  , which was supposed to be written as:
2 2
0 0
sin sin
4 4 4
x d x x d x
 
       
         
     
  . Another issue was that there were also 2 students having the wrong
approach
2 4
0 0
sin co s
4 4
x d x x d x
 
    
     
   
  while the correct answer would be:
2 2
0 0
sin co s
4 4
x d x x d x
 
    
     
   
  .
There are plenty of students (12 students) lacked the upper and lower values after finding
primitive
2
'
0
co s
4
sin
4
4
x
x d x
x
 


 
  
   
  
   
 
 
 . They should write:
2
2
'
0
0
co s
4
sin
4
4
x
x d x
x

 


 
  
   
  
   
 
 
 .
Other students also lacked dx in their paper. Here was the error of one learner in H15
2 2
0 0
sin co s
4 4
x d x x
 
    
     
   
  , while the right way is
2 2
0 0
sin co s
4 4
x d x x d x
 
    
     
   
  . The most serious
mistakes belonged to two participants who developed the wrong trigonometric formulas:
 sin sin sina b a b   . Meanwhile, there were 6 others who lacked derivative notation sign:
2
2
0
0
co s
4
sin
4
4
x
x d x
x

 


 
  
   
  
  
 
 
 . These students should write:
2
2
'
0
0
co s
4
sin
4
4
x
x d x
x

 


 
  
   
  
   
 
 
 . Some also
applied the wrong primitives formula
2
2
0 0
sin co s
4 4
x d x x


    
      
   
 , which should be understood as
 
2
2
0 0
sin 1 co s
4 4
x d x x


     
        
    
 . Finally, only one student did not find the differential when he changed
the variables in his paper. Overall, besides the majority of students who had the right solutions differently, the
rest ones still had difficulty in the given problem.
Table 3: Statistics of students according to their right solutions
Numbers of
solutions
0 1 2 3 4 5
Numbers of
students
0 (0%) 0 (0%) 0 (0%) 1 (3.2%) 7 (22.6%) 23 (74.2%)

Table 3 showed that students tried to find out different solutions to the problem. The strength of the experimental
group was that they all had 3 or more correct solutions. Among these, 74.2% accounted for 23 students with 5
right solutions. This showed that they had used pretty well the integral knowledge in current textbooks, thereby
contributing to the formation of their intellectual qualities. Besides, there were 7 students with 22.6%
respectively who had 4 correct methods, and most of them were not successful in using the arc formula. Among
31 experimental attendants, only 15 ones calculated integrals in three ways. The reason was that these students
did not solve the problems by transforming trigonometry method.
IV. CONCLUSIONS
The data analysis has pointed out the answers to the research questions: the students were active to solve the
problem in different ways. In particular, learners have really exploited the necessary knowledge in textbook to
give out reasonable explanation for their solutions, indirectly contributed to the training and improving learners’
critical thinking. Another notable conclusion is that there are still a lot of mistakes when students calculate the
integral relating to missing dx , lacking upper and lower values, using wrong primitive formulas, wrong values,
and not memorizing trigonometric formulas, etc. We believe that this is an interesting and essential research
topic which can be further studied in the near future.
REFERENCES
[1]. Hao, T. V (editor),. (2006). Calculus 12 (Giải tích 12), Hanoi: Publishing house Giáo dục. (in Vietnamese)
[2]. Polya, G. (1973). How to solve it. A new aspect of mathematical method. Princeton, NJ: Princeton University Press.
[3]. Silver, E. A., Ghousseini, H., Gosen, D., Charalambous, C., & Font Strawhun, B. T. (2005). Moving from rhetoric to praxis:
Issues faced by teachers in having students consider multiple solutions for problems in the mathematics classroom. Journal of
Mathematical Behavior, 24, 287–301.

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