# Approach to the Formulation of the Variable Change Theorem

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## Abstract

**:**

## 1. Introduction

^{n}(with n natural), up to the abscissa a. The methods used revealed the underlying concepts that were not be fully clarified until the nineteenth century: the limit and Riemann sums. The efficiency in the calculus of these early “integralists” is astonishing. However, Leibniz and Newton indeed gave decisive progress to the history of integration, organized around the Fundamental Theorem of Calculus (that of the reciprocity between integration and derivation) and a system of notations that allowed the integral calculus to access generality. In the eighteenth and nineteenth centuries, analysis developed in an extraordinary way, but before Cauchy, there was almost no concern to exactly define what was being talked about. However, the reflection of mathematicians returns to focus on the integral, regarding one of the great problems of the nineteenth century, that of the Fourier series deduced from a function. Therefore, Cauchy tried to define the integral over a segment of a continuous function with the help of “Riemann’s” sums (1823). Riemann, from his part, in 1854 investigated under what conditions the sums bearing his name converge when “the norm of the partition” tends to zero.

## 2. Theoretical–Methodological Foundations of the Research

- There will be an initial situation and a final situation.
- The route from one situation to another one must be unknown or cannot be accessed immediately.
- There must be a student who wants to solve it and has the necessary elements to seek relationships that allow him to transform the situation.

- Action 1. Approach to the problem.
- Action 2. Deepening the problem.
- Action 3. Selection of a work path.
- Action 4. Application of the selected route.
- Action 5. Assessment.

## 3. Didactic Strategy to Favor the Assimilation of Theorems

## 4. Application of the Didactic Strategy to Favor the Assimilation of the Variable Change Theorem

#### 4.1. Experimental Population

#### 4.2. Activity Design and Application Dynamics

#### Analysis and Discussions

- (a)
- Determine the value of a to verify that the area of the region under the curve f (x) = ax, above the horizontal axis, and between the lines x = 0 and x = 4, is 16 square units.
- (b)
- Can we construct another region bounded superiorly by the curve of f (x) = ax, inferiorly by the horizontal axis, and laterally by the lines x = x
_{1}, x = x_{2}, whose area is 16 square units? Argue your answer and determine, if possible, the values of a, x_{1}, and x_{2}. - (c)
- What relationships exist between the previously constructed regions, based on their shape and area, respectively? Discuss your answers.

**Objective:**To understand the change of variable in the integration regions and its relationship with the definite integral.

_{1}= 0, x

_{2}= 2. Then, when solving the definite integral ${\int}_{0}^{2}8xdx$, he determined the value 16. In this case, the student accepts that the region changed its shape, but according to the calculations made, the measure of its area is preserved. He stated that a transformation has occurred but did not argue the analytic relationship of the integrals ${\int}_{0}^{4}2xdx$ and ${\int}_{0}^{2}8xdx$. The rest established that the activity cannot be resolved since there is no function that meets the requirements. Others tended to proceed in the wrong way; after the teacher reoriented them, the difficulty of the activity was reduced.

- (a)
- How can you justify that, in fact, the areas of the regions are equal?
- (b)
- What are the implications of making a variable change?
- (c)
- What effects does the application of the Variable Change Theorem have on the resolution of the integral ${\int}_{0}^{2}\frac{1}{2}{\left(\frac{1}{2}x+1\right)}^{3}dx$?

**Objective:**To establish the scope and implications of the change of variable in the resolution of definite integrals.

- Student B: choose a suitable substitution for a variable in such a way that its derivative is related to the original function, and thus, it remains in simpler terms.
- Student F: to change a variable is to take a function to a simpler one. For this, we modify the evaluation interval that depends on the new variable that has been chosen.
- Student N: it helps us to make our work easier, but it is not always necessary to do so.
- Student P: transform the function in a certain interval into another function defined in another interval, of which it is easy to calculate the area delimited by the same function.

- (a)
- Make a graphical representation of the region and calculate its area without using integral calculus.
- (b)
- To calculate the area of the region above, the following integral $A\left(R\right)={\int}_{0}^{4}\left(x+1\right)dx$ is proposed. Applying the change of variable, u = x + 1, with du = dx, it follows that $A\left(R\right)={\int}_{0}^{4}\left(x+1\right)dx={\int}_{0}^{4}udu=\frac{{4}^{2}}{2}=8$. Discuss the obtained results.

**Objective:**to fix the theorem of the change of variable of the definite integral, in particular, to recognize the error associated with the region of integration.

- Student L: I do not see any errors.
- Student H: The figure shown is not a trapezoid.

**Objective:**to apply the change of variable theorem in the solution of the definite integral.

- Student E: in order to solve the integral ${\int}_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\mathrm{cos}x}{\mathrm{sin}x}dx$, we make $u=sinx$, $du=cosx\text{}dx$, evaluate $x=\frac{\pi}{2}$ → u $=$ 1 and $x=\frac{\pi}{4}$ → $u=$ $\frac{1}{\sqrt{2}}$, then$${\int}_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\mathrm{cos}x}{\mathrm{sin}x}dx={\int}_{\frac{1}{\sqrt{2}}}^{1}\frac{1}{u}du=ln\text{}\left(u\right)|\begin{array}{c}1\\ \frac{1}{\sqrt{2}}\end{array}=ln\text{}\left(1\right)\text{}-\text{}ln\text{}(\frac{1}{\sqrt{2}})=ln\text{}\left(1\right)-ln\text{}\left(1\right)+ln\left(\sqrt{2}\right)=ln\left(\sqrt{2}\right).$$

**Objective:**to apply the variable change theorem as a proof method.

- (a)
- Which is the relationship between ${\int}_{\frac{\pi}{4}}^{\frac{\pi}{2}}\raisebox{1ex}{$\mathrm{cos}x$}\!\left/ \!\raisebox{-1ex}{$\mathrm{sin}x$}\right.dx$ and ${\int}_{\frac{1}{\sqrt{2}}}^{1}\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$x$}\right.dx$?
- (b)
- While solving a definite integral using the change of variable theorem, what variations can you consider to occur in the main function for the given interval?

**Objective:**to assess the effects produced by the change of variable in the region of integration and in the calculation of the definite integral.

- Student B: The integral remains in terms of a function of easy integration; in addition, such change of variable must be adequate.
- Student C: The change of variable is used to put an integral in its simplest way to calculate, now when the change of variable is completed in a definite integral, the intervals of the function change.
- Student F: The change of variable is a method that allows us to solve integrals, this is completed in order to transform an integral to another simpler one, what the change of variable does is change a function for another simpler one; this implies that the evaluation intervals change with respect to the new variable.
- Student M: What makes the integral less complex by making it easier to integrate, since it changes the integration intervals and is less complex.
- Student Q: The change of variable modifies a function and its intervals of integration in such a way that it produces another function defined in a different interval, but that when integrating it turns out to have the same area as the function on which the change of variable was applied.

**Objective:**to evaluate necessary conditions for the application of the variable change theorem.

- Student H: If we make $u=\mathrm{tan}\left(\frac{x}{2}\right),$ in ${\int}_{0}^{2\pi \text{}}\frac{1}{2+cos\left(x\right)}\text{}dx$, then we obtain that the new limits of integration are $\mathrm{tan}\left(\frac{0}{2}\right)=0=\mathrm{tan}\left(\frac{2\pi}{2}\right)=\mathrm{tan}\left(\pi \right)=0$. It transforms a region with positive area into a point.
- Student K: If we transform the integral ${\int}_{0}^{2\pi \text{}}\frac{1}{2+cos\left(x\right)}\text{}dx$, to an undefined one, and solve it by using the change of universal variable, the integral gives us ${\int}^{\text{}}\frac{1}{2+cos\left(x\right)}\text{}dx$ = $\frac{2}{\sqrt{3}}arctan\left(\frac{u}{\sqrt{3}}\right)$, but if we derive it, then it does not fulfill the function $\frac{1}{2+cos\left(x\right)}$ over the entire interval [0,2π].

**Objective:**to propose an approach to the formulation of the change of variable theorem.

- Student B: given a function f(x), and the variable change theorem allows to find F(x), such that: $F\left(x\right)={\int}^{\text{}}f\left(x\right)dx$, then through a change of variable ${\int}^{\text{}}f\prime \left(u\right)\xb7u\prime dx=F\left(u\right)+c$, the change of variable is based on a suitable substitution in such a way that the integral is simpler.
- Student E: ${\int}_{a}^{b}f\left(x\right)dx={\int}_{a}^{b}\phi \left(x\right)dx$, where $u=\phi \left(x\right)\to du=dx$, when using the change of variable also changes the interval.
- Student F: theorem. If f(x) is continuous, then the integral ${\int}_{a}^{b}f\left(x\right)dx={\int}_{g\left(a\right)}^{g\left(b\right)}f\left(g\left(u\right)\right){g}^{\prime}\left(u\right)du$.
- Student J: let f(x) be an integrable function on the interval [a, b] and continuous. When solving for change of variable, we obtain: ${\int}_{a}^{b}f\left(x\right)dx={\int}_{u\left(a\right)}^{u\left(b\right)}udu$, the function and the limits of integration change, and we evaluate the function in the upper and lower limits obtained when changing the variable.
- Student Q: if a function f(x) is continuous on the interval [a, b], let u = φ (x) be differentiable function composed of f(u), then ${\int}_{a}^{b}f\left(x\right)dx={\int}_{u\left(a\right)}^{u\left(b\right)}f\left(u\right){u}^{\prime}du$.

## 5. Conclusions

## Author Contributions

## Funding

## Institutional Review Board Statement

## Informed Consent Statement

## Data Availability Statement

## Acknowledgments

## Conflicts of Interest

## References

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Theorems | Actions |
---|---|

Search for the theorem | -Search for assumptions -State a thesis -Formulate a theorem |

Search for the proof | -Find an idea of the proof -Come up with a demonstration plan -Develop or complete the proof |

Strategy for Solving Problems | Stages of the Assimilation Process | Operations |
---|---|---|

Approach to the problem | Comprehension | -Read the problem. -Are all the terms involved in formulating the problem familiar to you? -State the problem in your own words, specify what is given and what is sought, identify the knowledge that is related to what is given and what is sought. -Is it an intra-mathematical or extra-mathematical problem? -What field of knowledge can you associate the problem with? -Does it require the use of mathematical knowledge? -Have you seen any other one formulated in a similar way? -Is it a problem related to your sociocultural environment? |

Deepening the problem | Identification | -Determine the necessary knowledge to face the problem. -Have you seen any other one formulated in a similar way? -Establish analogies, experiment with other data, and propose regularities. -Have all possible cases been checked? -Underline the expressions that you consider to have the greatest semantic value in the problem. -Look for synonyms and antonyms of the terms that you consider fundamental. -Establishes the unknown(s), that is, what is being sought. -Determine the data that are given directly in the formulation of the problem. -Between what values should it be found? -In a second moment you can think of developing a scheme, diagram, table, etc. -Is there enough, or is there too much data? -Are there contradictory elements? |

Selecting a work path | Fixation | -Transform the problem into another equivalent one. -How can the data be related to the unknown(s)? -What inferences can be made from the data found? -Evaluate counterexamples. -Correctly use mathematical terminology and symbols. -Choose an appropriate language or a suitable notation. -Define what knowledge is related to the elements of the problem. -Justify ideas, judgments, and arguments. -Select the properties or definitions that may be useful to you. -Assume the problem is solved. -In which field of knowledge does the proposed problem move: arithmetic, algebraic, or geometric? -Define which knowledge system is related to the elements of the problem. -Which one of them is related to the premise or thesis of the problem? -Could a possible answer be given? |

Application of the selected route | Application | -Establish relationships in correspondence with the forms of work. -Use cognitive and metacognitive strategies. -Execute an action plan and consciously apply the procedures. -Make equivalent transformations in the premise and/or the thesis. -Analyze the possible ways of solution. -Consider particular and general cases. -Integrate the possible results so that they can be used in different contexts. -Can you apply that same work technique to another situation? -Can the method used to solve other problems be generalized? -Make assumptions based on possible solutions. |

Assessment. Consider and take a critical stance on the system of actions and operations carried out | Valoration | -What conjectures can you make? -Analyze the steps and actions taken; analyze errors and their possible causes; specify how to avoid errors. -Explain the reasoning process carried out and give details about it. -Think upon the procedures and methods used. -Choose an appropriate language or a suitable notation. -Are all the solutions found for the problem? -Explain with your words how you arrived at the solution. -Does the answer given make sense in relation to your experience? -Does the proposed solution really answer the problem in question? -What have you contributed from the social and/or mathematical point of view with the work on the problem? |

${\int}^{\text{}}{x}^{\alpha}dx=\frac{{x}^{\alpha +1}}{\alpha +1}+C,\alpha \ne -1$ | ${\int}^{\text{}}\frac{dx}{x}=\mathrm{ln}\left|x\right|+C$ | ${\int}^{\text{}}\mathrm{sin}xdx=-\mathrm{cos}x+C$ |

${\int}^{\text{}}\mathrm{cos}xdx=\mathrm{sin}x+C$ | ${\int}^{\text{}}\frac{dx}{{\mathrm{cos}}^{2}x}=\mathrm{tan}x+C$ | ${\int}^{\text{}}\frac{dx}{{\mathrm{sin}}^{2}x}=-\mathrm{cot}x+C$ |

${\int}^{\text{}}\mathrm{tan}xdx=-\mathrm{ln}\left|\mathrm{cos}x\right|+C$ | ${\int}^{\text{}}\mathrm{cot}xdx=\mathrm{ln}\left|\mathrm{sin}x\right|+C$ | ${\int}^{\text{}}{a}^{x}dx=\frac{{a}^{x}}{\mathrm{ln}a}+C$ |

${\int}^{\text{}}\frac{dx}{{a}^{2}+{x}^{2}}=\frac{1}{a}\mathrm{arctan}\frac{x}{a}+C$ | ${\int}^{\text{}}\frac{dx}{{a}^{2}-{x}^{2}}=\frac{1}{2a}\mathrm{ln}\left|\frac{a+x}{a-x}\right|+C$ | ${\int}^{\text{}}\frac{dx}{\sqrt{{a}^{2}-{x}^{2}}}=\mathrm{arcsin}\frac{x}{a}+C$ |

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## Share and Cite

**MDPI and ACS Style**

Carballo, A.M.; Locia Espinoza, E.; Sigarreta Almira, J.M.; Yero, I.G.
Approach to the Formulation of the Variable Change Theorem. *Educ. Sci.* **2021**, *11*, 357.
https://doi.org/10.3390/educsci11070357

**AMA Style**

Carballo AM, Locia Espinoza E, Sigarreta Almira JM, Yero IG.
Approach to the Formulation of the Variable Change Theorem. *Education Sciences*. 2021; 11(7):357.
https://doi.org/10.3390/educsci11070357

**Chicago/Turabian Style**

Carballo, Armando Morales, Edgardo Locia Espinoza, José M. Sigarreta Almira, and Ismael G. Yero.
2021. "Approach to the Formulation of the Variable Change Theorem" *Education Sciences* 11, no. 7: 357.
https://doi.org/10.3390/educsci11070357