Proof of integral property

In summary, we are trying to prove that the integral of a constant 1-form from point a to point c is equal to the sum of the integral from a to b and b to c. This follows from the transitive property, which states that the sum of two quantities is equal to the sum of their individual parts. However, this property only applies to constant 1-forms, as non-constant forms would not allow for such simplification.
  • #1
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Given a constant one-form [tex] k_1 \ dx + k_2 \ dy + k_3 \ dz [/tex] in [tex] \bold{R}^{3} [/tex], and three points [tex] \vec{a}, \ \vec{b}, \ \vec{c} [/tex] in [tex] \bold{R}^3 [/tex], prove that [tex] \int_{\vec{a}}^{\vec{c}} k_1 \ dx + k_2 \ dy + k_3 \ dz = \int_{\vec{a}}^{\vec{b}} k_1 \ dx + k_2 \ dy + k_3 \dz + \int_{\vec{b}}^{\vec{c}} k_1 \ dx + k_2 \ dy + k_3 \ dz [/tex].

So we want to show that [tex] k_{1}(c_1-a_1) + k_2(c_2-a_2) + k_3(c_3-a_3) = k_{1}(b_1-a_1) + k_2(b_2-a_2) + k_3(b_3-a_3) + k_1(c_1-b_1) + k_2(c_2-b_2) + k_3(c_3-a_3) [/tex].

Doesn't this follow from the transitive property, or the triangle inequality?
 
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  • #2
Since there is no inequality there, it certainly isn't a "triangle inequality"!

Perhaps if you stated the "transitive property" for integrals.
 
  • #3
Well I think the key idea is that we are using constant 1-forms. If it were not constant, then we could not use the transitive property.
 
  • #4
[tex] k_{1}(c_1-a_1) + k_2(c_2-a_2) + k_3(c_3-a_3) = k_{1}(b_1-a_1) + k_2(b_2-a_2) + k_3(b_3-a_3) + k_1(c_1-b_1) + k_2(c_2-b_2) + k_3(c_3-b_3) [/tex] You had an error here in the last parentese

if you collect the terms with same k you get

[tex] k_{1}(b_1-a_1) + k_2(b_2-a_2) + k_3(b_3-a_3) + k_1(c_1-b_1) + k_2(c_2-b_2) + k_3(c_3-b_3) = [/tex]

[tex]k_{1}(b_1-a_1 +c_1-b_1) + k_2(b_2-a_2+c_2-b_2)+k_3(b_3-a_3+c_3-b_3)= k_{1}(c_1-a_1) + k_2(c_2-a_2)+k_3(c_3-a_3)[/tex]

is that what you want?
 
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1. What is the proof of integral property?

The proof of integral property is a mathematical concept that proves the relationship between the definite integral and the corresponding function. It states that if a function is continuous on a closed interval, then the area under the curve of that function can be calculated using the definite integral.

2. Why is the proof of integral property important?

The proof of integral property is important because it provides a mathematical basis for calculating the area under a curve. This is a fundamental concept in many fields of science, including physics, engineering, and economics.

3. How is the proof of integral property used in real-world applications?

The proof of integral property is used in real-world applications to calculate the area under curves in various scenarios. For example, it can be used to calculate the work done by a force in physics or to calculate the total revenue of a company in economics.

4. What are the key components of the proof of integral property?

The key components of the proof of integral property include the fundamental theorem of calculus, which states that the definite integral of a function is equal to the difference between the values of the function at the upper and lower bounds of the integral, and the concept of Riemann sums, which are used to approximate the area under a curve.

5. Are there any limitations to the proof of integral property?

While the proof of integral property is a fundamental concept in mathematics, it does have some limitations. For example, it can only be applied to continuous functions on a closed interval, and it may not be applicable to more complex functions or scenarios. Additionally, it relies on the assumptions of the fundamental theorem of calculus, which may not hold in all cases.

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