Proof that Variation of Integral is Equal to Integral of the Variation

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Homework Help Overview

The discussion revolves around the mathematical proof that the variation of an integral is equal to the integral of the variation, focusing on the notation and interpretation of the delta symbol (##\delta##) in this context.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the meaning of the delta symbol in relation to functions and integrals, questioning its interpretation as a derivative versus a change in function. There are attempts to apply specific examples, such as ##F(x)=x^2##, to the general proof.

Discussion Status

The discussion is ongoing, with some participants providing clarifications about the use of the delta notation and its implications for the proof. There is a recognition of the need to differentiate between changes in functions and changes in variables.

Contextual Notes

Participants are navigating potential confusion regarding the application of the delta notation in different contexts, particularly in relation to integration and function variation.

Arman777
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Homework Statement
I need to proof that $$\delta(\int_a^b F(x)dx)=\int_a^b\delta F(x)dx$$
Relevant Equations
The variation comes from the calculus of variation. For a given path the extremum occcurs on $$\delta I=\delta \int_{x_1}^{x_2}f (y,y';x) dx=0$$
I actually don't know how to proceed.

I tried something like this

The left side of the equation equals to $$\delta(\int_a^b F(x)dx)=\delta f(x) |_{a}^{b}$$
where ##f'(x)=F(x)##

However $$\delta f(x) |_{a}^{b}=f'(x)\delta x dx|_{a}^{b} = \delta (F(b)-F(a))$$

where ##f'(x)=F(x)##. For the right side of the equation can I say that
$$\int_a^b\delta F(x)dx = \int_a^b F'(x)\delta x dx = \delta (F(b)-F(a))$$

Is this true ? Thanks
 
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I think you're confused. The ##\delta## doesn't mean a derivative, so you can't write ##\delta f = f' \delta x dx##.

The ##\delta## just means a change. Not a change as a function of ##x##, but a change in the entire function. Imagine that you have two functions, an original function, ##F(x)##, and a slightly different function ##\overline{F}(x)##. Then ##\delta F(x) = \overline{F}(x) - F(x)##. ##\delta ( \int F(x) dx) = \int \overline{F}(x) dx - \int F(x) dx##. So you're just asking to prove that:

##\int \overline{F}(x) dx - \int F(x) dx = \int (\overline{F}(x) - F(x)) dx##
 
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oh I see. In examples such as for ##F(x)=x^2## we are writing ##\delta F(x) = 2x\delta x## So I thought I can apply it here.

From your last equation it seems that prove is finished since its just the sum rule of the integration
 
Typically, the integration variable is not what is being varied. What is being varied is usually the function that the functional depends on.
 
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Arman777 said:
oh I see. In examples such as for ##F(x)=x^2## we are writing ##\delta F(x) = 2x\delta x## So I thought I can apply it here.

Well ##\delta## means a difference in two values. Sometimes it means the difference between the same function at two different points, and sometimes it means the difference between two different functions at the same point. The latter is what is meant here.
 
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