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

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The discussion centers on the proof that the variation of an integral is equal to the integral of the variation. The initial approach incorrectly interprets the symbol ##\delta## as a derivative, leading to confusion. It is clarified that ##\delta## represents a change in the entire function rather than a pointwise derivative. The participants conclude that the proof relies on the sum rule of integration, emphasizing that the variation pertains to the function itself, not the integration variable. Ultimately, the conversation reinforces the correct interpretation of variations in the context of integrals.
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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Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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