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

Gold Member
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 dont 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

stevendaryl
Staff Emeritus
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##

• Delta2, Arman777 and Orodruin
Gold Member
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

Orodruin
Staff Emeritus
Homework Helper
Gold Member
Typically, the integration variable is not what is being varied. What is being varied is usually the function that the functional depends on.

• Arman777
stevendaryl
Staff Emeritus
• 