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 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

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

Staff Emeritus