#### Arman777

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- Problem 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

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