- #1
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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
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