Proof that Variation of Integral is Equal to Integral of the Variation I need to proof that $$\delta(\int_a^b f(x)dx)=\int_a^b\delta f(x)dx$$
I actually dont know how to proceed. Can I use the 
Leibniz integral rule as an example for the derivation ? In that case he is using $$d/dx$$ but we have $$\delta$$ so I am not sure I can use that proof.
Any help. Thanks
The variation comes from the calculus of variation. For a given path the extremum occcurs on $$\delta I=\delta \int_{x_1}^{x_2}f (y,y';x) dx=0$$
 A: For variation of a function[al] we typically mean the small change in its value due to a class of variations of the argument. Hence, we must first agree on what is the varying argument here (and even how it is supposed to vary).
For example, the first equality you propose holds if you are interested in the variation of the definite integral for varying $f$ (called the dependent variable, for simplicity):
$$\delta \int \limits_{a}^{b}f(x)\mathrm{d}x=\int \limits _a^b\left(f(x)+\delta f(x)\right)\mathrm{d}x-\int \limits _a^bf(x)\mathrm{d}x=\int \limits _a^b\delta f(x)\mathrm{d}x\ .$$
On the other hand, a different form of the variation of the integral is obtained if the varying quantity is related to the independent variable, i.e. the extrema of integration.
Typically (but not always), in the calculus of variations we are interested in analysing the change a definite integral is subjected to when the dependent variables change. In these cases, it is true that for a continuous function $F$ $$\delta \int \limits_a^bF(x,y,y')\mathrm{d}x=\int \limits _a^b \delta F(x,y,y')\mathrm{d}x\ ,$$ where $\delta F=\frac{\partial F}{\partial y}\delta y+\frac{\partial F}{\partial y'}\delta y'$.
