General variation of a functional

Question

In Calculus of variations, I. M. Gelfand, S. V. Fomin, Section 13 they are showing how to variate functional with variable endpoints. How do they get from (4) to the next formula of the differential of the functional? How do they simplify the integrals which boundaries have pluses-the last two integrals in (4)? Edit: I have uploaded picture of that section, sorry for not typing it but it will take quite some time.

Answer 1

$$\small\int_{x_0}^{x_1}F(x,y+h,y'+h')-F(x,y,y')\,dx+\int_{x_1}^{x_1+\delta x_1}F(x,y+h,y'+h')\,dx-\int_{x_0}^{x_0+\delta x_0}F(x,y+h,y'+h')\,dx$$

The first integral is simplified with Taylor's theorem.
The second becomes $$\int_{x_1}^{x_1+\delta x_1}\color{red}{F(x,y,y')}+\color{blue}{hF_y(x,y,y')}+\color{green}{h'F_{y'}(x,y,y')}\,dx\\ \small=\color{red}{[F(x,y,y')]_{x_1}\delta x_1}+\color{blue}{\int_{x_1}^{x_1+\delta x_1}hF_y(x,y,y')\,dx}+\color{green}{\left[hF_{y'}(x,y,y')\right]_{x_1}^{x_1+\delta x_1}-\int_{x_1}^{x_1+\delta x_1}h\frac d{dx}\big(F_{y'}(x,y,y')\big)\,dx}$$ Explanation:
$\color{red}{\text{red}}$ - For $\delta x_1$ small, this is just a small strip of the function from $x_1$ to $x_1+\delta x_1$, and $F$ hardly changes as a result. So this integral is just the area of a very thing strip (a rectangle), so it becomes base $\times$ height, where base is $\delta x_1$ and height is $F(x,y,y')$ evaluated at $x_1$.
$\color{blue}{\text{blue}}$ - Nothing changes.
$\color{green}{\text{green}}$ - integration by parts, differentiate the function with $F$ and integrate the $h'$.

The third integral follows a similar process as this, but just has negative signs.

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Edit:

So the total integral becomes

$$\int_{x_0}^{x_1}hF_y(x,y,y')-h\frac d{dx}\big(F_{y'}(x,y,y')\big)\,dx\\ \small+{[F(x,y,y')]_{x_1}\delta x_1}+{\int_{x_1}^{x_1+\delta x_1}hF_y(x,y,y')\,dx}+{\left[hF_{y'}(x,y,y')\right]_{x_1}-\int_{x_1}^{x_1+\delta x_1}h\frac d{dx}\big(F_{y'}(x,y,y')\big)\,dx}\\\small-{[F(x,y,y')]_{x_0}\delta x_0}-{\int_{x_0}^{x_0+\delta x_0}hF_y(x,y,y')\,dx}-{\left[hF_{y'}(x,y,y')\right]_{x_0}+\int_{x_0}^{x_0+\delta x_0}h\frac d{dx}\big(F_{y'}(x,y,y')\big)\,dx}\\ =\int_{x_0+\delta x_0}^{x_1+\delta x_1}h\left(F_y-\frac d{dx}\big(F_{y'}\big)\right)\,dx+F\big|_{x_1}\delta x_1-F\big|_{x_0}\delta x_0+F_{y'}\cdot h\big|_{x_1}-F_{y'}\cdot h\big|_{x_0}$$