The Euler-Lagrange equation

Question

Here's an (edited) excerpt from my Calculus textbook. I've highlighted the part I didn't understand in boldface, so could anyone explain those bits? Let us consider the integral $$ I=\int_{a}^{b} F\left(y, y^{\prime}, x\right) d x $$ where $a, b$ and the form of the function $F$ are fixed by given considerations. The curve $y(x)$ is to be chosen so as to make stationary the value of $I$. Let us suppose that $y(x)$ is the function required to make $I$ stationary and consider making the replacement $$ y(x) \rightarrow y(x)+\alpha \eta(x) $$ where the parameter $\alpha$ is small and $\eta(x)$ is a nicely behaved arbitrary function. For the value of $I$ to be stationary with respect to these variations, we require $$ \left.\frac{d I}{d \alpha}\right|_{\alpha=0}=0 \quad \text { for all } \eta(x) $$ Expanding as a Taylor series in $\alpha$, we obtain $$ \begin{aligned} I(y, \alpha) &=\int_{a}^{b} F\left(y+\alpha \eta, y^{\prime}+\alpha \eta^{\prime}, x\right) d x \\ &=\int_{a}^{b} F\left(y, y^{\prime}, x\right) d x+\int_{a}^{b}\left(\frac{\partial F}{\partial y} \alpha \eta+\frac{\partial F}{\partial y^{\prime}} \alpha \eta^{\prime}\right) d x+\mathrm{O}\left(\alpha^{2}\right) \end{aligned} $$ Thus, $$ \left.\frac{d I}{d \alpha}\right|_{\alpha=0}= \delta I=\int_{a}^{b}\left(\frac{\partial F}{\partial y} \eta+\frac{\partial F}{\partial y^{\prime}} \eta^{\prime}\right) d x=0. $$ where $\delta I$ denotes the first-order variation in the value of $I$ due to the variation (22.2) in the function $y(x) .$ Integrating the second term by parts this becomes $$ \left[\eta \frac{\partial F}{\partial y^{\prime}}\right]_{a}^{b}+\int_{a}^{b}\left[\frac{\partial F}{\partial y}-\frac{d}{d x}\left(\frac{\partial F}{\partial y^{\prime}}\right)\right] \eta(x) d x=0. $$ Now, if we demand that the lower end-point $a$ is fixed, while we allow variation of the end-point $b$ along the curve $h(x, y)=0$, (Why/how is this?) we obtain through a similar analysis as above that the variation in the value of $I$ due to the arbitrary variation is given to first order by $$ \delta I=\left[\frac{\partial F}{\partial y^{\prime}} \eta\right]_{a}^{b}+\int_{a}^{b}\left(\frac{\partial F}{\partial y}-\frac{d}{d x} \frac{\partial F}{\partial y^{\prime}}\right) \eta d x+F(b) \Delta x $$ where $\Delta x$ is the displacement in the $x$ -direction of the upper end-point, and $F(b)$ is the value of $F$ at $x=b .$ We of course require the displacement $\Delta x$ to be small. We can show that $\Delta y=\eta(b)+y^{\prime}(b) \Delta x .$ Since the upper end-point must lie on $h(x, y)=0$ (Why/how is this?) we also require that, at $x=b$, $$ \frac{\partial h}{\partial x} \Delta x+\frac{\partial h}{\partial y} \Delta y=0. $$