Explanation of this proof of the Euler-Lagrange (Please note that I am a physics student, so please try to avoid rigorous mathematics in the explanation.)
I have the following proof for the Euler-Lagrange:
Consider the integral $$I=\int F(x,y,y').$$
Consider now a line $Y$ such that $Y=y+\epsilon h$ for some small $\epsilon$. Then $$\frac{dI}{d\epsilon}=\int_{x_1}^{x_2}\Big(\frac{\partial F}{\partial Y}\frac{\partial Y}{\partial \epsilon}+\frac{\partial F}{\partial Y'}\frac{\partial Y'}{\partial \epsilon}\Big)dx=\int_{x_1}^{x_2}\Big(\frac{\partial F}{\partial Y}h+\frac{\partial F}{\partial Y'}h'\Big).$$ Integrating by parts we get $$\int_{x_1}^{x_2}\frac{\partial F}{\partial Y'}h'dx=\Big[\frac{\partial F}{\partial Y'}h\Big]_{x_1}^{x_2}-\int_{x_1}^{x_2}h\frac{d}{dx}\Big(\frac{\partial F}{\partial Y'}\Big)dx.$$ Then $$\frac{dI}{d\epsilon}=\int_{x_1}^{x_2}\Big[\frac{\partial F}{\partial Y}h-h\frac{d}{dx}\Big(\frac{\partial F}{\partial Y'}\Big)\Big]dx=\int_{x_1}^{x_2}h\Big[\frac{\partial F}{\partial Y}-\frac{d}{dx}\Big(\frac{\partial F}{\partial Y'}\Big)\Big]dx.$$ $$\implies \frac{\partial F}{\partial y}-\frac{d}{dx}\Big(\frac{\partial F}{\partial y'}\Big)=0.$$ I understand everything up to the beginning of integration by parts. On the third line of equations (the first line after "Integrating by parts we get...") we must have $$\Big[\frac{\partial F}{\partial Y'}h\Big]_{x_1}^{x_2}=0$$ in order to get to the next line with the expression for $dI/d\epsilon$, but how is this so? I also don't understand how the last line implies the Euler-Lagrange.
I guess a helpful answer would be explaining everything from "Integration by parts" in detail.
 A: As for the last line implying Euler-Lagrange, it follows since $h$ is an arbitrary perturbation. See also the lemma of duBois Reymond: https://en.wikipedia.org/wiki/Fundamental_lemma_of_calculus_of_variations 
Also, I should add, we do want $dI/d\epsilon=0$ for that to happen, but that follows since we want to minimize the functional, so the usual calculus applies…
…and yes, Simple Art is right, we impose boundary conditions on $h$ (i.e. $h(x_1)=h(x_2)=0)$ to force $[h\frac{\partial F}{\partial Y'}]_{x_1}^{x_2}=0$. We can do that and still have the above hold. 
For a full proof with a little more details, I refer to the Wikipedia page: https://en.wikipedia.org/wiki/Euler%E2%80%93Lagrange_equation 
Look for "Derivation of one-dimensional Euler–Lagrange equation".
A: The previous answer is correct. I going to simplify the answer.
One consideration for $h$ is that $h(x_1)=h(x_2)=0$, then
\begin{eqnarray}
\left( h(x) \frac{\partial F(x)}{\partial Y'} \right|_{x_1}^{x_2} &=& \left(h(x_2) \frac{\partial F(x_2)}{\partial Y'} \right)-\left(h(x_1) \frac{\partial F(x_1)}{\partial Y'} \right)\\
&=& \left(0 \cdot \frac{\partial F(x_2)}{\partial Y'} \right)-\left(0\cdot \frac{\partial F(x_1)}{\partial Y'} \right)\\
&=& 0-0=0
\end{eqnarray}
