Minimum value of the integral I am struggling to find the solution to this problem. If anyone could help to explain how to solve this problem to me, it would be really appreciated.

What is the minimum value of the integral? y(x)=?

$\displaystyle\int_0^4 \left[y^2+\left(\frac{dy}{dx}\right)^2\right]\,dx$

$y(0)=0$ and $y(4)=1$
Euler-Lagrange equations: If

$$I = \int_0^4 dx \ F(y,y')$$

then $I$ is stationary with respect to $y$ and $y'$ when

$$\frac{d}{dt} \frac{\partial F}{\partial y'} = \frac{\partial F}{\partial y}$$

In this case, $F(y,y') = y^2+y'^2$.  Then the E-L equation implies that $y''=y$, so that

$$y(x) = A e^x + B e^{-x} $$

Given the boundary conditions, the solution is 

$$y(x) = \frac{\sinh{x}}{\sinh{4}} $$

For this solution in $y$ the value of $I$ is $\coth{4}$.  Comparing this with another function that satisfies the BC's, say, $y=x/4$, at which $I=19/12 &gt; \coth{4}$, we may conclude that the solution is a minimum.
Euler-Lagrange equations: If

$$I = \int_0^4 dx \ F(y,y')$$

then $I$ is stationary with respect to $y$ and $y'$ when

$$\frac{d}{dt} \frac{\partial F}{\partial y'} = \frac{\partial F}{\partial y}$$

In this case, $F(y,y') = y^2+y'^2$.  Then the E-L equation implies that $y''=y$, so that

$$y(x) = A e^x + B e^{-x} $$

Given the boundary conditions, the solution is 

$$y(x) = \frac{\sinh{x}}{\sinh{4}} $$

For this solution in $y$ the value of $I$ is $\coth{4}$.  Comparing this with another function that satisfies the BC's, say, $y=x/4$, at which $I=19/12 &gt; \coth{4}$, we may conclude that the solution is a minimum.