To find the minimum of $\int_0^1 (f''(x))^2dx$ I was trying to solve a question of an entrance exam. I am completely stuck in the problem. I am not able to find idea how to proceed. Please help me.

Let $A$ be the set of twice continuously differentiable functions on the interval $[0, 1]$ and let $B = \{f \in A : f(0) = f(1) = 0, f'(0) = 2\}$. Find the value of $$\text{min}_{f\in B} \displaystyle \int_0^1 (f''(x))^2dx.$$ 

I am really sorry for not showing some effort from my side but I can not find ant way to proceed. 
Please help me. Thnx in advance.
 A: Let $J = \int_0^1 L(x,f,f',f'') \ dx$ where $L(x,f,f',f'') = (f'')^2$.
Then by the Euler-Lagrange equation, $J$ has local extrema when
$$\cfrac{\partial L}{\partial f} - \cfrac{d \ }{dx}\left(\cfrac{\partial L}{\partial f'}\right) + \cfrac{d^2\ }{dx^2}\left(\cfrac{\partial L}{\partial f''}\right) = 0$$
Thus in this case, $$ 0 = \cfrac{d^2\ }{dx^2}(2f'') = 2f''''$$
and hence $f(x) = c_0 + c_1x + c_2x^2 + c_3x^3$.
Applying the conditions that define the set $B$:
$$f(0) = 0 \Rightarrow c_0 = 0. \quad\quad\quad f(1) = 0 \Rightarrow c_0 + c_1 + c_2 + c_3 = 0. \quad\quad\quad f'(0) = 2 \Rightarrow c_1 = 2$$
and we have 
$$f(x) = 2x + (\alpha - 2) x^2 - \alpha x^3 \ , \quad \text{ for some value of the parameter } \alpha$$
Now evaluate $J$ and minimize it as a function of $\alpha$.
$$J(\alpha) = \int_0^1 (2\alpha - 4 - 6\alpha x)^2 \ dx = 4(\alpha^2 + 2\alpha + 4) = 4\left((\alpha+1)^2 + 3\right)$$
and $J(\alpha=-1) = 12$ is the minimum.
Therefore
$$\min_{f \in B} \int_0^1 (f'')^2 \ dx = 12$$
with the minimum attained for $f(x) = x^3 - 3x^2 + 2x$.
