(a) Find the extremal x = x^*(t) for the following:
\int^{1}_{0} (4\dot{x}^2 + x^2) dt, x(0) = 0, x(1) = 1
i.e. satisfying
(1) \frac{d}{dt}(4\dot{x}^*(t)) = x^*(t)

(b) For the extremal x^*(t), show that
(2) \int^{1}_{0}[4[\dot{x}^*(t)]^2 + [x^*(t)]^2] dt = \int^{1}_{0}[4\dot{x}^*(t)\dot{x}(t) + x^*(t)x(t)]dt
for all x = x(t) \in C^2 with x(0) = 0, x(1) = 1.

(Hint: Multiplying equation (1) with x^*(t) - x(t), integrating in t yields

(3) \int^{1}_{0}\frac{d}{dt}[4\dot{x}^*(t)][x^*(t) - x(t)] = \int^1_0 x^*(t)[x^*(t) - x(t)] dt
and then integrating by parts in (3) with x^*(0) - x(0) = 0 and x^*(1) - x(1) = 0.

(c) prove that x = x^*(t) is a minimizing curve.
(Hint: use inequality ab \leq \frac{1}{2}(a^2 + b^2) for any two real number a, b \in \Re. Then you use the identity (2).)