Solution of Wave Equation using Reflection Principle A sample problem for an exam is as follows:

Consider the wave equation $U_{tt} = 4U_{xx}, 0 < x < 1$ with $U(0,t)= U(l,t)= 0$ and $U(x,0)= x(1-x)$, $U_t(x,0)= \pi$.
Find $U(1/4,1/4)$ and $U(1/2,1/2)$ using the reflection principle.

We went over the solution in lecture but I lost my notes and I'm unsure how to proceed.  If someone could provide a solution with some explanation I would greatly appreciate it.
Note: $U_x$ and $U_t$ denote the partial derivative with respect to $x$ or $t$ respectively.
 A: The reflection method consists in extending the problem to $\mathbb R\times [0,\infty)$ in such a way that the given boundary condition $u(0,t)=u(1,t)=0$ becomes a consequence of some symmetry of initial conditions $u(\cdot, 0)=g$ and $u_t(\cdot, 0)=h$. In order to make sure that there is no motion at $x=0$, we need $g(-x)=-g(x)$ and $h(-x)=-h(x)$. Similarly, to fix the point at $x=1$ we need $g(2-x)=-g(x)$ and $h(2-x)=-h(x)$. Thus, both $g$ and $h$ are $2$-periodic functions. Their period from $-1$ to $1$ is shown below: red is the initial position $g$, blue is the velocity $h$. 

Now it remains to apply d'Alembert's formula with propagation speed $c=2$. 
$$u(x,t) = \frac12 (g(x-2t) + g(x+2t)) + \frac{1}{4} \int_{x-2t}^{x+2t} h(s)\,ds$$
The piecewise nature of $g$ and $h$ would make it awkward to write out these integrals in general, but it's not hard to evaluate them at particular points $(x,t)$, which is what you asked to do. 
A: Here is a reference where the method is explained in detailes. Starting on page $19$.
