initial value problem with step function conditions 
Let $u(x,t)$ be the solution of the one-dimensional wave equation
  $$\frac{\partial^2u}{\partial t^2}-4\frac{\partial^2u}{\partial
 x^2}=0$$ where $-\infty<x<\infty$ and $t>0$ subject to the condition
  $$u(x,0)=\begin{cases}16-x^2 \ \ \text{if} \ \ |x|\leq 4 \\ 0 \ \ \ \
 \ \ \ \ \ \ \ \ \  \text{otherwise}\end{cases} \ \ \ \ \ \ , \ \ \ \
 \frac{\partial u}{\partial t}\bigg|_{(x,0)}=\begin{cases}1 \ \
 \text{if} \ \ |x|\leq 2 \\ 0 \ \  \text{otherwise}\end{cases}$$ For
  $1<t<3$, $u(2,t)$ is equal to
$(a)$
$\displaystyle\frac{1}{2}\bigg[16-(2-2t)^2\bigg]+\frac{1}{2}\bigg[1-\min
 \bigg\{1,t-1\bigg\}\bigg]$.
$(b)$ $\displaystyle\frac{1}{2}\bigg[32-(2-2t)^2-(2+2t)^2\bigg]+t$.
$(c)$ $\displaystyle\frac{1}{2}\bigg[32-(2-2t)^2-(2+2t)^2\bigg]+1$.
$(d)$ $\displaystyle\frac{1}{2}\bigg[16-(2-2t)^2\bigg]+\frac{1}{2}\bigg[1-\max
 \bigg\{1-t,-1\bigg\}\bigg]$.

If we have that $1<t<3$, so we have $4<2+2t<8$ and $-4<2-2t<0$. In D'Alembert's formula, if we denote $u(x,0)=f(x)$ and $u_t(x,0)=g(x)$, then $$u(x,t)=\frac{f(x+2t)+f(x-2t)}{2}+\frac{1}{4}\int_{x-2t}^{x+2t} g(z)dz \\ \implies u(2,t)=\frac{f(2+2t)+f(2-2t)}{2}+\frac{1}{4}\int_{2-2t}^{2+2t} g(z)dz$$ Now $f(2+2t)=0$ as the range of $2+2t$ lies outside $|x|\leq 4$ and $1<t<3$, as shown above, so only $f(2-2t)$ will contribute to the solution. Again, within $|x|\leq 2$, we have $g(x)=1$. So our solution should be $$u(2,t)=\displaystyle\frac{1}{2}\bigg[16-(2-2t)^2\bigg]+\frac{1}{4}\int_{2-2t}^{2+2t}dz \\ \implies \boxed{{u(2,t)=\displaystyle\frac{1}{2}\bigg[16-(2-2t)^2\bigg]+t}}$$ Now answer is given $(b)$, which is close to the solution, but I don't understand why $f(2+2t)=16-(2+2t)^2$ will contribute to the solution as this lies completely outside $|x|\leq 4$? Any help is appreciated.

Observe that in the second part of d'Alembert's formula, your interval of integration $[2-2t, 2+2t]$ covers the whole $[-2,2]$  when $t>2$ and you have to integrate between $-2$ and $2$. So for $t>2$ you get
$$
u(2,t)=\frac{1}{2}[16-(2-2t)^2]+\frac 1{4}\int_{-2}^2dz=\frac{1}{2}[16-(2-2t)^2]+1
$$
but when $t\in[1,2]$ the lower limit $2-2t>-2$ and you integrate between $2-2t$ and $2$ (your upper limit is always greater than 2). So for $t\in[1,2]$
$$
u(2,t)=\frac{1}{2}[16-(2-2t)^2]+\frac 1{4}\int_{2-2t}^2dz=\frac{1}{2}[16-(2-2t)^2]+t/2
$$
which is answer $(d)$.
