Additional term in solution for wave equation Given the wave equation:
$u_{tt} = c^2u_{xx} \quad x \in \mathbb{R}, t> 0$
$ u(x,0) = \phi(x), x \in \mathbb{R}$
$ u_{t}(x,0) = \psi (x), x \in \mathbb{R}$
The solution is:
$u(x,t) = \frac{1}{2} ( \phi(x - ct) + \phi(x + ct)) + \frac{1}{2c} \underbrace{\int_{x-ct}^{x+ct} \psi(\zeta) d\zeta}_{=:D}$
The first two term are very intuitive. These are simple the waves propagating with a speed of $c$. But what is the term marked with $D$? I've never seen it when solving the wave equation in physics. It there any effect physical that corresponds to this term?
Is there an intuitive explanation for this term? 
Edit: to see what actually happens I plotted an example:
$g(x) = \frac{1}{\sigma \sqrt{2 \pi}} e^{\frac{-1}{2} (x/ \sigma)^2}$
$f(x,t) := \frac{1}{(t+1)} g(x)$
$\frac{d}{dt} f(x,t) = -\frac{1}{(t+1)^2} g(x)$

I've never seen anything like this. In real waves, even though $\psi $ is generally not 0
 A: You can understand the term containing $\psi$ as arising in a similar way to the terms containing $\phi$, through a superposition argument and  integration in time: We can write the full solution as a superposition
$$ u = v+w, $$
where $v$ and $w$ satisfy
$$ v_{tt}=c^2 v_{xx} \ \ (x\in\mathbb{R}, t>0), \quad v(x,0)=\phi(x), \quad v_t(x,0)=0,$$
$$ w_{tt}=c^2w_{xx} \ \ (x\in\mathbb{R}, t>0), \quad w(x,0)=0, \quad w_t(x,0)=\psi(x).$$
Naturally
$$v(x,t)=\frac12(\phi(x-ct)+\phi(x+ct)).$$
Notice now, that the time derivative $z=w_t$ also satisfies the same wave equation, with
$$ z_{tt}=c^2 z_{xx} \ \ (x\in\mathbb{R}, t>0), \quad z(x,0)=\psi(x), \quad z_t(x,0)=0 \ \ (=c^2 w_{xx}(x,0)).$$
Naturally then, the time derivative is the sum of right- and left-going waves with $$w_t(x,t)=\frac12(\psi(x-ct)+\psi(x+ct)).$$
The solution for $w$ at any given point $(x,t)$ is then due to the integrated effect of these derivative waves passing $x$ over the time interval $[0,t]$: Because $w(x,0)=0$, 
$$ w(x,t) = \int_0^t w_t(x,s)\,ds = \frac12\int_0^t \psi(x-cs)\,ds+\frac12\int_0^t\psi(x+cs)\,ds .$$
This gives exactly the term you marked $D$ after
using the substitutions $\zeta=x-cs$ and $\zeta=x+cs$ in the two integrals.
As for how this relates to the real world, the behavior of solutions of the wave equation (e.g., the Huyghens phenomenon) is pretty different in different space dimensions. I suspect it might be rare to encounter really 1D behavior in the 3D world. 
A: Because you said you wanted an intuition for what it is:
What you are seeing is the d'Alembert solution, and the extra term D comes from your second initial condition: $u_t(x,0)=ψ(x), x ∈ R$. For ease let's say that if $u(x,0)$ refers to the initial height of points on a string, which you prescribe as $ϕ(x)$, then $u_t(x,0)$ refers to the initial velocity of the points that string, which you prescribe as $ψ(x)$. 
The way that I like to understand it is that the integral D refers to the height response due to your initial velocity 'kick' $ψ(x)$. It affects between $x+ct$ and $x-ct$, which is as far as the information can propagate in both directions from $x$ through your medium at speed $c$ given time $t$. This is very similar intuitively to the way that $\frac{1}{2}(ϕ(x−ct)+ϕ(x+ct))$ describes two waves moving in opposite directions, each with same shape but half the amplitude of the initial condition $u(x)$. 
Sorry, but I'm not sure what you're doing with your functions f and g so can't really comment on that but I'll try. It looks like you're setting $u(x,0) = ϕ(x) = f(x,t)$ and $u_t(x,0) = ψ(x) = f_t(x,t)$ because of the differentiation wrt $t$. To be clear: $ϕ(x)$ and $ψ(x)$ do not need to be related in any way. $ϕ(x)$ is just the initial height of $u(x)$, and $ψ(x)$ is the initial velocity. Just like you can have a string at rest ($u_t(x,0)=0$) with an initial shape/height $ϕ(x)$, and "let it go" to achieve $\frac{1}{2}(ϕ(x−ct)+ϕ(x+ct))$, you can have a string starting with no deflections ($u(x,0)=0$) with an initial velocity $ψ(x)$ to achieve $D$. If you want a better feel for the effect of $D$, set $u(x,0)=0$ and play around with different known functions for $u_t(x)$, like $\sin(x)$, $\exp(x)$  and see what happens. 
Chris Tisdell has a few YouTube lectures on how you can get to this solution for yourself: https://www.youtube.com/watch?v=NM4GTf57w2c.
