This is a problem from a PDE class.
(i) Show that $$v(x, t)=\frac{1}{2c}\int_{0}^{t}\left(\int_{x-c(t-\tau)}^{x+c(t-\tau)}f(\xi, \tau) d\xi\right)d\tau$$ is a solution of the Cauchy problem: $$v_{tt}-c^2v_{xx}=f(x, t), x\in (-\infty, \infty), t>0, v(x, 0)=0, v_t(x, 0)=0$$
The question has nothing to do with the PDE; I'm having trouble differentiating each side with respect to $t$ to verify that $v(x, t)$ solves the PDE.
I know that we have $$\frac{d}{dt}\int_{a(t)}^{b(t)}G(\xi, t)d\xi=G(b(t), t)b'(t)-G(a(t), t)a'(t)+\int_{a(t)}^{b(t)}\frac{\partial}{\partial t}G(\xi, t)d\xi$$ but I'm not sure how to apply this rule to above. Let $$G(\tau, t)=\int_{x-c(t-\tau)}^{x+c(t-\tau)}f(\xi, \tau) d\xi$$, i.e the integral inside the brackets. Then, we have $v(x, t)=\frac{1}{2c}\int_0^t G(\tau, t)d\tau$, and $$v_t(x, t)=\frac{1}{2c}\left(G(t, t)+\int_0^t G_t(\tau, t)d\tau\right)=\frac{1}{2c}\int_0^t G_t(\tau, t)d\tau \\ =\frac{1}{2c}\int_0^t\left(f(x+c(t-\tau), \tau)\cdot c+f(x-c(t-\tau), \tau)\cdot c+\int_{x-c(t-\tau)}^{x+c(t-\tau)}\frac{\partial}{\partial t}f(\tau, t)dt\right)\\ =\frac{1}{2}\int_0^t \left(f(x+c(t-\tau), \tau)+f(x-c(t-\tau), \tau\right)+\color{red}{\frac{1}{2c}\int_{x-c(t-\tau)}^{x+c(t-\tau)}\frac{\partial}{\partial t}f(\tau, t)dt}$$ The solution manual says that the red term shouldn't appear, but I can't see why. Please enlighten me.
when you say : $$ G(\tau,t) = \int_{x-c(t-\tau)}^{x+c(t-\tau)}f(\xi,\tau)d\xi $$ function $f(\xi,\tau)$ isn't function of $t$! only functionality is in limit of definite integral. so there it's like : $$ \frac{d}{dt} \int_{a(t)}^{b(t)} G(x)dx = G(b(t))b'(t)-G(a(t))a'(t) $$ because $G(x,t)$ in your formula doesn't explicitly depend on $t$. in this problem $\frac{\partial}{\partial t} f(\xi,\tau) =0$.
maybe its the second line that mislead you to this problem when you suddenly put $t$ instead of $\tau$ in second argument of $f$.