Consider wave equation:
$u_{tt} = c^2u_{xx}$
with initial conditions:
$u(x,0) = f(x)$ , $u_t(x,0) = g(x)$
Using change of variables $\mu = x + ct$ and $\eta = x-ct$, we get $u_{\mu\eta} = 0$ and then $u(\mu , \eta) = F(\mu) + G(\eta)$.
$u(x,t) = F(x+ct) + G(x-ct) \implies u_t(x,t) = cF'(x+ct) - cG'(x-ct)$
where primes denote derivative with respect to $x+ct$ and $x-ct$, respectively.
Choosing $t=0$, we get:
$u_t(x,0) = cF'(x) - cG'(x)$
Here is the problem. If we integrate with respect to $x$ then $\int_{x_0}^{x}F'(\lambda)d\lambda = F(x) - F(x_0)$? We know that $F'$ means derivative with respect to $x+ct$ not $x$. So why that is true? I spent two hours and only got confused more and more.
I think I now understand the confusion. So you know that $$u(t,x) = F(t+cx) + G(t-cx)$$ and \begin{align} \partial_t u(t,x) &= cF'(x + ct) - cG'(x - ct) \end{align} We then integrate with respect $x$ and evaluate at $t=0$, hence \begin{align} \int_{x_0}^x \partial_tu(t,y)dy|_{t=0} = \int_{x_0}^x \partial_tu(0,y)dy &= \int_{x_0}^x [cF'(y + ct) - cG'(y - ct)]dy|_{t=0} \\ &= \int_{x_0}^x cF'(y) - cG'(y)dy \\ &= c[F(x) - F(x_0)] - c[G(x) - G(x_0)] \end{align} Since you are integrating in $x$ it does not matter when you evaluate $t=0$. If you evaluate it after the integration then \begin{align} \int_{x_0}^x \partial_tu(t,y)dy|_{t=0} &= \int_{x_0}^x [cF'(y + ct) - cG'(y - ct)]dy|_{t=0} \\ &=\bigg(c\int_{x_0}^x F'(y + ct)dy - c\int_{x_0}^xG'(y - ct)dy\bigg)|_{t=0} \\ &=\bigg(c\int_{x_0+ct}^{x+ct} F'(y)dy - c\int_{x_0-ct}^{x-ct}G'(y)dy\bigg)|_{t=0} \\ &=\bigg(c[F(x+ct) - F(x_0 + ct)] - c[G(x-ct) - G(x_0 - ct)]\bigg)|_{t=0} \\ &=c[F(x) - F(x_0)] - c[G(x) - G(x_0)] \end{align}
Which is the same thing. Let me know if this was the issue.