I am trying to understand a part of the proof for the following proposition:
The initial value problem $\frac{dx}{dt}=f(t,x),x(\tau)=\xi$ is equivalent to the integral equation $x(t)=\xi+\int_{\tau}^{t}f(s,x(s))ds$.
$(=>)$ Let $\frac{dx}{dt}=f(t,x),x(\tau)=\xi$ be an initial value problem.
$\frac{dx}{dt}=f(t,x)$
$\int_{\tau}^{t}x'(s)ds = \int_{\tau}^{t}f(s,x(s))ds$
How did this $\int_{\tau}^{t}f(s,x(s))ds$ become? I am not sure how the $x(s)$ came about. If I integrate the vector field $f(t,x)$ with respect to $s$, where does $x(s)$ come from? I'm not very expert on Multivariable Calculus, sorry!
$x(t)-x(\tau) = \int_{\tau}^{t}f(s,x(s))ds$
$x(t)-\xi = \int_{\tau}^{t}f(s,x(s))ds$
$x(t) =\xi+ \int_{\tau}^{t}f(s,x(s))ds$
$(<=)$ Let $x(t) =\xi+ \int_{\tau}^{t}f(s,x(s))ds$.
$x(t) =\xi+ \int_{\tau}^{t}f(s,x(s))ds$
$x'(t)=(\xi)'+(\int_{\tau}^{t}f(s,x(s))ds)'$
$x'(t)=0+f(t,x)=f(t,x)$
Similarly, how come $(\int_{\tau}^{t}f(s,x(s))ds)'=f(t,x)$?