I'm having a tough time understanding the proof that Cauchy problem is equivalent to integral equation: $$x(t)=x_0+\int^t_{t_0}f(s,x(s))ds, t \in I = [t_0-\delta,t_0+\delta]. $$
1) If $x(t)$ is a solution to Cauchy problem, then $x'(t)=f(t,x(t))$. Why is it so? If a function would be written as this for example $s^2+s,$ then integrating it we would get a function with argument $t$, but why is it a function of $x$?
Then $$\int^t_{t_0}x'(s)ds=\int^t_{t_0}f(s,x(s))ds$$
$$x(t)-x(t_0)=\int^t_{t_0}f(s,x(s))ds$$
2) as I understand in this instance $x(t)$ simply denotes some function, that results after integrating $x(s)$? But if it is some other function, then why only argument changes and not the function itself, why isn't it $g(t)$ for example?
Then $$x(t)=x(t_0)+\int^t_{t_0}f(s,x(s))ds$$
And after differentiation we get $$x'(t)=f(t,x(t))$$
Here I assume $x(t_0)$ disappears because it is a constant?
Then: $$x(t_0)=x_0+\int^{t_0}_{t_0}f(s,x(s))ds=x_0$$
This is I assume, because the function is continuous? But I don't understand why it is needed to include $\int^{t_0}_{t_0}$? What is it's function in the proof?
Then to proof the theorem it must be proved that integral equation has only one solution in the Interval $I$.