Constant $C$ when integrating a differential equation. Consider the differential equation $y'(x)=f(x)$
Let's suppose we integrate to find $y(x)$ an thus we have
$$y(x)=\int f(x)\ dx + c$$
Would it be the same as writing $$y(x)=\int_0^x f(s)\ ds$$
I'm thinking this because the first expression (let $F(x)$ be the antiderivative of $f(x)$) could be rewritten as
$$y(x)=F(x)+C$$
and the second as
$$y(x)=F(x)-F(0)$$
and indeed subtracting the 2 equations we could take that $C=F(0)$
 A: The solution can be written as $y(x)=F(x)+C$, where $F(x)$ is any anti-derivative, not necessarily $\int_0^x f(t)\,dt$. Then you can say $y(0)=F(0)+C$ so $C=y(0)-F(0)$. It is not necessarily true that $F(0)=0$ unless you define the anti-derivative as $F(x)=\int_0^x f(t)\,dt$. Hence, one way to write the solution is 
$$
y(x)=\int_0^x f(t)\,dt + y(0).
$$
Equivalently, from $y'(x)=f(x)$, you can say $\int_0^x y'(t)\,dt=\int_0^x f(t)\,dt$ so $y(x)-y(0)=\int_0^x f(t)\,dt$ and $$y(x)=\int_0^x f(t)\,dt+y(0),$$
which is the same thing.
A: In this example certain boundary conditions are required to "solve" the equation. It is easily determined that the solution to $y'(x) = f(x)$ is
$$y(x) = \int^{x} f(u) \, du + c_{0}.$$
Without some range for $x$ it is generally taken to be 
$$y(x) = \int_{- \infty}^{x} f(u) \, du + c_{0}.$$
Given at least one condition, say initial condition $y(x_{0}) = y_{0}$, then
$$y(x_{0}) = y_{0} = \int_{- \infty}^{x_{0}} f(u) \, du + c_{0}$$
and yields
$$c_{0} = y_{0} - \int_{- \infty}^{x_{0}} f(u) \, du.$$
Now,
\begin{align}
y(x) &= \int_{- \infty}^{x} f(u) \, du + c_{0} \\
&= y_{0} + \int_{- \infty}^{x} f(u) \, du - \int_{- \infty}^{x_{0}} f(u) \, du \\
y(x) &= y_{0} + \int_{x_{0}}^{x} f(u) \, du.
\end{align}
This is the general solution involving an initial condition. 
If $x_{0} = y_{0} = 0$ then
$$y(x) = \int_{0}^{x} f(u) \, du.$$
