What is the difference between $\int g(x) dx$ and $\int_0^x g(\xi) d\xi$ when solving DEs? Please point me to the question if this has already been asked.
What is the difference between
$$\int g(x) dx\quad \text{and}\quad \int_0^x g(\xi) d\xi$$
When do you use one in favour of the other?
Background:
I am solving a differential equation
$\frac{du}{dx}=g(x)$. I thought to use the indefinite integral and introduce a constant of integration. My professor wrote the definite integral as above. They give different answers, but of course you can just absorb the parts that the $0$ terminal introduces into the integration constant of the indefinite integral when equating the two.
I don't understand why you would use the definite integral.
 A: A differential equation can be solved by both definite and indefinite integrals. Suppose the differential equation is $\frac{du}{dx}=g(x)$, as you say and you have been provided with the condition $u(0)=k$ where $k$ is a real number.


*

*Definite Integral


$$\frac{du}{dx}=g(x)$$
or $$du = g(x)dx$$
or $$\int_{u(0)}^u du =\int_0^x g(x)dx$$
or $$u-u(0) =\int_0^x g(x)dx$$
or $$u =\int_0^x g(x)dx + k$$


*Indefinite Integral


$$\frac{du}{dx}=g(x)$$
or $$du = g(x)dx$$
or $$\int du =\int g(x)dx$$
or $$u =\int g(x)dx + c$$ where $c$ is an integration constant
Now we know that $u(0)=k$ i.e. at $x=0$ , $u=k$.
So we have $$k =\int_0 g(x)dx + c$$
or $$c = k - \int_0 g(x)dx$$
Hence the solution is $$u =\int g(x)dx + k - \int_0 g(x)dx$$
or $$u =\int_0^x g(x)dx + k$$
Thus we can see the difference between the solutions: Nothing.
A: The indefinite integral is going to tell you about the anti-derivative and have a constant. You can solve for this constant using the initial values (if given). The definite integral is going to incorporate the initial value automatically through the limits of integration. Either should work, just do what your professor wants.
