# What is the difference between $\int g(x) dx$ and $\int_0^x g(\xi) d\xi$ when solving DEs?

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.

• Definite integral won't have a constant. – ultrainstinct Dec 15 '15 at 4:51
• Can you please explain the downvotes? – Phill Dec 15 '15 at 4:52
• Maybe you were given an initial condition that $u(0)=0$ along with the equation $u'(x)=g(x).$ If so the definite integral gives the answer without further need to determine the value of the added constant. – coffeemath Dec 15 '15 at 4:56
• @Phill I do not understand the downvotes. This is a good question. The answer is that there isn't a major difference. The definite integral might be used more in practice since you don't have to worry about solving for your constant of integration and instead the lower limit in the definite integral will take care of it. – Cameron Williams Dec 15 '15 at 4:59
• So what if the limit is at, say, $x=L$? Do you take the lower limit from that? When would you reverse the terminals so it was from $x$ to $0$? – Phill Dec 15 '15 at 5:06

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.

1. 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 $$u =\int_0^x g(x)dx + k$$

1. 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.

• @Phill You're welcome. – SchrodingersCat Dec 15 '15 at 5:14

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.