Antiderivative.

If I want to show that $F(x)$ is antiderivative of $f(x),$ then usually I find $F'(x).$ And if $F'(x)=f(x)$, then $F$ is antiderivative, otherwise not.

My question is if I want to show that $F$ is antiderivative, can I also say that $$\int f(x)\, dx$$, and if $$\int f(x)\, dx = F(x),$$ then the answer is yes, otherwise not. Is not there a problem with a constant in this method. So for me it seems like that I cannot use it. Am I right?

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Well, you already have the solution. What we mean by $\int f(x)\ dx$ is really a family of functions that differ only by a constant. So what you do is you take this family of antiderivatives of $f$ and pick one. Say we call it $G$. So then if $G - F$ is a constant (technically locally a constant), then $F$ is an antiderivative of $f$.
But of course, the easiest method is to simply differentiatie $F$.
...if the difference $G-F$ is locally constant, then... – Emanuele Paolini Feb 27 '13 at 19:59
If $f(x)=x^2$ and $F(x)=1/3*x^3$. I want to show that F is antiderivative of f. So to do the other way I find that $\int f(x)\, dx = 1/3x^3+k$. So then what is my conclusion? Can I conclude that F is one of the antiderivatives of f? – Reader Feb 27 '13 at 20:03
@Reader: You know that for every $k \in \mathbb{R}$, $x^3/3 + k$ is an antiderivative of $x^2$. You pick one $k$, say $k=2$ (this would work for any $k$). Take $G(x) = x^3/3+2$. $G(x) - F(x) = 2$, which is a constant, so you can conclude that $F$ is an antiderivative of $f$. But again, this is a roundabout way to do something pretty simple. – Javier Feb 27 '13 at 20:07