Let $I$ be an open interval of $\mathbb{R}$ and $f:I \longrightarrow \mathbb{R}$ a real function.

A function $F:I \longrightarrow \mathbb{R}$ is called an antiderivative or a primitive of $f$ if $F$ is differentiable and its derivative is equal to $f$ , i.e.

   for all

Note that there are an infinite number of antiderivatives for any function $f$ since any constant can be added or subtracted from any valid antiderivative to yield another equally valid antiderivative.

To account for this, we express the general antiderivative, or indefinite integral, as follows: $$ \int f(x)\ dx = F+C$$ where $C$ is an arbitrary constant called the constant of integration. The $dx$ portion means ``with respect to $x$ '', because after all, our functions $F$ and $f$ are functions of $x$ .

There is no loss in generality with this notation since in fact all antiderivatives of $f$ take this form as the following theorem demonstrates:

Theorem. Let $F, G$ be two antiderivatives of a given function $f$ defined on an open interval $I$ . Then $F-G = {const}$ .

Proof. Since $F'(x) = f(x)$ and $G'(x) = f(x)$ , we have $F'(x)-G'(x) = 0$ on the whole $I$ . Thus, by the fundamental theorem of integral calculus, $F(x)-G(x) = {const}$ . $\square$

This is no longer true if the domain of the function $f$ is not an open interval (is not connected). For that scenario, the following more general result holds:

Theorem. Let $U \subset \mathbb{R}$ be an open set (not necessarily an interval). Suppose $F, G$ are antiderivatives of a given function $f:U \longrightarrow \mathbb{R}$ . Then $F-G$ is constant in each connected component of $U$ (each interval in $U$ ).

For example, consider the function $f:\mathbb{R}\!\smallsetminus\!\{0\} \longrightarrow \mathbb{R}$ given by $f(x) = \frac{1}{x}$ . Notice that the domain of $f$ is not an interval, but the union of the disjoint intervals $(-\infty,\, 0)$ and $(0,\,+\infty)$ . Then, all the antiderivatives of $f$ take the form

$if$ \;\; x < 0\\ \log(x) + C_2, & $if$ \;\; x > 0 \end{cases}\end{displaymath}">

Remarks

• For complex functions, the definition of antiderivative is exactly the same and the above results also hold (one just needs to consider ``connected open subsets'' instead of ``open intervals'').