antiderivativeAntiderivative2002-02-01 23:52:432009-03-30 15:54:41Definitionfundamental theorem of integral calculusconstant of integration% this is the default PlanetMath preamble. as your knowledge
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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 \emph{antiderivative} or a \emph{primitive} of $f$ if $F$ is differentiable and its derivative is equal to $f$, i.e.
\begin{displaymath}
F'(x) = f(x) \quad \text{for all}\; x \in I.
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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 \emph{general antiderivative}, or \emph{indefinite integral}, as follows:
$$ \int f(x)\ dx = F+C$$
where $C$ is an arbitrary constant called the \emph{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 \emph{all} antiderivatives of $f$ take this form as the following theorem demonstrates:\\
{\bf Theorem.} \emph{Let $F, G$ be two antiderivatives of a given function $f$ defined on an open interval $I$. Then $F-G = \textrm{const}$.}
\emph{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,\, $F(x)-G(x) = \textrm{const}$.\, $\square$\\
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:\\
{\bf Theorem.} \emph{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
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\begin{cases}
\log(-x) + C_1, & $if$\;\; x < 0\\
\log(x) + C_2, & $if$\;\; x > 0
\end{cases}
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\subsection{Remarks}
\begin{itemize}
\item 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'').
\end{itemize}