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Version 14 |
| Let $I$ be an open interval of $\mathbb{R}$ and\, $f:I \longrightarrow \mathbb{R}$\, a real function. |
Let $I$ be an open interval of $\mathbb{R}$ and\, $f:I \longrightarrow \mathbb{R}$\, a real function. |
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A function \,$F:I \longrightarrow \mathbb{R}$\, is called an \emph{antiderivative} of $f$ if $F$ is differentiable and its derivative is equal to $f$, i.e.
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A function \,$F:I \longrightarrow \mathbb{R}$\, is called an {\bf antiderivative} of $f$ if $F$ is differentiable and its derivative is equal to $f$, i.e.
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| \begin{displaymath} |
\begin{displaymath} |
| F'(x) = f(x) \quad \text{for all}\; x \in I. |
F'(x) = f(x) \quad \text{for all}\; x \in I. |
| \end{displaymath} |
\end{displaymath} |
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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.
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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.
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To account for this, we express the \emph{general antiderivative}, or \emph{indefinite integral}, as follows:
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To account for this, we express the {\bf general antiderivative}, or {\bf indefinite integral}, as follows:
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| $$ \int f(x)\ dx = F+C$$ |
$$ \int f(x)\ dx = F+C$$ |
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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 $x$.
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where $C$ is an arbitrary constant called the {\bf constant of integration}. The $dx$ portion means ``with respect to $x$'', because after all, our functions $F$ and $f$ are functions of $x$.
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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:
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There is no loss in generality with this notation since, in fact, \emph{all} antiderivatives of $f$ take this form:
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| {\bf Theorem.} Let $F, G$ be two antiderivatives of a given function $f$ defined on an open interval $I$. Then $F-G = \textrm{const}$. |
{\bf Theorem.} Let $F, G$ be two antiderivatives of a given function $f$ defined on an open interval $I$. Then $F-G = \textrm{const}$. |
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\emph{Proof.} Since $F'(x) = f(x)$ and $G'(x) = f(x)$, we have $F'(x)-G'(x) = 0$. Thus, $F(x)-G(x) = \textrm{const}$. $\square$
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\emph{Proof.} Since $F'(x) = f(x)$ and $G'(x) = f(x)$, then $F'(x)-G'(x) = 0$ and so $F(x)-G(x) = \textrm{const}$. $\square$
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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:
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This is no longer true if the domain of the function $f$ is not an open interval (is not connected). For that, the following more general result holds:
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| {\bf 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 $U$). |
{\bf 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 $U$). |
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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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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{displaymath} |
\begin{displaymath} |
| \begin{cases} |
\begin{cases} |
| \log(-x) + C_1, & $if$\;\; x < 0\\ |
\log(-x) + C_1, & $if$\;\; x < 0\\ |
| \log(x) + C_2, & $if$\;\; x > 0 |
\log(x) + C_2, & $if$\;\; x > 0 |
| \end{cases} |
\end{cases} |
| \end{displaymath} |
\end{displaymath} |
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| \subsection{Remarks} |
\subsection{Remarks} |
| \begin{itemize} |
\begin{itemize} |
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\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'').
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\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").
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| \end{itemize} |
\end{itemize} |
