Example 2.1.
Since \(x^2\) is an anti-derivative of \(2x\) (whose domain is an interval), it follows from the discussion above that
\begin{equation*}
\int 2x\ dx = \{x^2 + k \colon k \in \mathbb{R} \} = x^2 + C.
\end{equation*}
Section 2.1 Anti-derivatives
So far we have been concerning ourselves with the problem of find the derivative of a given a function \(f(x)\text{.}\) In this section, we study the ``inverse’’ problem, namely find a function \(F(x)\) whose derivative is a given function \(f(x)\text{.}\)
We say that a function \(F(x)\) an anti-derivative of \(f(x)\) if \(F'(x) = f(x)\text{.}\) First note that, a function can have at most one derivative but if it has an antiderivative at all, it will have a class of anti-derivatives. For example, \(x^2\) and \(x^2 +1\) are two anti-derivatives of \(2x\text{.}\) We use the notation
\begin{equation*}
\int f(x)\ dx
\end{equation*}
to denote the class of all anti-derivatives of \(f(x)\). This class of functions is also called the indefinite integral of \(f(x)\) with respect to \(x\text{.}\) Second, if \(F_1(x)\) and \(F_2(x)\) are anti-derivatives of a function \(f(x)\) then
\begin{equation*}
(F_1(x) - F_2(x))' = F_1'(x) - F_2'(x) = f(x) - f(x) = 0.
\end{equation*}
According to the mean value theorem, the functions with 0 derivatives are precisely the locally constant functions. In particular, if the domain of \(f\) is simply an interval then these are simply the constant functions on that interval. The class of locally constant functions (with the domain understood) is customarily denoted by \(C\) and is known as the integration constant.
Example 2.2.
Let us find the indefinite integral of the function \(f(x)=1/x\text{.}\) We know that the derivative of \(\ln x\) is \(1/x\text{,}\) so \(\int 1/x\
dx\) should be \(\ln x +C\text{.}\) However, care must be paid to the domains of these functions. When we say that \((\ln x)' = 1/x\text{,}\) we are making a statement about the function \(\ln x\) whose domain is \(x >0\text{.}\) However, when we talk about the indefinite integral of \(1/x\text{,}\) we are making a statement about \(1/x\) whose domain is \(x \neq 0\text{.}\) And \(\ln x\) is only ``half’’ of an anti-derivative of \(1/x\) (graph them and you will see). What is the other half then? By symmetry, it is easy to see that the derivative of \(\ln (-x)\) is \(1/x\) on \(x \lt 0\text{.}\) Indeed one checks that this is the case by differentiation. Thus, \(\ln |x|\) is an anti-derivative of \(1/x\text{.}\) Hence,
\begin{equation*}
\int \frac{1}{x}\ dx = \ln |x| +C. \quad (x \neq 0)
\end{equation*}
Note also that \(x \neq 0\) is the union of two disjoint open interval. So, \(C\) the class of locally constant functions on \(x
\neq 0\) is the class of functions of the form
\begin{equation*}
\begin{cases}
c_1 & x \gt 0 \\ c_2 & x \lt 0
\end{cases}
\end{equation*}
where \(c_1, c_2\) are constants but not necessarily the same. For example, the function
\begin{equation*}
F(x) =
\begin{cases}
\ln(x) + 2 & x \gt 0 \\ \ln(-x) -3 & x \lt 0
\end{cases}
\end{equation*}
is also an anti-derivative of \(1/x\text{.}\)
