Suppose we are given a continuous, real function $f(x)$ defined on an interval $\Delta \subseteq \mathbb{R}$ and let $a \in \Delta$ be a fixed point.

Any other function $F(x)$, with domain $D_{F}= \Delta$ will be called

$$

F'(x)=f(x)

$$

(notice that $F$ is by definition differentiable (and thus continuous) in $\Delta$.)

The above definition implies that:

\begin{equation} \notag

G'(x)=F'(x)=f(x)

\end{equation}

will also be an antiderivative function. In such a case $F(x)$ and $G(x)$ will differ by a constant:

\begin{equation} \notag

G(x)=F(x)+c

\end{equation}

for some $c \in \mathbb{R}$. (this comes from a well known theorem of elementary calculus). We can thus now lay the following

\begin{equation} \notag

\begin{array}{r}

\int f(x)dx = \{F | F'(x)=f(x), \ x \in \Delta \} = \\

\\

= \{G(x)+c |\textrm{for all } c \in \mathbb{R} \} \ \ \ \ \ \ \ \

\end{array}

\end{equation}

where in the last equality $G$ is an antiderivative function (actually

Now it can be proved (the proof can be found on standard calculus texts and i will -hopefully- post it here later) that:

\begin{equation} \notag

\int_{a}^{x} f(t)dt

\end{equation}

In other words: $(\int_{a}^{x} f(t)dt )'=f(x)$ for all $x \in \Delta$.

\begin{equation} \notag

\int f(x)dx = \int_{a}^{x} f(t)dt + c

\end{equation}

for all $c \in \mathbb{R}$.

If $f(x)=2x$ and $\Delta = \mathbb{R}$, then for any real $a$ we have

\begin{equation} \notag

\int_{a}^{x} 2tdt = [t^{2}]_{a}^{x}=x^{2}-a^{2}

\end{equation}

But, the family of functions $\{F(x)=x^{2}-a^{2}, \ x \in \mathbb{R} | \textrm{for all } a \in \mathbb{R} \}$ does not include for example the function $G(x)=x^{2}+1$, which is an obvious antiderivative function of $f(x)=2x$.

Any other function $F(x)$, with domain $D_{F}= \Delta$ will be called

**an antiderivative function**of $f$ if$$

F'(x)=f(x)

$$

(notice that $F$ is by definition differentiable (and thus continuous) in $\Delta$.)

The above definition implies that:

*the antiderivative function is not uniquely determined, but rather there is a family of functions satisfying the above relation*. Actually, any other function $G(x)$, $D_{G}= \Delta$ with the property\begin{equation} \notag

G'(x)=F'(x)=f(x)

\end{equation}

will also be an antiderivative function. In such a case $F(x)$ and $G(x)$ will differ by a constant:

\begin{equation} \notag

G(x)=F(x)+c

\end{equation}

for some $c \in \mathbb{R}$. (this comes from a well known theorem of elementary calculus). We can thus now lay the following

__Definition:__We will call**antiderivative**or**indefinite integral**of $f$, and we will denote it by $\int f(x)dx$ the set of all functions satisfying the above property, thus:\begin{equation} \notag

\begin{array}{r}

\int f(x)dx = \{F | F'(x)=f(x), \ x \in \Delta \} = \\

\\

= \{G(x)+c |\textrm{for all } c \in \mathbb{R} \} \ \ \ \ \ \ \ \

\end{array}

\end{equation}

where in the last equality $G$ is an antiderivative function (actually

__any__antiderivative function) of $f$.Now it can be proved (the proof can be found on standard calculus texts and i will -hopefully- post it here later) that:

**one of these functions belonging in the above set (thus, one of the antiderivative functions of $f$ or equivalently: one of the indefinite integrals of $f$) is the function**__Proposition:__\begin{equation} \notag

\int_{a}^{x} f(t)dt

\end{equation}

In other words: $(\int_{a}^{x} f(t)dt )'=f(x)$ for all $x \in \Delta$.

__Remarks:__**Notice that the above proposition readily implies that $\int_{a}^{x} f(t)dt$ is**__(1).____differentiable__(and thus__continuous__) for any $x \in \Delta$. Of course the $\ ' \ $ symbol indicates differentiation with respect to the variable $x$.**Thus: the definite integral $\int_{a}^{x} f(t)dt $ with variable upper limit of integration, is an antiderivative function of $f$. Consequently, we can write**__(2).__\begin{equation} \notag

\int f(x)dx = \int_{a}^{x} f(t)dt + c

\end{equation}

for all $c \in \mathbb{R}$.

__(3).__What the above proposition actually tells us is that:**any function $f$ which is continuous on an interval $\Delta \subseteq \mathbb{R}$, has an antiderivative function given by $\int_{a}^{x} f(t)dt $ for $a,x \in \Delta$**.__(4).__It is worth noticing the meaning of the number $a \in \Delta$: Varying the value of $a \in \Delta$ produces different antiderivative functions (because the variation of $a \in \Delta$ simply alters the value of the constant of integration $c$). However, we cannot hope that varying the value of $a \in \Delta$ "covers" all possible antiderivatives of a given (continuous function $f$). In other words, this means that:**although the above theorem tells us that**$\int_{a}^{x} f(t)dt$**is an antiderivative function of**$f$,**not all antiderivative functions of**$f$**can necessarily be expressed as**$\int_{a}^{x} f(t)dt$**for some**$a \in \Delta$. This can be clearly seen in the following example:If $f(x)=2x$ and $\Delta = \mathbb{R}$, then for any real $a$ we have

\begin{equation} \notag

\int_{a}^{x} 2tdt = [t^{2}]_{a}^{x}=x^{2}-a^{2}

\end{equation}

But, the family of functions $\{F(x)=x^{2}-a^{2}, \ x \in \mathbb{R} | \textrm{for all } a \in \mathbb{R} \}$ does not include for example the function $G(x)=x^{2}+1$, which is an obvious antiderivative function of $f(x)=2x$.

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