Theoretical Remarks #4

   In Monday's 6/Jan/2014 post we mentioned a proof for the existence of an antiderivative for any function continuous on an interval.
   In today's post, i am going to supply an alternative proof for the same proposition. For the reader's convenience, i am repeating at this point the statement of the proposition:
 Proposition: Let a real function $f$, continuous on an interval $\Delta$ and let $a \in \Delta$ be a fixed point. Then the function $F(x)=\int_{a}^{x}f(t)dt$ is an antiderivative function of $f$ in $\Delta$. In other words:
$$
F'(x) = \big( \int_{a}^{x}f(t)dt \big)' = f(x)
$$
for all $x \in \Delta$.
Prooof: (alternative)
   It is sufficient to show that for any fixed point $x_{0} \in \Delta$ we have $F'(x_{0})=f(x_{0})$. Let $x_{0}, x_{0}+h \in \Delta$ with $h \neq 0$. Then we can compute
$$
\begin{array}{c}
F(x_{0}+h) - F(x_{0}) = \int_{a}^{x_{0}+h}f(t)dt - \int_{a}^{x_{0}}f(t)dt = \\
     \\
\bigg( \int_{a}^{x_{0}}f(t)dt + \int_{x_{0}}^{x_{0}+h}f(t)dt \bigg)- \int_{a}^{x_{0}}f(t)dt =
 \int_{x_{0}}^{x_{0}+h}f(t)dt
\end{array}
$$
and since $h \neq 0$, this implies that
\begin{equation} \label{diff}
\frac{F(x_{0}+h) - F(x_{0})}{h} = \frac{1}{h} \int_{x_{0}}^{x_{0}+h}f(t)dt
\end{equation}
In order to proceed, we will distinguish between two cases:

• $h > 0$   $\rightsquigarrow$  (I)
• $h < 0$   $\rightsquigarrow$  (II)

(I). $h > 0$: Since $[x_{0},x_{0}+h] \subseteq \Delta$, $f$ is continuous on $[x_{0},x_{0}+h]$ and the Extreme value theorem applies: there are numbers $c,d \in [x_{0},x_{0}+h]$ such that $f(c)=m$ and $f(d)=M$ are the absolute minimum and absolute maximum values respectively of $f$ in $[x_{0},x_{0}+h]$. Consequently
$$
\begin{array} {c}
mh \leq \int_{x_{0}}^{x_{0}+h}f(t)dt \leq Mh \Leftrightarrow f(c)h \leq \int_{x_{0}}^{x_{0}+h}f(t)dt \leq f(d)h \Leftrightarrow \\ \\
   \\
\Leftrightarrow f(c) \leq \frac{1}{h} \int_{x_{0}}^{x_{0}+h}f(t)dt \leq f(d) \stackrel{\eqref{diff}}{\Leftrightarrow} f(c) \leq \frac{F(x_{0}+h) - F(x_{0})}{h} \leq f(d)
\end{array}
$$

So we have concluded that
\begin{equation} \label{sand1}
 f(c) \leq \frac{F(x_{0}+h) - F(x_{0})}{h} \leq f(d)
\end{equation}
At this point, we have to observe the following thing: by the application of the extreme value theorem on the continuous function $f$ on the interval $[x_{0},x_{0}+h]$ it follows that both $c$ and $d$ depend in general on the value of $h > 0$. It is easy to see that their values are actually functions of the positive $h$: So we can write $c(h)$ and $d(h)$. Not much needs to be said about these functions; their behaviour may be complicated in general (for example, you can provide an argument to show that $c(h), \ d(h)$ need not even be continuous in general!). However we have:
\begin{equation} \label{concomplim1}
\begin{array}{c}
\lim_{h \rightarrow 0^{+}} c(h) = x & ,   &  \lim_{h \rightarrow 0^{+}} d(h) = x
\end{array}
\end{equation}
\eqref{concomplim1} can be proved as a simple application of the $(\varepsilon, \delta)$-definition of the limit. Readers are adviced to show that explicitly for practise!
   Taken that $f$ is continuous on $\Delta$ and thus on $[x_{0},x_{0}+h]$, \eqref{concomplim1} imply that
\begin{equation} \label{concomplim2}
\begin{array}{c}
\lim_{h \rightarrow 0^{+}} f(c) = \lim_{h \rightarrow 0^{+}} f(c(h)) = f(x) \\
    \\
\lim_{h \rightarrow 0^{+}} f(d) = \lim_{h \rightarrow 0^{+}} f(d(h)) = f(x)
\end{array}
\end{equation}
Now combining \eqref{sand1} together with \eqref{concomplim2} and applying the squeeze theorem from the right, we get
\begin{equation} \label{from the right}
\lim_{h \rightarrow 0^{+}} \frac{F(x_{0}+h) - F(x_{0})}{h} = f(x)
\end{equation}

(II). $h < 0$:  In this case $[x_{0}+h,x_{0}] \subseteq \Delta$ and we proceed again following exactly the same steps as before keeping however in mind that now $h < 0$. We leave the intermediate details to the reader. We finally end up in
 \begin{equation} \label{from the left}
\lim_{h \rightarrow 0^{-}} \frac{F(x_{0}+h) - F(x_{0})}{h} = f(x)
\end{equation}
   Combining \eqref{from the right}, \eqref{from the left} we get the result
$$
\lim_{h \rightarrow 0} \frac{F(x_{0}+h) - F(x_{0})}{h} = F'(x_{0}) = f(x_{0})
$$
which finally concludes the proof!

Theoretical Remarks #3

We come in today's post to supply a proof for a well known Calculus proposition:  In friday's 27/Dec/2013 post we mentioned (without proof), the following proposition:
Proposition: Let a real function $f$, continuous on an interval $\Delta$ and let $a \in \Delta$ be a fixed point. Then the function $F(x)=\int_{a}^{x}f(t)dt$ is an antiderivative function of $f$ in $\Delta$. In other words:
$$
F'(x) = \big( \int_{a}^{x}f(t)dt \big)' = f(x)
$$
for all $x \in \Delta$.

Proof: 
   It is sufficient to show that for any fixed point $x_{0} \in \Delta$ we have $F'(x_{0})=f(x_{0})$.

   Let us first study the difference quotient $\frac{F(x)-F(x_{0})}{x-x_{0}}$ whose limit at $x \rightarrow x_{0}$, $x \neq x_{0}$ defines the value of $F'(x_{0})$:
$$
\begin{array}{c}
\frac{F(x)-F(x_{0})}{x-x_{0}}= \frac{1}{x-x_{0}}\bigg( \int_{a}^{x}f(t)dt - \int_{a}^{x_{0}}f(t)dt \bigg) =  \\
    \\
= \frac{1}{x-x_{0}}\bigg(  \int_{x_{0}}^{a}f(t)dt + \int_{a}^{x}f(t)dt  \bigg) =  \frac{1}{x-x_{0}} \int_{x_{0}}^{x}f(t)dt
\end{array}
$$
thus
\begin{equation} \label{diff*}
\frac{F(x)-F(x_{0})}{x-x_{0}}=\frac{1}{x-x_{0}} \int_{x_{0}}^{x}f(t)dt
\end{equation}
and since
\begin{equation} \label{fun}
f(x_{0}) = \frac{1}{x-x_{0}}(x-x_{0})f(x_{0}) = \frac{1}{x-x_{0}}\int_{x_{0}}^{x}f(x_{0}) dt
\end{equation}
combining \eqref{diff*} and \eqref{fun}, we readily get the following relation:
\begin{equation} \label{diffun}
\frac{F(x)-F(x_{0})}{x-x_{0}}-f(x_{0}) = \frac{1}{x-x_{0}} \int_{x_{0}}^{x} \big( f(t) - f(x_{0}) \big) dt
\end{equation}
   Since $f$ is continuous at $x_{0} \in \Delta$, for any $\varepsilon > 0$ there is a $\delta > 0$ such that: for any $t \in \Delta$ with $|t-x_{0}| < \delta$ we will have $|f(t)-f(x_{0})| < \varepsilon$.

   Thus, for any $x \in \Delta$ with $0 < |x-x_{0}| < \delta$, using \eqref{diffun} we get:
$$
\begin{array}{c}
\bigg| \frac{F(x)-F(x_{0})}{x-x_{0}}-f(x_{0})  \bigg| = \frac{1}{|x-x_{0}|} \bigg| \int_{x_{0}}^{x} \big( f(t) - f(x_{0}) \big) dt \bigg|  \leq \\
      \\
\leq \frac{1}{|x-x_{0}|} \bigg| \int_{x_{0}}^{x} \big| f(t) - f(x_{0}) \big| dt \bigg| < \frac{1}{|x-x_{0}|} \big| \int_{x_{0}}^{x} \varepsilon dt \big| =  \\
    \\
= \frac{1}{|x-x_{0}|} \varepsilon |x-x_{0}| = \varepsilon
\end{array}
$$
But the above means -according to the $(\varepsilon, \delta)$ definition of the limit- that
$$
F'(x_{0}) = \lim_{x \rightarrow x_{0}} \bigg( \frac{F(x)-F(x_{0})}{x-x_{0}} \bigg) = f(x_{0})
$$
which finally concludes the proof.

Question of the week #4 - the answer

Question of the week #4: Let a continuous real function $f$ and $f(x)=e^{\int_{0}^{x}f(t)dt}$  for all $x<1$.  Find the formula of the function $f$.

Solution: First we note that $f(0)=e^{\int_{0}^{0}f(t)dt}=e^{0}=1$. Moreover, we can see that
$$
f(x)>0, \ \  \forall x<1
$$
   For any $x<1$ we have:
$$
\begin{array}{c}
f^{\prime}(x)=\bigg( e^{\int_{0}^{x}f(t)dt} \bigg) ^{\prime}=e^{\int_{0}^{x}f(t)dt} \big( \int_{0}^{x}f(t)dt \big)^{\prime}=f(x) \cdot f(x) \Leftrightarrow \\
      \\
\Leftrightarrow f^{\prime}(x)=f^{2}(x)  \Leftrightarrow \frac{f^{\prime}(x)}{f^{2}(x)} =1 \Leftrightarrow \\
    \\
\Leftrightarrow \bigg( -\frac{1}{f(x)} \bigg)^{\prime}=(x)^{\prime} \Leftrightarrow  -\frac{1}{f(x)}=x+c
\end{array}
$$
where $c \in \mathbb{R}$ is the constant of integration.
But the above relation readily implies (for $x=0$) that $c=-\frac{1}{f(0)}=-1$.
So we finally get:
$$
f(x)=\frac{1}{1-x}, \ \  \forall x<1
$$

Theoretical Remarks #2 - some insight into the function $\int_{a}^{x}f(t)dt$

   Let us first recall the proposition mentioned (without proof) in yesterday's post:
Proposition: Let a real function $f$, with domain $D_{f}$ and continuous on an interval $\Delta \subseteq D_{f} \subseteq  \mathbb{R}$ and let $a \in \Delta$ be a fixed point of $\Delta$. Then the function $F(x)=\int_{a}^{x}f(t)dt$ is an antiderivative function of $f$. In other words:
$$
F'(x) = \big( \int_{a}^{x}f(t)dt \big)' = f(x)
$$
for all $x \in \Delta$.

   If $x \in \Delta$ is a variable point,
then the value of the definite  integral  $F(x)= \int_{a}^{x}f(t)dt$ represents the area of the shaded  region shown in the figure:
Remark: Notice that the function $F(x)$ is defined on any interval $\Delta \subseteq D_{f}$ in which:

• $a \in \Delta$
• f is continuous on $\Delta$

and (according to the preceding proposition) is differentiable on that $\Delta$.

   Considering $h$ to be infinitesimal, we can now compute
$$
\begin{array}{c}
\Delta F(x) = F(x+h)-F(x) = \int_{a}^{x+h}f(t)dt - \int_{a}^{x}f(t)dt = \\
     \\
 = \int_{x}^{x+h}f(t)dt = E(\Omega) \approx h \cdot  f(x)
\end{array}
$$
In the above, the change $\Delta F(x)$ in the value of the function $F(x)=\int_{a}^{x}f(t)dt$ when $x$ changes to $x+h$ is denoted by $E(\Omega)$ and equals the area of the shaded strip $\Omega$, displayed in the next figure:

Consequently, $f(x) \approx \frac{E(\Omega)}{h}$ and since $h$ is considered to be infinitesimal, we can write:
$$
f(x)=\lim_{h \rightarrow 0}\frac{F(x+h)-F(x)}{h}=F'(x) = \frac{d \big(\int_{a}^{x}f(t)dt \big)}{dx}
$$
Remarks:
(1). The above should not be considered to be a rigorous proof. It should rather be taken as an intuitive  line of thinking, aiming to shed some light into "what is really going on" inside the function $\int_{a}^{x}f(t)dt$
(2). Let a real function $f$, continuous on an interval $\Delta \subseteq  \mathbb{R}$ and let $a \in \Delta$ be a fixed point of $\Delta$. Let another real function $g(x)$ with domain $D_{g}$ and differentiable (and thus: continuous) on an interval $\Delta_{1} \subseteq D_{g}$. We can then consider the composite function $G=F \circ g$:
$$
G(x) = \int_{a}^{g(x)}f(t)dt
$$
The domain $D_{G}$ of $G$ will be:
$$
D_{G} = \{ x \in D_{g} \ , \ g(x) \in \Delta \}
$$
   The composite function $G=F \circ g$ is differentiable on $\Delta_{2} = \Delta_{1} \cap D_{G}$. Its derivative can be computed by combining the above differentiation rule with the chain rule of differentiation (for differentiating composite functions). We readily get the following formula:
$$
\Big( \int_{a}^{g(x)}f(t)dt \Big)' = f\big( g(x) \big) \cdot g'(x)
$$
for all $x \in \Delta_{2}$.

Theoretical Remarks #1 - indefinite integrals, antiderivatives and the function $\int_{a}^{x}f(t)dt$

   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 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:
Proposition: 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
\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:
(1). Notice that the above proposition readily implies that $\int_{a}^{x} f(t)dt$ is differentiable (and thus continuous) for any $x \in \Delta$. Of course the $\ ' \ $ symbol indicates differentiation with respect to the variable $x$.
(2). 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
\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 $f(x)=2x$.

Question of the week #4

This week's posting will have to do with  integral calculus and more specifically with integral equations and antiderivatives of continuous functions (recall that the antiderivative is another name for the indefinite integral).
Let a continuous real function $f$ satisfying $f(x)=e^{\int_{0}^{x}f(t)dt}$ for all $x<1$. Find the formula of the function $f$.