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\begin{document}


\title{On-line Math 21}


\email{david.johnson@lehigh.edu}


\urladdr{http://www.lehigh.edu/\textasciitilde{}dlj0}

\maketitle

\subsection{The Fundamental Theorem of Calculus}

For a function \( f \), defined on \( [a,b] \), define \( G(x) \) to be the
definite integral of \( f \) from \( a \) to \( x \). So, now the upper ``limit
of integration'', the upper value of the variable that forms the end of the
region, becomes the \textit{variable} for a function. 

Note how I kept away from naming the variable of \( f \). That's because of
our repeated use of \( x \) as the variable all the time. Now, we can't write\[
G(x):=\int _{a}^{x}f(x)dx,\]
even though that's what we're thinking, because each letter is supposed to only
stand for one thing. Is \( x \) the variable for \( f \), or is it the upper
limit of integration? Now, a rose is a rose and all that, it doesn't matter
what we call the variable of \( f \). But we can't use \( x \), since it's
already spoken for as the variable of \( G \), so we write \( t \) as the
variable for \( f \), and we have: \[
G(x):=\int _{a}^{x}f(t)dt.\]


\vspace{0.3cm}
{\par\centering \includegraphics{G.ps} \par}
\vspace{0.3cm}

For example, the function \( g \) defined by: \[
G(x):=\int _{1}^{x}\frac{dt}{t}\]


\vspace{0.3cm}
{\par\centering \includegraphics{log.ps} \par}
\vspace{0.3cm}

is the area under the curve \( y=1/t \) from \( t=1 \) to \( t=x \). It is
quite easy now to find values of this function for any \( x \), and so numerically
this function is simple to understand. For each \( x \), you just find the
area. For \( x<1 \) you have to remember that the limits are backwards, so
that the integral is really negative. 


\subsubsection{Preliminary result, another MVT}

In order to prove the FTC, we need a slightly different version of the MVT (Mean
Value Theorem) that applies to integrals. 

\begin{prop}
\textbf{\emph{{[}MVT, integral form{]}}} If \( f(x) \) is continuous on \( [a,b] \),
then there is a \( c\in [a,b] \) so that \[
f(c)=\frac{\int _{a}^{b}f(x)dx}{b-a}.\]

\end{prop}
\begin{rem}
If you are used to thinking about the definite integral as an antiderivative,
then this is really just the old MVT applied to the antiderivative of \( f \),
but we're not there, just yet. This is a slightly different statement from the
regular MVT, but is even easier to prove. We are assuming that any continuous
function is integrable, which we proved earlier.
\end{rem}
\textbf{\emph{Proof:}}


\subsubsection{The Theorem }

The way most texts describe it these days, there are two ``Fundamental Theorems''
of calculus. The first one will show that the general function \( G(x) \) defined
as \[
G(x):=\int _{a}^{x}f(t)dt\]
 has derivative \( G'(x)=f(x) \). That is really the most serious part of this
theorem. What it shows is not so much that \( f \) has an antiderivative, although
that is theoretically useful, but it shows that the definite integral is related
to that antiderivative.  

The second part of the Fundamental Theorem uses that information to show the
``easy way'' to compute a definite integral, by evaluating an antiderivative
of the function (which you find by integration techniques).  

\begin{thm}
\textbf{\textup{{[}Fundamental Theorem of Calculus, Part 1{]}.}} Let \( f \)
be a continuous function on an open interval containing \( [a,b] \). Set\[
G(x):=\int _{a}^{x}f(t)dt.\]
 Then, \( G(x) \) is differentiable on \( [a,b] \) and its derivative is \( f \),
that is, \( G'(x)=f(x). \)
\end{thm}
\textbf{\emph{Proof:}}

\begin{example}
Find an antiderivative of \[
f(x)=\sqrt{2+\sin ^{2}x}.\]

\end{example}

\subsubsection*{Real Stupid Solution}

\begin{example}
Find the derivative of \[
F(x)=\int _{0}^{x^{2}}\sqrt{1+t^{6}}dt.\]

\end{example}

\subsubsection*{Solution}

\begin{xca}
Find\[
\frac{d}{dx}\left( \int _{x}^{x^{2}}\sin (t^{2})dt\right) =\]

\end{xca}
\begin{thm*}
\textbf{\textup{{[}Second Fundamental Theorem of Calculus{]}.}} Let \( f \)
be a continuous function on an open interval containing \( [a,b] \). Let \( F(x) \)
be \textbf{any} antiderivative of \( f \). Then:\[
\int _{a}^{b}f(x)dx=F(b)-F(a):=\left. F(x)\right| ^{b}_{a}.\]

\end{thm*}
\begin{rem*}
The notation on the last line, \[
\left. F(x)\right| ^{b}_{a},\]
 is \textit{defined} by \[
F(b)-F(a):=\left. F(x)\right| ^{b}_{a}.\]
 It just makes it easier when you are working out an integration problem, to
have the chance to write down the indefinite integral \( F \) \textit{before}
you evaluate it at the endpoints. 
\end{rem*}
\textbf{\emph{Proof:}}

\begin{example}
\[
\int _{0}^{\pi /2}\cos (x)dx.\]
 
\end{example}

\subsubsection*{Solution }

\begin{xca}
\[
\int _{1}^{3}\frac{1}{x^{3}}dx=\]

\end{xca}
\begin{example}
\[
\int _{-1}^{2}|x|dx.\]
 
\end{example}

\subsubsection*{Solution }

\begin{example}
\[
\int _{2}^{5}f(x)dx.\]
 where \[
f(x)=\left\{ \begin{array}{cc}
3x-4, & \mathrm{if}\, x\geq 4\\
x+2, & \mathrm{if}\, x<4
\end{array}\right. \]
 
\end{example}

\subsubsection*{Solution}

\begin{example}
\[
\int _{0}^{2}x\sqrt{4-x^{2}}dx\]
 
\end{example}

\subsubsection*{Solution}

\begin{example}
\[
\int _{0}^{2}\sqrt{4-x^{2}}dx\]
 
\end{example}

\subsubsection*{Solution}
\end{document}
