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\begin{document}


\title{On-line Math 21}


\email{david.johnson@lehigh.edu}


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

\maketitle
Integration has its theoretical basis in the idea of finding areas of regions
in the plane. That is, to find, for example, the area bounded by the graphs\[
y=x^{2},\, y=0,\, \mathrm{and}\, x=3.\]
The process begins by finding a way to approximate that area. We know the area
of a rectangle, so we use that. But, approximating that region by a rectangle
is rather a poor approximation, so we instead approximate the pieces of the
region obtained by slicing it by the lines, say, \( x=1\, \mathrm{and}\, x=2 \),
that is, we approximate the area by three rectangles, not one. That's not a
very good approximation either, but it's a start. 

\vspace{0.3cm}
{\par\centering \resizebox*{3in}{3in}{\rotatebox{-90}{\includegraphics{riemann-1.ps}}} \par}
\vspace{0.3cm}

Now, what should be the heights of the rectangles in each piece? The tallest
point on the curve \( y=x^{2} \)? That's the way it's pictured. Or, perhaps
the lowest point? Somewhere in between? You can choose any one. Obviously, choosing
the tallest point gives a rectangle with \textit{more} area than the original
region, while choosing the lowest gives less area. If you are very lucky, choosing
rectangles in between will give exactly the right area. Let's do some estimates.
 

Using the highest point on each piece (the right-edge of each region gives the
height as \( f(x) \), where \( x \) is at the right edge), we get:\[
Area\approx 1\times 1+4\times 1+9\times 1=14,\]
while using the lowest point gives:\[
Area\approx 0\times 1+1\times 1+4\times 1=5.\]
 

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

The real area is somewhere in between. 

We'll try it again: now, let's approximate this area using 20 rectangles. The
picture is (using the highest points, the \textit{circumscribed} rectangles):

\vspace{0.3cm}
{\par\centering \resizebox*{3in}{3in}{\rotatebox{-90}{\includegraphics{riemann-2.ps}}} \par}
\vspace{0.3cm}

The approximate area of this is 9.68625. I got that using Maple, which has a
special package for these calculations. Using the lowest points in each subinterval,
that is, using the \textit{inscribed} rectangles, gives an area of 8.33625 for
20 rectangles. 

\textbf{\emph{How do we do this calculation?}}

The reality is that, for most curves, it doesn't matter whether you use the
right edge, or the left, or anywhere in between. You also don't have to take
the subintervals to all be the same length, just so they get smaller, to get
a better and better approximation of the real area. 

We will be spending quite a bit of time dealing with integration as a summation
process, and it will get awkward writing down all these sums. We need a better
notation. In addition, we need some specific formulas for certain sums, which
you may not have seen before. Look through this topic now:

\textbf{\emph{Summation notation and special formulas}}


\subsection{The definition of the Riemann Integral}

Let's look at the process in general, for a function \( f(x) \) over an interval
\( [a,b] \). Choose any old set of points \( \{x_{0},x_{1},\dots ,x_{n}\} \)
to chop the interval \( [a,b] \) up into subintervals. Set \( a=x_{0} \) and
set \( b=x_{n} \) so that we refer to them all the same way. For each of those
subintervals, find a point \[
x_{i}^{*}\in [x_{i-1},x_{i}]\]
 (this is the \( i^{th} \) subinterval). 

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

Then, \( f(x_{i}^{*}) \) is the height of the curve at that point, so that
height gives the height of a rectangle whose top crosses the curve. Well, if
\( f(x_{i}^{*})<0 \), it's below the axis, so there is a minus sign there.
The area of the \( i^{th} \) rectangle is then \[
f(x_{i}^{*})\Delta x_{i},\]
where where \( \Delta x_{i}:=x_{i}-x_{i-1} \) is the width of the \( i^{th} \)
subinterval. This {}``area{}'' is negative when \( f(x_{i}^{*})<0 \), but
even then the absolute value is the area of the rectangle. 

Then, the {}``area under the curve{}'' is approximated by the sum\[
\sum _{i=1}^{n}f(x_{i}^{*})\Delta x_{i},\]


The reason the {}``area under the curve{}'' is in quotation marks is that
we will still use the idea of this limit of sums, even if the function is \textit{negative}
somewhere, and so the {}``area under the curve{}'' doesn't quite make sense.
However, if you count the areas of those regions where \( f(x)<0 \) as \textit{negative
area}, it does give the {}``area under the curve{}'' as this limit of sums.

\begin{defn}
A \textit{partition} \( P \) of an interval \( [a,b] \) is a decomposition
of the interval into smaller pieces, as above, \( a=x_{0},x_{1},\ldots ,x_{n}=b \).
We usually just refer to the numbers \( x_{0},\ldots ,x_{n} \) as the \textit{partition}
\( P \) of \( [a,b] \). 

The \textit{mesh} \( \left\Vert P\right\Vert  \) of a partition \( P \) is
the length of the longest section (or \textit{subinterval}) of the partition,
\( \left\Vert P\right\Vert :=max\{x_{1}-x_{0},x_{2}-x_{1},\ldots ,x_{n}-x_{n-1}\} \).
\end{defn}
Now comes the formal definition of an integral. 

\begin{defn}
\( f \) is some function, defined for \( x\in [a,b] \) (you actually can leave
out ``a few'' points without causing any problems). For each number \( \delta >0 \),
choose a partition \( P=x_{0},\ldots ,x_{n} \) of mesh \( \left\Vert P\right\Vert  \),
and choose (randomly) points \( x_{i}^{*}\in [x_{i-1},x_{i}] \). The \textit{Riemann
sum} of the function \( f \) corresponding to that partition and set of sampling
numbers is the sum:\[
\sum _{i=1}^{n}f(x_{i}^{*})\Delta x_{i}.\]
The number you get from that process depend on the partition and the sampling
numbers, as well as the function itself. However, \textbf{if} as \( \left\Vert P\right\Vert \rightarrow 0 \)
(as the mesh gets finer without bound), the limit:\[
\lim _{\left\Vert P\right\Vert \rightarrow 0}\sum _{i=1}^{n}f(x_{i}^{*})\Delta x_{i}:=\int _{a}^{b}f(x)dx\]
exists, \textit{independently} of the partitions (as long as the mesh is getting
smaller), and of the sampling numbers, then we say that the function \( f \)
is \textit{integrable} over the interval \( [a,b] \), and the \textit{integral}
is the limit value, written as \[
\int _{a}^{b}f(x)dx.\]
 
\end{defn}
Integration involves a limit of sums. But, the limit is ``sloppy'' in this case
in that so many things are not determined: the subintervals themselves, the
point you pick in each subinterval, the number of subintervals. Still, rather
remarkably, ``most'' functions are integrable in this sense, even ones that
are not continuous. 

\begin{example}
If \[
f(x)=\left\{ \begin{array}{ll}
1, & \mathrm{if}\, 0\leq x\leq 1\\
1/2, & \mathrm{if}\, 1<x\leq 2
\end{array}\right. ,\]
 then \[
\int _{0}^{2}f(x)dx=3/2.\]
 
\end{example}

\subsubsection*{Solution}

\begin{example}
If \( f(x)=3x \), then \[
\int _{1}^{2}f(x)dx=9/2.\]
 
\end{example}

\subsubsection*{Solution}

\begin{example}
\[
\int _{0}^{1}x^{2}dx=\frac{1}{3}.\]
 
\end{example}

\subsubsection*{Solution}

\begin{xca}
Show that \[
\int _{-1}^{2}x^{3}dx=\frac{15}{4}.\]
 

Hint: Use the formulas for the sum of the first \( n \) cubes in the review
section, and, as before, presume that the limit exists, so that you can use
uniform subintervals and circumscribed rectangles.

\textbf{\emph{Answer:}}
\end{xca}
\begin{prop}
Rules  
\begin{enumerate}
\item \[
\int _{a}^{b}c\, dx=c(b-a).\]

\item \[
\int _{a}^{b}f(x)dx=-\int _{b}^{a}f(x)dx.\]
 Actually, this rule is a definition. We hadn't had a definition for the right-hand
side (at least when \( a<b \)); this helps for some later formulas. 
\item \[
\int _{a}^{a}f(x)dx=0.\]
 As before, this is really a definition. 
\item \[
\int _{a}^{b}cf(x)dx=c\int _{a}^{b}f(x)dx.\]
 
\item \[
\int _{a}^{b}f(x)\pm g(x)dx=\int _{a}^{b}f(x)dx\pm \int _{a}^{b}g(x)dx.\]
 
\item If \( f(x)\geq 0 \) for all \( x\in [a,b] \), \( a<b \), then \[
\int _{a}^{b}f(x)dx\geq 0.\]

\item If \( f(x)\geq g(x) \) for all \( x\in [a,b] \), \( a<b \), then \[
\int _{a}^{b}f(x)dx\geq \int _{a}^{b}g(x)dx.\]

\item For all \( a \), \( b \), and \( c \), \[
\int _{a}^{c}f(x)dx+\int _{c}^{b}f(x)dx=\int _{a}^{b}f(x)dx.\]

\item If \( m\leq f(x)\leq M \) for all \( x\in [a,b] \), \( a<b \), then \[
m(b-a)\leq \int _{a}^{b}f(x)dx\leq M(b-a).\]

\item \[
\left| \int _{a}^{b}f(x)dx\right| \leq \int _{a}^{b}\left| f(x)\right| dx.\]

\end{enumerate}
\end{prop}
\textbf{\emph{Proofs:}}

\begin{thm}
If \( f \) is continuous on \( [a,b] \), then \[
\int _{a}^{b}f(x)dx\]
 exists.
\end{thm}
\textbf{\emph{Proof:}}

Use the rules and theorems above to sove the following exercises.

\begin{xca}
Find \[
\int _{-2}^{3}\left| x\right| dx.\]

\end{xca}
\textbf{\emph{Hint:}} Split the integral up into easier pieces, and evaluate
the integral of each piece by high-school geometry.

\begin{xca}
Show that \[
\int _{0}^{\pi /4}\sin ^{3}xdx\leq \int _{0}^{\pi /4}\sin ^{2}xdx.\]

\end{xca}
%

\begin{example}
Show that \[
\int _{0}^{2\pi }\sin ^{2}xdx=\pi .\]
 The same is true for \( \cos ^{2}x \), \[
\int _{0}^{2\pi }\cos ^{2}xdx=\pi .\]

\end{example}

\subsubsection*{Solution}
\end{document}
