Mathematics for Applied Sciences (Osnabrück 2023-2024)/Part I/Lecture 18/latex
\setcounter{section}{18}
In the following lectures, we will be concerned with \stichwort {integration theory} {,} i.e. we want to study and to compute the area of a surface which is bounded by the graph of a function
\zusatzklammer {the \stichwort {integrand} {}} {} {}
\mathdisp {f \colon [a,b] \longrightarrow \R} { }
and the $x$-axis. At the same time, there is a direct relation with finding a \stichwort {primitive function} {} of $f$, these are functions such that their derivative equals $f$. The concept of the area of a surface itself is problematic, which is understood thoroughly within \stichwort {measure theory} {.} However, this concept can be understood from an intuitive perspective, and we will only use some basic facts. These are only used for motivation, and not as arguments. The starting point is that the area of a rectangle is just the product of the side lengths, and that the area of a surface, which one can \stichwort {exhaust} {} with rectangles, equals the sum of the areas of these rectangles. We will work with the \stichwort {Riemann integral} {,} which provides a satisfactory theory for continuous functions. Here, all rectangles are parallel to the coordinate system, their width
\zusatzklammer {on the $x$-axis} {} {}
can vary and their height
\zusatzklammer {length} {} {}
is in relation to the value of the function over the base. By this method, the functions are approximated by so-called \stichwort {step functions} {.}
\zwischenueberschrift{Step functions}
\inputdefinition
{ }
{
Let $I$ be a
real interval
with endpoints
\mavergleichskette
{\vergleichskette
{ a,b
}
{ \in }{\R
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{.}
Then a
function
\mathdisp {t \colon I \longrightarrow \R} { }
is called a \definitionswort {step function}{,} if there exists a partition
\mavergleichskettedisp
{\vergleichskette
{ a
}
{ =} { a_0
}
{ <} { a_1
}
{ <} { a_2
}
{ <} { \cdots
}
}
{
\vergleichskettefortsetzung
{ <} { a_{n-1}
}
{ <} { a_n
}
{ =} { b
}
{ } {}
}{}{}
of $I$ such that $t$ is
constant
}
This definition does not require a certain value at the partition points. We call the interval \mathl{]a_{i-1},a_i[}{} the $i$-th interval of the partition, and \mathl{a_i-a_{i-1}}{} is called the length of this interval. If the lengths of all intervals are constant, then the partition is called an \stichwort {equidistant partition} {.}
\inputdefinition
{ }
{
Let $I$ be a
real interval
with endpoints
\mavergleichskette
{\vergleichskette
{ a,b
}
{ \in }{\R
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{,}
and let
\mathdisp {t \colon I \longrightarrow \R} { }
denote a
step function
for the partition
\mavergleichskette
{\vergleichskette
{a
}
{ = }{ a_0
}
{ < }{ a_1
}
{ < }{ a_2
}
{ < }{ \cdots
}
}
{
\vergleichskettefortsetzung
{<}{ a_{n-1}
}
{<}{ a_n
}
{}{b
}
{}{}
}{}{,}
with the values
\mathbed {t_i} {}
{i=1 , \ldots , n} {}
{} {} {} {.}
Then
\mavergleichskettedisp
{\vergleichskette
{ T
}
{ \defeq} { \sum_{i = 1}^n t_i (a_i - a_{i-1})
}
{ } {
}
{ } {
}
{ } {
}
}
{}{}{}
}
We denote the step integral also by \mathl{\int_{ a }^{ b } t ( x) \, d x}{.} If we have an equidistant partition of interval length \mathl{\frac{b-a}{n}}{,} then the step integral equals \mathl{\frac{b-a}{n} { \left( \sum_{i = 1}^n t_i \right) }}{.} The step integral does not depend on the partition chosen. As long as we have a step function with respect to the partition, one can pass to a refinement of the partition.
\inputdefinition
{ }
{
Let $I$ denote a
bounded interval,
and let
\mathdisp {f \colon I \longrightarrow \R} { }
denote a
function.
Then a
step function
\mathdisp {t \colon I \longrightarrow \R} { }
is called a \definitionswort {step function from above}{} for $f$, if
\mavergleichskette
{\vergleichskette
{ t(x)
}
{ \geq }{ f(x)
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{}
holds for all
\mavergleichskette
{\vergleichskette
{ x
}
{ \in }{ I
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{.}
A step function
\mathdisp {s \colon I \longrightarrow \R} { }
is called a \definitionswort {step function from below}{} for $f$, if
\mavergleichskette
{\vergleichskette
{ s(x)
}
{ \leq }{ f(x)
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{}
holds for all
\mavergleichskette
{\vergleichskette
{ x
}
{ \in }{ I
}
{ }{
}
{ }{
}
{ }{
}
}
}
A step function from above \zusatzklammer {below} {} {} for $f$ exists if and only if $f$ is bounded from above \zusatzklammer {from below} {} {.}
\inputdefinition
{ }
{
Let $I$ denote a
bounded interval,
and let
\mathdisp {f \colon I \longrightarrow \R} { }
denote a
function.
For a
step function from above
\mathdisp {t \colon I \longrightarrow \R} { }
of $f$, with respect to the partition
\mathbed {a_i} {}
{i=0 , \ldots , n} {}
{} {} {} {,}
and values
\mathbed {t_i} {}
{i=1 , \ldots , n} {}
{} {} {} {,}
the
step integral
\mavergleichskettedisp
{\vergleichskette
{ T
}
{ \defeq} { \sum_{i = 1}^n t_i { \left( a_i - a_{i-1} \right) }
}
{ } {
}
{ } {
}
{ } {
}
}
{}{}{}
}
\inputdefinition
{ }
{
Let $I$ denote a
bounded interval,
and let
\mathdisp {f \colon I \longrightarrow \R} { }
denote a
function.
For a
step function from below
\mathdisp {t \colon I \longrightarrow \R} { }
of $f$, with respect to the partition
\mathbed {a_i} {}
{i=0 , \ldots , n} {}
{} {} {} {,}
and values
\mathbed {s_i} {}
{i=1 , \ldots , n} {}
{} {} {} {,}
the
step integral
\mavergleichskettedisp
{\vergleichskette
{ S
}
{ \defeq} { \sum_{i = 1}^n s_i { \left( a_i - a_{i-1} \right) }
}
{ } {
}
{ } {
}
{ } {
}
}
{}{}{}
}
Different step functions from above yield different step integrals from above.
For further integration concepts, we need the following definitions which refer to arbitrary subsets of the real numbers.
\inputdefinition
{ }
{
For a nonempty subset
\mavergleichskette
{\vergleichskette
{M
}
{ \subseteq }{\R
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{,}
an
upper bound
$T$ of $M$ is called the \definitionswort {supremum}{} of $M$, if
\mavergleichskette
{\vergleichskette
{ T
}
{ \leq }{ S
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{}
}
\inputdefinition
{ }
{
For a nonempty subset
\mavergleichskette
{\vergleichskette
{M
}
{ \subseteq }{\R
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{,}
a
lower bound
$t$ of $M$ is called the \definitionswort {infimum}{} of $M$, if
\mavergleichskette
{\vergleichskette
{ t
}
{ \geq }{ s
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{}
}
The existence of infimum and supremum follows from the completeness of the real numbers.
\inputfaktbeweisnichtvorgefuehrt
{Real numbers/Bounded subset has supremum/Fact}
{Theorem}
{}
{
\faktsituation {}
\faktvoraussetzung {Every nonempty subset of the real numbers, which is
bounded from above,}
\faktfolgerung {has a
supremum
in $\R$.}
\faktzusatz {}
\inputdefinition
{ }
{
Let $I$ denote a
bounded interval,
and let
\mathdisp {f \colon I \longrightarrow \R} { }
denote a
function,
which is bounded from above. Then the
infimum
of all
step integrals
of
step functions from above
}
\inputdefinition
{ }
{
Let $I$ denote a
bounded interval,
and let
\mathdisp {f \colon I \longrightarrow \R} { }
denote a
function,
which is bounded from below. Then the
supremum
of all
step integrals
of
step functions from below
}
The boundedness from below makes sure that there exists at all a step function from below, so that the set of step integrals from below is not empty. This condition alone does not guarantee that a supremum exists. However, if the function is bounded from both sides, then the upper integral and the lower integral exist. If a partition is given, then there exists a smallest step function from above \zusatzklammer {a largest from below} {} {} which is given by the suprema \zusatzklammer {infima} {} {} of the function on the intervals of the partition. For a continuous function on a closed interval, these are maxima and minima. To compute the integral, we have to look at all step functions for all partitions.
\zwischenueberschrift{Riemann-integrable functions}
In the following, we will talk about compact interval, which is just a bounded and closed interval, hence of the form
\mavergleichskette
{\vergleichskette
{I
}
{ = }{[a,b]
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{}
with
\mavergleichskette
{\vergleichskette
{a,b
}
{ \in }{ \R
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{.}
\inputdefinition
{ }
{
Let $I$ denote a
compact interval
and let
\mathdisp {f \colon I \longrightarrow \R} { }
denote a
function.
Then $f$ is called \definitionswort {Riemann-integrable}{} if the
upper integral
and the
lower integral
}
It might by historically more adequate to call this \stichwort {Darboux-integrable} {.}
\inputdefinition
{ }
{
Let
\mavergleichskette
{\vergleichskette
{ I
}
{ = }{ [a,b]
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{}
denote a
compact interval.
For a
Riemann-integrable
function
\mathdisp {f \colon I = [a,b] \longrightarrow \R
, t \longmapsto f(t)} { , }
we call the
upper integral
of $f$
\zusatzklammer {which by definition coincides with the
lower integral} {} {}
the \definitionswort {definite integral}{} of $f$ over $I$. It is denoted by
\mathdisp {\int_{ a }^{ b } f ( t) \, d t \text{ or by } \int_{ I }^{ } f ( t) \, d t} { . }
}
The computation of such integrals is called to \stichwort {integrate} {.} Don't think too much about the symbol $dt$. It expresses that we want to integrate with respect to this variable. The name of the variable is not relevant, we have
\mavergleichskettedisp
{\vergleichskette
{ \int_{ a }^{ b } f(t) \, d t
}
{ =} { \int_{ a }^{ b } f(x) \, d x
}
{ } {
}
{ } {
}
{ } {
}
}
{}{}{.}
\inputfaktbeweis
{Riemann integral/Step functions with equal limit/Integral/Fact}
{Lemma}
{}
{
\faktsituation {Let $I$ denote a
compact interval,
and let
\mathdisp {f \colon I \longrightarrow \R} { }
denote a
function.
Suppose that there exists a
sequence
of
lower step functions
\mathbed {{ \left( s_n \right) }_{ n \in \N }} {with}
{s_n \leq f} {}
{} {} {} {}
and a sequence of upper step functions
\mathbed {{ \left( t_n \right) }_{ n \in \N }} {with}
{t_n \geq f} {}
{} {} {} {.}}
\faktvoraussetzung {Suppose furthermore that the corresponding sequences of step integrals
converge
to the same real number.}
\faktfolgerung {Then $f$ is
Riemann-integrable,
and the
definite integral
equals this limit, so
\mavergleichskettedisp
{\vergleichskette
{ \lim_{n \rightarrow \infty} \int_{ a }^{ b } s_n ( x) \, d x
}
{ =} { \int_{ a }^{ b } f ( x) \, d x
}
{ =} { \lim_{n \rightarrow \infty} \int_{ a }^{ b } t_n ( x) \, d x
}
{ } {
}
{ } {
}
}
{}{}{.}}
\faktzusatz {}
{See Exercise 18.12 . }
\inputbeispiel{}
{
We consider the function
\mathdisp {f \colon [0,1] \longrightarrow \R
, t \longmapsto t^2} { , }
which is
strictly increasing
in this interval. Hence, for a subinterval
\mavergleichskette
{\vergleichskette
{ [a,b]
}
{ \subseteq }{ [0,1]
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{,}
the value \mathl{f(a)}{} is the
minimum,
and \mathl{f(b)}{} is the
maximum
of the function on this subinterval. Let $n$ be a positive natural number. We partition the interval \mathl{[0,1]}{} into the $n$ subintervals
\mathbed {\left[ i { \frac{ 1 }{ n } } , (i+1) { \frac{ 1 }{ n } } \right]} {}
{i=0 , \ldots , n-1} {}
{} {} {} {,}
of length ${ \frac{ 1 }{ n } }$. The
step integral
for the corresponding
lower step function
is
\mavergleichskettedisp
{\vergleichskette
{ \sum_{i = 0}^{n-1} \frac{1}{n} { \left(i \frac{1}{n}\right) }^2
}
{ =} { \frac{1}{n^3} \sum_{i = 0}^{n-1} i^2
}
{ =} { \frac{1}{n^3} { \left( \frac{1}{3} n^3 - \frac{1}{2}n^2 + \frac{1}{6} n\right) }
}
{ =} { \frac{1}{3} - \frac{1}{2n} + \frac{1}{6n^2}
}
{ } {
}
}
{}{}{}
\zusatzklammer {see
Exercise 2.10
for the formula for the sum of the squares} {} {.}
Since the
sequences
\mathkor {} {{ \left( 1/2n \right) }_{ n \in \N }} {and} {{ \left( 1/6n^2 \right) }_{ n \in \N }} {}
converge
to $0$, the
limit
for \mathl{n \rightarrow \infty}{} of these step integrals equals ${ \frac{ 1 }{ 3 } }$. The
step integral
for the corresponding
step function from above
is
\mavergleichskettedisp
{\vergleichskette
{ \sum_{i = 0}^{n-1} \frac{1}{n} { \left((i+1) \frac{1}{n}\right) }^2
}
{ =} { \frac{1}{n^3} \sum_{i = 0}^{n-1} (i+1)^2
}
{ =} { \frac{1}{n^3} \sum_{j = 1}^{n} j^2
}
{ =} { \frac{1}{n^3} { \left(\frac{1}{3} n^3 + \frac{1}{2}n^2 + \frac{1}{6} n\right) }
}
{ =} { \frac{1}{3} + \frac{1}{2n} + \frac{1}{6n^2}
}
}
{}{}{.}
The limit of this sequence is again ${ \frac{ 1 }{ 3 } }$. By
Lemma 18.13
,
the
upper integral
and the
lower integral
coincide, hence the function is
Riemann-integrable,
and for the
definite integral
we get
\mavergleichskettedisp
{\vergleichskette
{ \int_{ 0 }^{ 1 } t^2 \, d t
}
{ =} { { \frac{ 1 }{ 3 } }
}
{ } {
}
{ } {
}
{ } {
}
}
{}{}{.}
}
\inputfaktbeweis
{Compact interval/Real function/Riemann integrable on partition/Fact}
{Lemma}
{}
{
\faktsituation {Let
\mavergleichskette
{\vergleichskette
{ I
}
{ = }{ [a,b]
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{}
be a compact interval, and let
\mathdisp {f \colon I \longrightarrow \R} { }
be a function.}
\faktuebergang {Then the following statements are equivalent.}
\faktfolgerung {\aufzaehlungdrei {The function $f$ is
Riemann-integrable.
} {There exists a partition
\mavergleichskette
{\vergleichskette
{ a
}
{ = }{ a_0
}
{ < }{ a_1
}
{ < }{ \cdots
}
{ < }{ a_n
}
}
{
\vergleichskettefortsetzung
{}{ b
}
{}{}
{}{}
{}{}
}{}{,}
such that the restrictions
\mavergleichskette
{\vergleichskette
{f_i
}
{ \defeq }{ f |_{[a_{i-1},a_i]}
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{}
are Riemann-integrable.
} {For every partition
\mavergleichskette
{\vergleichskette
{ a
}
{ = }{ a_0
}
{ < }{ a_1
}
{ < }{ \cdots
}
{ < }{ a_n
}
}
{
\vergleichskettefortsetzung
{}{ b
}
{}{}
{}{}
{}{}
}{}{,}
the restrictions
\mavergleichskette
{\vergleichskette
{f_i
}
{ \defeq }{ f |_{[a_{i-1},a_i]}
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{}
are Riemann-integrable.
}}
\faktzusatz {In this situation, the equation
\mavergleichskettedisp
{\vergleichskette
{ \int_{ a }^{ b } f ( t) \, d t
}
{ =} { \sum_{i = 1}^n \int_{ a_{i-1} }^{ a_i } f_i ( t) \, d t
}
{ } {
}
{ } {
}
{ } {
}
}
{}{}{}
holds.}
{See Exercise 18.14 . }
\inputdefinition
{ }
{
Let
$f \colon I \rightarrow \R$
be a function on a
real interval.
Then $f$ is called \definitionswort {Riemann-integrable}{,} if the
restriction
of $f$ to every
compact
interval
\mavergleichskette
{\vergleichskette
{ [a,b]
}
{ \subseteq }{ I
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{}
is
}
Due to this lemma, both definitions coincide for a compact interval \mathl{[a,b]}{.} The integrability of a function $f \colon \R \rightarrow \R$ does not mean that \mathl{\int_\R f(x) dx}{} has a meaning or exists.
\zwischenueberschrift{Riemann-integrability of continuous functions}
\inputfaktbeweisnichtvorgefuehrt
{Interval/Continuous function/Riemann integrable/Fact}
{Theorem}
{}
{
\faktsituation {Let
$f \colon I \rightarrow \R$
denote a
continuous function.}
\faktfolgerung {Then $f$ is
Riemann-integrable.}
\faktzusatz {}
}
{Interval/Continuous function/Riemann integrable/Fact/Proof
\inputfaktbeweis
{Riemann integrability/Elementary properties/Fact}
{Lemma}
{}
{
\faktsituation {Let
\mavergleichskette
{\vergleichskette
{ I
}
{ = }{ [a,b]
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{}
denote a
compact interval,
and let
$f,g \colon I \rightarrow \R$
denote
Riemann-integrable
functions.}
\faktuebergang {Then the following statements hold.}
\faktfolgerung {\aufzaehlungsieben {If
\mavergleichskette
{\vergleichskette
{m
}
{ \leq }{f(x)
}
{ \leq }{M
}
{ }{
}
{ }{
}
}
{}{}{}
holds for all
\mavergleichskette
{\vergleichskette
{x
}
{ \in }{I
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{,}
then
\mavergleichskette
{\vergleichskette
{ m(b-a)
}
{ \leq }{ \int_{ a }^{ b } f ( t) \, d t
}
{ \leq }{ M(b-a)
}
{ }{
}
{ }{
}
}
{}{}{}
holds.
} {If
\mavergleichskette
{\vergleichskette
{ f(x)
}
{ \leq }{ g(x)
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{}
holds for all
\mavergleichskette
{\vergleichskette
{x
}
{ \in }{I
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{,}
then
\mavergleichskette
{\vergleichskette
{ \int_{ a }^{ b } f ( t) \, d t
}
{ \leq }{ \int_{ a }^{ b } g ( t) \, d t
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{}
holds.
} {The sum \mathl{f+g}{} is Riemann-integrable, and the identity
\mavergleichskette
{\vergleichskette
{ \int_{ a }^{ b } (f+g)(t) \, d t
}
{ = }{ \int_{ a }^{ b } f ( t) \, d t + \int_{ a }^{ b } g ( t) \, d t
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{}
holds.
} {For
\mavergleichskette
{\vergleichskette
{c
}
{ \in }{\R
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{}
we have
\mavergleichskette
{\vergleichskette
{ \int_{ a }^{ b } (cf)(t) \, d t
}
{ = }{ c \int_{ a }^{ b } f ( t) \, d t
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{.}
} {The functions \mathl{{\max { \left( f , g \right) } }}{} and \mathl{{\min { \left( f , g \right) } }}{} are Riemann-integrable.
} {The function \mathl{\betrag { f }}{} is Riemann-integrable.
} {The product \mathl{fg}{} is Riemann-integrable.
}}
\faktzusatz {}
}
{
For (1) to (4) see
Exercise 18.14
.
For (5) see
Exercise 18.17
.
(6) follows directly from (5), because of
\mavergleichskette
{\vergleichskette
{ \betrag { f }
}
{ = }{ {\max { \left( f,-f , \right) } }
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{.}
For (7), see
Exercise 18.18
.