# 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 \mathla_{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.
\faktsituation \faktvoraussetzung {Every nonempty subset of the real numbers, which is bounded from above,} \faktfolgerung {has a supremum in $\R$.} \faktzusatz } {Real numbers/Bounded subset has supremum/Fact/Proof
\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
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 } { } { } { } { } { } { } }{.}
\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
\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 } } } { } { } { } { } { } { } }{.}
\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.}
\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}
\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
\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 .