Mathematics for Applied Sciences (Osnabrück 2023-2024)/Part I/Lecture 18

In the following lectures, we will be concerned with 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 (the integrand)

$f\colon [a,b]\longrightarrow \mathbb {R}$

and the ${}x$-axis. At the same time, there is a direct relation with finding a 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 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 exhaust with rectangles, equals the sum of the areas of these rectangles. We will work with the Riemann integral, which provides a satisfactory theory for continuous functions. Here, all rectangles are parallel to the coordinate system, their width (on the ${}x$-axis) can vary and their height (length) is in relation to the value of the function over the base. By this method, the functions are approximated by so-called step functions.

Step functions

A step function. In the statistic context, it is also called a histogram or a bar chart.

Definition

Let ${}I$ be a real interval with endpoints ${}a,b\in \mathbb {R}$. Then a function

$t\colon I\longrightarrow \mathbb {R}$

is called a step function, if there exists a partition

${}a=a_{0}<a_{1}<a_{2}<\cdots <a_{n-1}<a_{n}=b\,$

of ${}I$ such that ${}t$ is constant

on every open interval ${}]a_{i-1},a_{i}[$.

This definition does not require a certain value at the partition points. We call the interval ${}]a_{i-1},a_{i}[$ the ${}i$-th interval of the partition, and ${}a_{i}-a_{i-1}$ is called the length of this interval. If the lengths of all intervals are constant, then the partition is called an equidistant partition.

Definition

Let ${}I$ be a real interval with endpoints ${}a,b\in \mathbb {R}$, and let

$t\colon I\longrightarrow \mathbb {R}$

denote a step function for the partition ${}a=a_{0}<a_{1}<a_{2}<\cdots <a_{n-1}<a_{n}=b$, with the values ${}t_{i}$, ${}i=1,\ldots ,n$. Then

${}T:=\sum _{i=1}^{n}t_{i}(a_{i}-a_{i-1})\,$

is called the step integral of ${}t$ on ${}I$.

We denote the step integral also by ${}\int _{a}^{b}t(x)\,dx$. If we have an equidistant partition of interval length ${}{\frac {b-a}{n}}$, then the step integral equals ${}{\frac {b-a}{n}}{\left(\sum _{i=1}^{n}t_{i}\right)}$. 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.

Definition

Let ${}I$ denote a bounded interval, and let

$f\colon I\longrightarrow \mathbb {R}$

denote a function. Then a step function

$t\colon I\longrightarrow \mathbb {R}$

is called a step function from above for ${}f$, if ${}t(x)\geq f(x)$ holds for all ${}x\in I$. A step function

$s\colon I\longrightarrow \mathbb {R}$

is called a step function from below for ${}f$, if ${}s(x)\leq f(x)$ holds for all

${}x\in I$.

A step function from above (below) for ${}f$ exists if and only if ${}f$ is bounded from above (from below).

Definition

Let ${}I$ denote a bounded interval, and let

$f\colon I\longrightarrow \mathbb {R}$

denote a function. For a step function from above

$t\colon I\longrightarrow \mathbb {R}$

of ${}f$, with respect to the partition ${}a_{i}$, ${}i=0,\ldots ,n$, and values ${}t_{i}$, ${}i=1,\ldots ,n$, the step integral

${}T:=\sum _{i=1}^{n}t_{i}{\left(a_{i}-a_{i-1}\right)}\,$

is called a step integral from above for ${}f$ on ${}I$.

Definition

Let ${}I$ denote a bounded interval, and let

$f\colon I\longrightarrow \mathbb {R}$

denote a function. For a step function from below

$t\colon I\longrightarrow \mathbb {R}$

of ${}f$, with respect to the partition ${}a_{i}$, ${}i=0,\ldots ,n$, and values ${}s_{i}$, ${}i=1,\ldots ,n$, the step integral

${}S:=\sum _{i=1}^{n}s_{i}{\left(a_{i}-a_{i-1}\right)}\,$

is called a step integral from below for ${}f$ on ${}I$.

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.

Definition

For a nonempty subset ${}M\subseteq \mathbb {R}$, an upper bound ${}T$ of ${}M$ is called the supremum of ${}M$, if ${}T\leq S$

holds for all upper bounds ${}S$ of ${}M$.

Definition

For a nonempty subset ${}M\subseteq \mathbb {R}$, a lower bound ${}t$ of ${}M$ is called the infimum of ${}M$, if ${}t\geq s$

holds for all lower bounds ${}s$ of ${}M$.

The existence of infimum and supremum follows from the completeness of the real numbers.

Theorem

Every nonempty subset of the real numbers, which is bounded from above, has a supremum in ${}\mathbb {R}$.

Proof

This proof was not presented in the lecture.

$\Box$

Definition

Let ${}I$ denote a bounded interval, and let

$f\colon I\longrightarrow \mathbb {R}$

denote a function, which is bounded from above. Then the infimum of all step integrals of step functions from above

of ${}f$ is called the upper integral of ${}f$.

Definition

Let ${}I$ denote a bounded interval, and let

$f\colon I\longrightarrow \mathbb {R}$

denote a function, which is bounded from below. Then the supremum of all step integrals of step functions from below

of ${}f$ is called the lower integral of ${}f$.

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 (a largest from below) which is given by the suprema (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.

Riemann-integrable functions

In the following, we will talk about compact interval, which is just a bounded and closed interval, hence of the form ${}I=[a,b]$ with ${}a,b\in \mathbb {R}$.

A step function from above and one from below. The green area is a step integral from below and the yellow area (partly covered) is a step function from above.

Definition

Let ${}I$ denote a compact interval and let

$f\colon I\longrightarrow \mathbb {R}$

denote a function. Then ${}f$ is called Riemann-integrable if the upper integral and the lower integral

of ${}f$ exist and coincide.

It might by historically more adequate to call this Darboux-integrable.

Definition

Let ${}I=[a,b]$ denote a compact interval. For a Riemann-integrable function

$f\colon I=[a,b]\longrightarrow \mathbb {R} ,t\longmapsto f(t),$

we call the upper integral of ${}f$ (which by definition coincides with the lower integral) the definite integral of ${}f$ over ${}I$. It is denoted by

$\int _{a}^{b}f(t)\,dt{\text{ or by }}\int _{I}^{}f(t)\,dt.$

The computation of such integrals is called to 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

${}\int _{a}^{b}f(t)\,dt=\int _{a}^{b}f(x)\,dx\,.$

Lemma

Let ${}I$ denote a compact interval, and let

$f\colon I\longrightarrow \mathbb {R}$

denote a function. Suppose that there exists a sequence of lower step functions ${}{\left(s_{n}\right)}_{n\in \mathbb {N} }$ with ${}s_{n}\leq f$ and a sequence of upper step functions ${}{\left(t_{n}\right)}_{n\in \mathbb {N} }$ with ${}t_{n}\geq f$. Suppose furthermore that the corresponding sequences of step integrals converge to the same real number. Then ${}f$ is Riemann-integrable, and the definite integral equals this limit, so

${}\lim _{n\rightarrow \infty }\int _{a}^{b}s_{n}(x)\,dx=\int _{a}^{b}f(x)\,dx=\lim _{n\rightarrow \infty }\int _{a}^{b}t_{n}(x)\,dx\,.$

Proof

See Exercise 18.12 .

$\Box$

Example

We consider the function

$f\colon [0,1]\longrightarrow \mathbb {R} ,t\longmapsto t^{2},$

which is strictly increasing in this interval. Hence, for a subinterval ${}[a,b]\subseteq [0,1]$, the value ${}f(a)$ is the minimum, and ${}f(b)$ is the maximum of the function on this subinterval. Let ${}n$ be a positive natural number. We partition the interval ${}[0,1]$ into the ${}n$ subintervals ${}\left[i{\frac {1}{n}},(i+1){\frac {1}{n}}\right]$, ${}i=0,\ldots ,n-1$, of length ${}{\frac {1}{n}}$. The step integral for the corresponding lower step function is

${}\sum _{i=0}^{n-1}{\frac {1}{n}}{\left(i{\frac {1}{n}}\right)}^{2}={\frac {1}{n^{3}}}\sum _{i=0}^{n-1}i^{2}={\frac {1}{n^{3}}}{\left({\frac {1}{3}}n^{3}-{\frac {1}{2}}n^{2}+{\frac {1}{6}}n\right)}={\frac {1}{3}}-{\frac {1}{2n}}+{\frac {1}{6n^{2}}}\,$

(see Exercise 2.10 for the formula for the sum of the squares). Since the sequences ${}{\left(1/2n\right)}_{n\in \mathbb {N} }$ and ${}{\left(1/6n^{2}\right)}_{n\in \mathbb {N} }$ converge to ${}0$, the limit for ${}n\rightarrow \infty$ of these step integrals equals ${}{\frac {1}{3}}$. The step integral for the corresponding step function from above is

${}\sum _{i=0}^{n-1}{\frac {1}{n}}{\left((i+1){\frac {1}{n}}\right)}^{2}={\frac {1}{n^{3}}}\sum _{i=0}^{n-1}(i+1)^{2}={\frac {1}{n^{3}}}\sum _{j=1}^{n}j^{2}={\frac {1}{n^{3}}}{\left({\frac {1}{3}}n^{3}+{\frac {1}{2}}n^{2}+{\frac {1}{6}}n\right)}={\frac {1}{3}}+{\frac {1}{2n}}+{\frac {1}{6n^{2}}}\,.$

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

${}\int _{0}^{1}t^{2}\,dt={\frac {1}{3}}\,.$

Lemma

Let ${}I=[a,b]$ be a compact interval, and let

$f\colon I\longrightarrow \mathbb {R}$

be a function. Then the following statements are equivalent.

1. The function ${}f$ is Riemann-integrable.
2. There exists a partition ${}a=a_{0}<a_{1}<\cdots <a_{n}=b$, such that the restrictions ${}f_{i}:=f|_{[a_{i-1},a_{i}]}$ are Riemann-integrable.
3. For every partition ${}a=a_{0}<a_{1}<\cdots <a_{n}=b$, the restrictions ${}f_{i}:=f|_{[a_{i-1},a_{i}]}$ are Riemann-integrable.

In this situation, the equation

${}\int _{a}^{b}f(t)\,dt=\sum _{i=1}^{n}\int _{a_{i-1}}^{a_{i}}f_{i}(t)\,dt\,$

holds.

Proof

See Exercise 18.14 .

$\Box$

Definition

Let ${}f\colon I\rightarrow \mathbb {R}$ be a function on a real interval. Then ${}f$ is called Riemann-integrable, if the restriction of ${}f$ to every compact interval ${}[a,b]\subseteq I$ is

Riemann-integrable.

Due to this lemma, both definitions coincide for a compact interval ${}[a,b]$. The integrability of a function ${}f\colon \mathbb {R} \rightarrow \mathbb {R}$ does not mean that ${}\int _{\mathbb {R} }f(x)dx$ has a meaning or exists.

Riemann-integrability of continuous functions

Theorem

Let ${}f\colon I\rightarrow \mathbb {R}$ denote a continuous function. Then ${}f$ is Riemann-integrable.

Proof

This proof was not presented in the lecture.

$\Box$

Lemma

Let ${}I=[a,b]$ denote a compact interval, and let ${}f,g\colon I\rightarrow \mathbb {R}$ denote Riemann-integrable

functions. Then the following statements hold.

1. If ${}m\leq f(x)\leq M$ holds for all ${}x\in I$, then ${}m(b-a)\leq \int _{a}^{b}f(t)\,dt\leq M(b-a)$ holds.
2. If ${}f(x)\leq g(x)$ holds for all ${}x\in I$, then ${}\int _{a}^{b}f(t)\,dt\leq \int _{a}^{b}g(t)\,dt$ holds.
3. The sum ${}f+g$ is Riemann-integrable, and the identity ${}\int _{a}^{b}(f+g)(t)\,dt=\int _{a}^{b}f(t)\,dt+\int _{a}^{b}g(t)\,dt$ holds.
4. For ${}c\in \mathbb {R}$ we have ${}\int _{a}^{b}(cf)(t)\,dt=c\int _{a}^{b}f(t)\,dt$.
5. The functions ${}{\max {\left(f,g\right)}}$ and ${}{\min {\left(f,g\right)}}$ are Riemann-integrable.
6. The function ${}\vert {f}\vert$ is Riemann-integrable.
7. The product ${}fg$ is Riemann-integrable.

Proof

For (1) to (4) see Exercise 18.14 . For (5) see Exercise 18.17 . (6) follows directly from (5), because of ${}\vert {f}\vert ={\max {\left(f,-f,\right)}}$. For (7), see Exercise 18.18 .

$\Box$

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