Mathematics for Applied Sciences (Osnabrück 2023-2024)/Part I/Lecture 19/refcontrol

Mean value theorem for integrals

For a Riemann-integrable MDLD/Riemann-integrable function MDLD/function ${}f\colon [a,b]\rightarrow \mathbb {R}$ , one may consider

{\frac {\int _{a}^{b}f(t)\,dt}{b-a}}

as the mean height of the function, since this value, multiplied with the length ${}b-a$ of the interval, yields the area below the graph of ${}f$ . The Mean value theorem for definite integrals claims that, for a continuous function, this mean value is in fact obtained by the function somewhere.

Theorem Create referencenumber

Suppose that ${}[a,b]$ is a compact interval,MDLD/compact interval (R) and let

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

be a continuous function.MDLD/continuous function (R) Then there exists some ${}c\in [a,b]$ such that

{}\int _{a}^{b}f(t)\,dt=f(c)(b-a)\,.

Proof

On the compact interval,MDLD/compact interval (R) the function ${}f$ is bounded from above and from below, let ${}m$ and ${}M$ denote the minimumMDLD/minimum (global R) and the maximumMDLD/maximum (global R) of the function. Due to Theorem 11.13 , they are both obtained. Then, in particular, ${}m\leq f(x)\leq M$ for all ${}x\in [a,b]$ , and so

{}m(b-a)\leq \int _{a}^{b}f(t)\,dt\leq M(b-a)\,.

Therefore, ${}\int _{a}^{b}f(t)\,dt=d(b-a)$ with some ${}d\in [m,M]$ . Due to the Intermediate value theorem, there exists a ${}c\in [a,b]$ such that ${}f(c)=d$ .

\Box

The Fundamental theorem of calculus

It is useful to allow bounds for an integral, where the lower bound is larger than the upper bound. For ${}a<b$ and an integrable function ${}f\colon [a,b]\rightarrow \mathbb {R}$ , we define

{}\int _{b}^{a}f(t)\,dt:=-\int _{a}^{b}f(t)\,dt\,.

Definition

Let ${}I$ denote a real interval, let

f\colon I\longrightarrow \mathbb {R}

denote a Riemann-integrable function,MDLD/Riemann-integrable function and let ${}a\in I$ . Then the function

I\longrightarrow \mathbb {R} ,x\longmapsto \int _{a}^{x}f(t)\,dt,

is called the integral function for

{}f

for the starting point

{}a

.

This function is also called the indefinite integral.

The following statement is called Fundamental theorem of calculus.

TheoremTheorem 19.3 change

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

f\colon I\longrightarrow \mathbb {R}

denote a continuous function.MDLD/continuous function (R) Let ${}a\in I$ , and let

{}F(x):=\int _{a}^{x}f(t)\,dt\,

denote the corresponding integral function.MDLD/integral function Then ${}F$ is differentiable,MDLD/differentiable (R) and the identity

{}F'(x)=f(x)\,

holds for all

{}x\in I

.

Proof

Let ${}x$ be fixed. The difference quotient MDLD/difference quotient (R) is

{}{\frac {F(x+h)-F(x)}{h}}={\frac {1}{h}}{\left(\int _{a}^{x+h}f(t)\,dt-\int _{a}^{x}f(t)\,dt\right)}={\frac {1}{h}}\int _{x}^{x+h}f(t)\,dt\,.

We have to show that for ${}h\rightarrow 0$ , the limit MDLD/limit (function R) exists and equals ${}f(x)$ . Because of the Mean value theorem for definite integrals, for every ${}h$ , there exists a ${}c_{h}\in [x,x+h]$ with

{}f(c_{h})\cdot h=\int _{x}^{x+h}f(t)dt\,,

and therefore

{}f(c_{h})={\frac {\int _{x}^{x+h}f(t)dt}{h}}\,.

For ${}h\rightarrow 0$ , ${}c_{h}$ converges to ${}x$ , and because of the continuity of ${}f$ , also ${}f(c_{h})$ converges to ${}f(x)$ .

\Box

Primitive functions

Definition

Let ${}I\subseteq \mathbb {R}$ denote an interval,MDLD/interval (R) and let

f\colon I\longrightarrow \mathbb {R}

denote a function.MDLD/function A function

F\colon I\longrightarrow \mathbb {R}

is called a primitive function for ${}f$ , if ${}F$ is differentiable MDLD/differentiable (R) on ${}I$ and if ${}F'(x)=f(x)$ holds for all

{}x\in I

.

A primitive function is also called an antiderivative. The fundamental theorem of calculus might be rephrased, in connection with Theorem 18.17 , as an existence theorem for primitive functions.

Corollary Create referencenumber

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

f\colon I\longrightarrow \mathbb {R}

denote a continuous function.MDLD/continuous function (R) Then ${}f$ has a

primitive function.MDLD/primitive function (R)

Proof

Let ${}a\in I$ be an arbitrary point. Due to Theorem 18.17 , there exists the function

{}F(x)=\int _{a}^{x}f(t)\,dt\,,

and because of the Fundamental theorem, the identity ${}F'(x)=f(x)$ holds. This means that ${}F$ is a primitive function for ${}f$ .

\Box

LemmaLemma 19.6 change

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

f\colon I\longrightarrow \mathbb {R}

denote a function.MDLD/function Suppose that ${}F$ and ${}G$ are primitive functionsMDLD/primitive functions of ${}f$ . Then ${}F-G$ is a

constant function.MDLD/constant function

Proof

We have

{}(F-G)'=F'-G'=f-f=0\,.

Therefore, due to Corollary 15.6 , the difference ${}F-G$ is constant.

\Box

The following statement is also a version of the fundamental theorem, it is called the Newton-Leibniz-formula.

Corollary Create referencenumber

Let ${}I$ denote a real interval,MDLD/real interval and let

f\colon I\longrightarrow \mathbb {R}

denote a continuous function.MDLD/continuous function (R) Suppose that ${}F$ is a primitive function MDLD/primitive function (R) for ${}f$ . Then for ${}a,b\in I$ , the identity

{}\int _{a}^{b}f(t)\,dt=F(b)-F(a)\,

holds.

Proof

Due to Theorem 18.17 , the integral exists. With the integral functionMDLD/integral function

{}G(x):=\int _{a}^{x}f(t)\,dt\,,

we have the relation

{}\int _{a}^{b}f(t)\,dt=G(b)=G(b)-G(a)\,.

Because of Theorem 19.3 , the function ${}G$ is differentiable MDLD/differentiable (R) and

{}G'(x)=f(x)\,

holds. Hence ${}G$ is a primitive function for ${}f$ . Due to Lemma 19.6 , we have ${}F(x)=G(x)+c$ . Therefore,

{}\int _{a}^{b}f(t)\,dt=G(b)-G(a)=F(b)-c-F(a)+c=F(b)-F(a)\,.

\Box

Since a primitive function is only determined up to an additive constant, we sometimes write

{}\int _{}^{}f(t)\,dt=F+c\,.

Here ${}c$ is called a constant of integration. In certain situations, in particular in relation with differential equations, this constant is determined by further conditions.

Notation

Let ${}I$ denote a real interval,MDLD/real interval and

F\colon I\longrightarrow \mathbb {R}

a primitive function MDLD/primitive function for a function ${}f\colon I\rightarrow \mathbb {R}$ . Suppose that ${}a,b\in I$ . Then one sets

{}F|_{a}^{b}:=F(b)-F(a)=\int _{a}^{b}f(t)\,dt\,.

This notation is basically used for computations, in particular, when we want to determine definite integrals.

Using known results about the derivatives of differentiable functions, we obtain a list of primitive functions for some important functions. In general however, it is difficult to find a primitive function.

The primitive function of ${}x^{a}$ , where ${}x\in \mathbb {R} _{+}$ and ${}a\in \mathbb {R}$ , ${}a\neq -1$ , is ${}{\frac {1}{a+1}}x^{a+1}$ .

Example Create referencenumber

Suppose that the distance between two masses (thought of as mass points) ${}M$ and ${}m$ is ${}R_{0}$ . Because of gravitation, this system contains a certain potential energy. How is this potential energy changing, when we move these masses to a distance ${}R_{1}\geq R_{0}$ ?

The needed energy is force times path, where the force itself depends on the distance between the masses. Due to the gravitation law, the force, given the distance ${}r$ between the masses, equals

{}F(r)=\gamma {\frac {Mm}{r^{2}}}\,,

where ${}\gamma$ denotes the constant of gravitation. Therefore, the energy needed to increase the distance from ${}R_{0}$ to ${}R_{1}$ , equals

{}E=\int _{R_{0}}^{R_{1}}\gamma {\frac {Mm}{r^{2}}}\,dr=\gamma Mm\int _{R_{0}}^{R_{1}}{\frac {1}{r^{2}}}\,dr=\gamma Mm{\left(-{\frac {1}{r}}|_{R_{0}}^{R_{1}}\right)}=\gamma Mm{\left({\frac {1}{R_{0}}}-{\frac {1}{R_{1}}}\right)}\,.

Hence it is possible to assign a value to the difference between the potential energies for the two distances ${}R_{0}$ and ${}R_{1}$ , though it is not possible to assign an absolute value to the potential energy for a given distance.

The primitive function of the function ${}{\frac {1}{x}}$ is the natural logarithm.

The primitive function of the exponential function is the exponential function itself.

The primitive function of ${}\sin x$ is ${}-\cos x$ , the primitive function of ${}\cos x$ is ${}\sin x$ .

The primitive function of ${}{\frac {1}{1+x^{2}}}$ is ${}\arctan x$ , due to Theorem 16.20 (3).

The primitive function of ${}{\frac {1}{1-x^{2}}}$ (for ${}x\in {]{-1},1[}$ ) is ${}{\frac {1}{2}}\ln {\frac {1+x}{1-x}}$ , because we have

{}{\begin{aligned}{\left({\frac {1}{2}}\cdot \ln {\frac {1+x}{1-x}}\right)}^{\prime }&={\frac {1}{2}}\cdot {\frac {1-x}{1+x}}\cdot {\frac {(1-x)+(1+x)}{(1-x)^{2}}}\\&={\frac {1}{2}}\cdot {\frac {2}{(1+x)(1-x)}}\\&={\frac {1}{(1-x^{2})}}.\end{aligned}}

Caution! Integration rules are only applicable for functions, which are defined on the whole interval. In particular, the following is not true

{}\int _{-a}^{a}{\frac {dt}{t^{2}}}\,dt=-{\frac {1}{x}}|_{-a}^{a}=-{\frac {1}{a}}-{\frac {1}{a}}=-{\frac {2}{a}}\,,

since we integrate over a point where the function is not defined.

Example Create referencenumber

We consider the function

f\colon \mathbb {R} \longrightarrow \mathbb {R} ,t\longmapsto f(t),

given by

{}f(t):={\begin{cases}0{\text{ for }}t=0,\\{\frac {1}{t}}\sin {\frac {1}{t^{2}}}{\text{ for }}t\neq 0\,.\end{cases}}\,

This function is not Riemann-integrable,MDLD/Riemann-integrable because it it neither bounded from above nor from below. Hence, there exist no upper step functionsMDLD/upper step functions for ${}f$ . However, ${}f$ still has a primitive function.MDLD/primitive function To see this, we consider the function

{}H(t):={\begin{cases}0{\text{ for }}t=0,\\{\frac {t^{2}}{2}}\cos {\frac {1}{t^{2}}}{\text{ for }}t\neq 0\,.\end{cases}}\,

This function is differentiable.MDLD/differentiable For ${}t\neq 0$ , the derivative is

{}H'(t)=t\cos {\frac {1}{t^{2}}}+{\frac {1}{t}}\sin {\frac {1}{t^{2}}}\,.

For ${}t=0$ , the difference quotient MDLD/difference quotient (R) is

{}{\frac {{\frac {h^{2}}{2}}\cos {\frac {1}{h^{2}}}}{h}}={\frac {h}{2}}\cos {\frac {1}{h^{2}}}\,.

For ${}h\mapsto 0$ , the limit MDLD/limit (function R) exists and equals ${}0$ , so that ${}H$ is differentiable everywhere (but not continuously differentiable). The first summand in ${}H'$ is continuous,MDLD/continuous (R) and therefore, due to Theorem 18.17 , it has a primitive function ${}G$ . Hence ${}H-G$ is a primitive function for ${}f$ . This follows for ${}t\neq 0$ from the explicit derivative and for ${}t=0$ from

{}H'(0)-G'(0)=0-0=0\,.

Primitive functions for power series

We recall that the derivative of a convergent power series is obtained by derivating the summands.

LemmaLemma 19.11 change

Let ${}f=\sum _{n=0}^{\infty }a_{n}x^{n}$ denote a power series MDLD/power series (R) which convergesMDLD/converges (power series R) on ${}]-r,r[$ . Then the power series

\sum _{n=1}^{\infty }{\frac {a_{n-1}}{n}}x^{n}

converges also on ${}]-r,r[$ , and represents a primitive function MDLD/primitive function (R)

for

{}f

.

Proof

This proof was not presented in the lecture.

\Box

With the help of this statement, one can sometimes find the Taylor polynomial (or Taylor series) of a function by using the Taylor polynomial of the derivative. We give a typical example.

Example Create referencenumber

We would like to determine the Taylor series of the natural logarithm MDLD/natural logarithm in the point ${}1$ . The derivative MDLD/derivative (R) of the natural logarithm equals ${}1/x$ , due to Corollary 16.6 . This function has the power series expansion

{}{\frac {1}{x}}=\sum _{k=0}^{\infty }(-1)^{k}(x-1)^{k}\,,

due to Theorem 9.13 (which converges for ${}\vert {x-1}\vert <1$ ). Therefore, because of Lemma 19.11 , the power series expansion of the natural logarithm is

\sum _{k=1}^{\infty }{\frac {(-1)^{k-1}}{k}}(x-1)^{k}.

Setting ${}z=x-1$ , we may write this series as

z-{\frac {z^{2}}{2}}+{\frac {z^{3}}{3}}-{\frac {z^{4}}{4}}+{\frac {z^{5}}{5}}-\ldots .

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