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

Mean value theorem for integrals

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

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

as the mean height of the function, since this value, multiplied with the length ${\displaystyle {}b-a}$ of the interval, yields the area below the graph of ${\displaystyle {}f}$. 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

Suppose that ${\displaystyle {}[a,b]}$ is a compact interval, and let

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

be a continuous function. Then there exists some ${\displaystyle {}c\in [a,b]}$ such that

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

### Proof

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

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

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

${\displaystyle \Box }$

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 ${\displaystyle {}a and an integrable function ${\displaystyle {}f\colon [a,b]\rightarrow \mathbb {R} }$, we define

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

## Definition

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

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

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

${\displaystyle I\longrightarrow \mathbb {R} ,x\longmapsto \int _{a}^{x}f(t)\,dt,}$
is called the integral function for ${\displaystyle {}f}$ for the starting point ${\displaystyle {}a}$.

This function is also called the indefinite integral.

The following statement is called Fundamental theorem of calculus.

## Theorem

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

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

denote a continuous function. Let ${\displaystyle {}a\in I}$, and let

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

denote the corresponding integral function. Then ${\displaystyle {}F}$ is differentiable, and the identity

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

holds for all ${\displaystyle {}x\in I}$.

### Proof

Let ${\displaystyle {}x}$ be fixed. The difference quotient is

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

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

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

and therefore

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

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

${\displaystyle \Box }$

Primitive functions

## Definition

Let ${\displaystyle {}I\subseteq \mathbb {R} }$ denote an interval, and let

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

denote a function. A function

${\displaystyle F\colon I\longrightarrow \mathbb {R} }$

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

${\displaystyle {}x\in I}$.

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

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

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

denote a continuous function. Then ${\displaystyle {}f}$ has a primitive function.

### Proof

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

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

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

${\displaystyle \Box }$

## Lemma

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

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

denote a function. Suppose that ${\displaystyle {}F}$ and ${\displaystyle {}G}$ are primitive functions of ${\displaystyle {}f}$. Then ${\displaystyle {}F-G}$ is a constant function.

### Proof

We have

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

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

${\displaystyle \Box }$

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

## Corollary

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

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

denote a continuous function. Suppose that ${\displaystyle {}F}$ is a primitive function for ${\displaystyle {}f}$. Then for ${\displaystyle {}a,b\in I}$, the identity

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

holds.

### Proof

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

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

we have the relation

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

Because of Theorem 19.3 , the function ${\displaystyle {}G}$ is differentiable and

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

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

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

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

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

Here ${\displaystyle {}c}$ 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 ${\displaystyle {}I}$ denote a real interval, and

${\displaystyle F\colon I\longrightarrow \mathbb {R} }$

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

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

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

## Example

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

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 ${\displaystyle {}r}$ between the masses, equals

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

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

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

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

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

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

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

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

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

{\displaystyle {}{\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

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

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

## Example

We consider the function

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

given by

$cases}"): {\displaystyle {{}} f(t) := \begin{cases} 0 \text{ for } t [[Category:Wikiversity soft redirects|Mathematics for Applied Sciences (Osnabrück 2023-2024)/Part I/Lecture 19]] __NOINDEX__ 0, \\ \frac{1}{t} \sin \frac{1}{t^2} \text{ for } t \neq 0 \, .\end{cases} \,$

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

$cases}"): {\displaystyle {{}} H(t) := \begin{cases} 0 \text{ for } t [[Category:Wikiversity soft redirects|Mathematics for Applied Sciences (Osnabrück 2023-2024)/Part I/Lecture 19]] __NOINDEX__ 0, \\ \frac{ t^2}{2} \cos \frac{1}{t^2} \text{ for } t \neq 0 \, .\end{cases} \,$

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

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

For ${\displaystyle {}t=0}$, the difference quotient is

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

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

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

Primitive functions for power series

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

## Lemma

Let $\displaystyle {{}} f = \sum_{n [[Category:Wikiversity soft redirects|Mathematics for Applied Sciences (Osnabrück 2023-2024)/Part I/Lecture 19]] __NOINDEX__ 0}^\infty a_n x^n$ denote a power series which converges on ${\displaystyle {}]-r,r[}$. Then the power series

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

converges also on ${\displaystyle {}]-r,r[}$, and represents a primitive function for ${\displaystyle {}f}$.

### Proof

This proof was not presented in the lecture.
${\displaystyle \Box }$

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

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

$\displaystyle {{}} \frac{ 1 }{ x } = \sum_{k [[Category:Wikiversity soft redirects|Mathematics for Applied Sciences (Osnabrück 2023-2024)/Part I/Lecture 19]] __NOINDEX__ 0}^\infty (-1)^k (x-1)^k \, ,$

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

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

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

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

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