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

Mean value theorem for integrals

For a Riemann-integrable 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

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

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

be a continuous function. 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, the function ${}f$ is bounded from above and from below, let ${}m$ and ${}M$ denote the minimum and the maximum 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, 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 ${}x$ from the theorem is ${}x_{0}$ in the animation, and ${}x+h$ in the theorem is the moving ${}x$ in the animation. The moving point ${}z$ in the animation is a point which exists by the mean value theorem of definite integrals, applied to ${}x_{0}$ and ${}x$.

The following statement is called Fundamental theorem of calculus.

Theorem

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

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

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

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

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

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

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

Proof

Let ${}x$ be fixed. The difference quotient 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 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, and let

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

denote a function. A function

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

is called a primitive function for ${}f$, if ${}F$ is differentiable 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

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

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

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

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 ${}f$.

$\Box$

Lemma

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

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

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

Proof

We have

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

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

$\Box$

Isaac Newton (1643-1727)

Gottfried Wilhelm Leibniz (1646-1716)

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

Corollary

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

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

denote a continuous function. Suppose that ${}F$ is a primitive function 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 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 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, and

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

a 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

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 .

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

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, because it it neither bounded from above nor from below. Hence, there exist no upper step functions for ${}f$. However, ${}f$ still has a 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. 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 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 exists and equals ${}0$, so that ${}H$ is differentiable everywhere (but not continuously differentiable). The first summand in ${}H'$ is continuous, 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.

Lemma

Let ${}f=\sum _{n=0}^{\infty }a_{n}x^{n}$ denote a power series which converges 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 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

We would like to determine the Taylor series of the natural logarithm in the point ${}1$. The derivative 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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