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

Mean value theorem for integrals

For a Riemann-integrable function $f:[a,b]\to ℝ$, 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

$f:[a,b]⟶ℝ$

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, $m\le f(x)\le M$ for all $x\in [a,b]$, and so

$m(b-a)\le {\int }_a^{b}f(t)dt\le M(b-a).$

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 $f(c)=d$.

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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 $f:[a,b]\to ℝ$, we define

${\int }_b^{a}f(t)dt:=-{\int }_a^{b}f(t)dt.$

Definition

Let $I$ denote a real interval, let

$f:I⟶ℝ$

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

$I⟶ℝ,x⟼{\int }_a^{x}f(t)dt,$

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

$f:I⟶ℝ$

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

$F^{\prime }(x)=f(x)$

holds for all

$x\in I$.

Proof

Let $x$ be fixed. The difference quotient is

$\frac{F(x+h)-F(x)}{h}=\frac{1}{h}({\int }_a^{x+h}f(t)dt-{\int }_a^{x}f(t)dt)=\frac{1}{h}{\int }_x^{x+h}f(t)dt.$

We have to show that for $h\to 0$, 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 $h\to 0$, $c_h$ converges to $x$, and because of the continuity of $f$, also $f(c_h)$ converges to $f(x)$.

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Primitive functions

Definition

Let $I\subseteq ℝ$ denote an interval, and let

$f:I⟶ℝ$

denote a function. A function

$F:I⟶ℝ$

is called a primitive function for $f$, if $F$ is differentiable on $I$ and if $F^{\prime }(x)=f(x)$ 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

$f:I⟶ℝ$

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 $F^{\prime }(x)=f(x)$ holds. This means that $F$ is a primitive function for $f$.

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Lemma

Let $I$ denote a real interval, and let

$f:I⟶ℝ$

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

constant function.

Proof

We have

$(F-G{)}^{\prime }=F^{\prime }-G^{\prime }=f-f=0.$

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

$◻$

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

$f:I⟶ℝ$

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

$G^{\prime }(x)=f(x)$

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

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

$◻$

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

$F:I⟶ℝ$

a primitive function for a function $f:I\to ℝ$. 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 $x\in {ℝ}_{+}$ and $a\in ℝ$, $a\ne -1$, 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 $R_1\ge R_0$?

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

$E={\int }_{R_0}^{R_1}\gamma \frac{Mm}{r^2}dr=\gamma Mm{\int }_{R_0}^{R_1}\frac{1}{r^2}dr=\gamma Mm(-\frac{1}{r}{|}_{R_0}^{R_1})=\gamma Mm(\frac{1}{R_0}-\frac{1}{R_1}).$

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

$\begin{array}{cc}{(\frac{1}{2}\cdot \ln \frac{1+x}{1-x})}^{\prime }& =\frac{1}{2}\cdot \frac{1-x}{1+x}\cdot \frac{(1-x)+(1+x)}{(1-x{)}^2}\\ & =\frac{1}{2}\cdot \frac{2}{(1+x)(1-x)}\\ & =\frac{1}{(1-x^2)}.\end{array}$

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

$f:ℝ⟶ℝ,t⟼f(t),$

given by

$f(t):=\{\begin{array}{c}0\text{ for }t=0,\\ \frac{1}{t}\sin \frac{1}{t^2}\text{ for }t\ne 0.\end{array}$

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

$H(t):=\{\begin{array}{c}0\text{ for }t=0,\\ \frac{t^2}{2}\cos \frac{1}{t^2}\text{ for }t\ne 0.\end{array}$

This function is differentiable. For $t\ne 0$, the derivative is

$H^{\prime }(t)=t\cos \frac{1}{t^2}+\frac{1}{t}\sin \frac{1}{t^2}.$

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 $H^{\prime }$ 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 $t\ne 0$ from the explicit derivative and for $t=0$ from

$H^{\prime }(0)-G^{\prime }(0)=0-0=0.$

Primitive functions for power series

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

Lemma

Let $f=\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n$ denote a power series which converges on $]-r,r[$. Then the power series

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

converges also on $]-r,r[$, and represents a primitive function

for $f$.

Proof

This proof was not presented in the lecture.

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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

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

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

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

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

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

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