Average theorem of integral calculus

The mean theorem of integral calculus (also called Cauchy's mean theorem ) is an important theorem in analysis . It allows integrals to be estimated without calculating the actual value and provides a simple proof of the fundamental theorem of analysis .

statement

For the geometric interpretation of the mean value theorem for . ${\displaystyle g=1}$

The Riemann integral is considered here. The statement reads:

Let be a continuous function as well as integrable and either or (i.e. without a change of sign ). Then one exists such that ${\displaystyle f\colon [a,b]\to \mathbb {R} }$ ${\displaystyle g\colon [a,b]\to \mathbb {R} }$ ${\displaystyle g\geq 0}$ ${\displaystyle g\leq 0}$ ${\displaystyle \xi \in [a,b]}$

${\displaystyle \int \limits _{a}^{b}{f(x)g(x)dx}=f(\xi )\int \limits _{a}^{b}{g(x)dx}}$

applies. Some authors refer to the above statement as the extended mean theorem and the statement for as the mean theorem or first mean theorem . For you get the important special case: ${\displaystyle g=1}$ ${\displaystyle g=1}$

${\displaystyle \int \limits _{a}^{b}{f(x)dx}=f(\xi )(b-a)}$,

which can be easily interpreted geometrically: The area under the curve between and is equal to the content of a rectangle of medium height . ${\displaystyle a}$ ${\displaystyle b}$

proof

Be on the interval . The other case can be traced back to this by going to. ${\displaystyle g(x)\geq 0}$ ${\displaystyle [a,b]}$ ${\displaystyle -g}$

Because of continuity, in takes on a minimum and a maximum according to the principle of minimum and maximum . With and is ${\displaystyle f}$ ${\displaystyle [a,b]}$ ${\displaystyle k}$ ${\displaystyle K}$ ${\displaystyle k\leq f(x)\leq K}$ ${\displaystyle g(x)\geq 0}$

${\displaystyle kg(x)\leq f(x)g(x)\leq Kg(x)}$;

with the monotony and linearity of the Riemann integral

${\displaystyle k\int \limits _{a}^{b}{g(x)\,{\rm {d}}x}\leq \int \limits _{a}^{b}{f(x)g(x)\,{\rm {d}}x}\leq K\int \limits _{a}^{b}{g(x)\,{\rm {d}}x}}$.

With therefore applies ${\displaystyle I:=\int \limits _{a}^{b}{g(x)\,{\rm {d}}x}}$

${\displaystyle kI\leq \int \limits _{a}^{b}{f(x)g(x)\,{\rm {d}}x}\leq KI}$ (1) .

A distinction must now be made between the following cases:

Case I: ${\displaystyle I\neq 0}$ . - Then the assertion has the equivalent form

${\displaystyle f(\xi )={\frac {1}{I}}\cdot \int \limits _{a}^{b}{f(x)g(x)\,{\rm {d}}x}}$;

the right-hand side of this equation is a number, and it has to be shown that one takes this number as its value (2) . ${\displaystyle f}$ ${\displaystyle \xi \in [a,b]}$

Because is , and (1) has after division by the form ${\displaystyle g(x)\geq 0}$ ${\displaystyle I>0}$ ${\displaystyle I}$

${\displaystyle k\leq {\frac {1}{I}}\cdot \int \limits _{a}^{b}{f(x)g(x)\,{\rm {d}}x}\leq K}$;

from this follows (2) with the intermediate value theorem for continuous functions, q. e. d.

Case II: ${\displaystyle I=0}$ . - Then it follows from (1):

${\displaystyle \int \limits _{a}^{b}{f(x)g(x)\,{\rm {d}}x}=0}$,

and the assertion takes the form that is valid for everything ${\displaystyle \xi \in [a,b]}$

${\displaystyle f(\xi )\cdot 0=0}$, q. e. d.

Condition on g

The condition that or applies is important. In fact, for functions without this condition , the mean theorem generally does not hold , as the following example shows: For and is ${\displaystyle g\geq 0}$ ${\displaystyle -g\geq 0}$ ${\displaystyle g}$ ${\displaystyle [a,b]=[-1,1]}$ ${\displaystyle f(x)=g(x)=x}$

${\displaystyle \int \limits _{a}^{b}f(x)\cdot g(x)\mathrm {d} x={\tfrac {2}{3}}}$,

however

${\displaystyle f(\xi )\cdot \int \limits _{a}^{b}g(x)\mathrm {d} x=0}$for everyone . ${\displaystyle \xi \in [a,b]}$

Second mean value theorem of integral calculus

Be functions, monotonous and continuous. Then one exists such that ${\displaystyle f,g\colon [a,b]\to \mathbb {R} }$ ${\displaystyle f}$ ${\displaystyle g}$ ${\displaystyle \xi \in [a,b]}$

${\displaystyle \int \limits _{a}^{b}{f(x)g(x)dx}=f(a)\int \limits _{a}^{\xi }{g(x)dx}+f(b)\int \limits _{\xi }^{b}{g(x)dx}}$.

In the case that it is even continuously differentiable, one can choose. The proof requires partial integration , the fundamental theorem of analysis, and the above theorem. ${\displaystyle f}$ ${\displaystyle \xi \in (a,b)}$

See also

• Integral calculus #means of continuous functions
• Mean #mean of a function
• Mean value theorem of differential calculus

Web links

Wikibooks: Math for non-freaks: mean theorem for integrals  - learning and teaching materials

literature

• Otto Forster : Analysis 1. Differential and integral calculus of a variable. 7th edition. Vieweg, Braunschweig 2004, ISBN 3-528-67224-2 .
• Harro Heuser: Textbook of Analysis . Part 1. 8th edition. BG Teubner, Stuttgart 1990, ISBN 3-519-12231-6 .