First mean theorem

The first mean value theorem is one of the theorems on a definite integral .

Wording

Let the function $f(x)$ ${\ displaystyle f (x)}$ integrable on the segment $[a;b]$ ${\ displaystyle [a; b]}$ , and is limited by numbers on it $m$ ${\ displaystyle m}$ and $M$ ${\ displaystyle M}$ so that $m\leq f(x)\leq M$ ${\ displaystyle m \ leq f (x) \ leq M}$ . Then there is such a number $\mu$ ${\ displaystyle \ mu}$ , $m\leq \mu \leq M$ ${\ displaystyle m \ leq \ mu \ leq M}$ , what

\int \limits _{a}^{b}f(x)dx=\mu (b-a)

{\ displaystyle \ int \ limits _ {a} ^ {b} f (x) dx = \ mu (ba)}

.

Proof

From inequality $m\leq f(x)\leq M$ ${\ displaystyle m \ leq f (x) \ leq M}$ by the property of the monotonicity of the integral, we have

m(b-a)\leq \int \limits _{a}^{b}f(x)dx\leq M(b-a)

{\ displaystyle m (ba) \ leq \ int \ limits _ {a} ^ {b} f (x) dx \ leq M (ba)}

.

Marking $\mu ={\frac {1}{b-a}}\int _{a}^{b}f(x)dx$ ${\ displaystyle \ mu = {\ frac {1} {ba}} \ int _ {a} ^ {b} f (x) dx}$ , we obtain the required statement. So certain number $\mu$ ${\ displaystyle \ mu}$ called the average value of the function $f(x)$ ${\ displaystyle f (x)}$ on the segment $[a;b]$ ${\ displaystyle [a; b]}$ , whence the name of the theorem.

Note

If the function $f(x)$ ${\ displaystyle f (x)}$ continuous on $[a;b]$ ${\ displaystyle [a; b]}$ then as $m$ ${\ displaystyle m}$ and $M$ ${\ displaystyle M}$ we can take its largest and smallest values (which, by the Weierstrass theorem , are achieved), then by the well-known theorem there exists such a point $c\in [a;b]$ ${\ displaystyle c \ in [a; b]}$ , what $f(c)=\mu$ ${\ displaystyle f (c) = \ mu}$ , therefore, the statement of the theorem can be rewritten in the form

\int \limits _{a}^{b}f(x)dx=f(c)(b-a)

{\ displaystyle \ int \ limits _ {a} ^ {b} f (x) dx = f (c) (ba)}

.

If we use the Newton-Leibniz formula , then this equality can be written as

F(b)-F(a)=F'(c)\;(b-a)

{\ displaystyle F (b) -F (a) = F '(c) \; (ba)}

,

Where $F(x)$ ${\ displaystyle F (x)}$ - antiderivative function $f(x)$ ${\ displaystyle f (x)}$ , which is nothing but the Lagrange formula for the function $F(x)$ ${\ displaystyle F (x)}$ .

Summary

Let functions $f(x)$ ${\ displaystyle f (x)}$ and $g(x)$ ${\ displaystyle g (x)}$ integrable on the segment $[a;b]$ ${\ displaystyle [a; b]}$ , and still $m\leq f(x)\leq M$ ${\ displaystyle m \ leq f (x) \ leq M}$ , and the second of them does not change sign (that is, either everywhere is non-negative: $g(x)\geq 0$ ${\ displaystyle g (x) \ geq 0}$ or everywhere positive $g(x)\leq 0$ ${\ displaystyle g (x) \ leq 0}$ ) Then there is such a number $\mu$ ${\ displaystyle \ mu}$ , $m\leq \mu \leq M$ ${\ displaystyle m \ leq \ mu \ leq M}$ , what

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

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

.

Proof

Let be $g(x)$ ${\ displaystyle g (x)}$ non-negative, then we have

mg(x)\leq f(x)g(x)\leq Mg(x)

{\ displaystyle mg (x) \ leq f (x) g (x) \ leq Mg (x)}

,

whence, in view of the monotonicity of the integral

m\int \limits _{a}^{b}g(x)dx\leq \int \limits _{a}^{b}f(x)g(x)dx\leq M\int \limits _{a}^{b}g(x)dx

{\ displaystyle m \ int \ limits _ {a} ^ {b} g (x) dx \ leq \ int \ limits _ {a} ^ {b} f (x) g (x) dx \ leq M \ int \ limits _ {a} ^ {b} g (x) dx}

.

If $\int _{a}^{b}g(x)dx=0$ ${\ displaystyle \ int _ {a} ^ {b} g (x) dx = 0}$ , then from this inequality it follows that $\int _{a}^{b}f(x)g(x)dx=0$ ${\ displaystyle \ int _ {a} ^ {b} f (x) g (x) dx = 0}$ , and the statement of the theorem holds for any $\mu$ ${\ displaystyle \ mu}$ . Otherwise, put

\mu ={\frac {\int \limits _{a}^{b}f(x)g(x)dx}{\int \limits _{a}^{b}g(x)dx}}

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

.

The generalization is proved. If the function $f(x)$ ${\ displaystyle f (x)}$ continuous, it can be argued that there is a point $c\in [a;b]$ ${\ displaystyle c \ in [a; b]}$ such that

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

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

(similar to the previous one).

Literature

Fichtenholtz G. M. The course of differential and integral calculus. - M .: Nauka, 1969 .-- T. II.
Zorich V.A. Mathematical analysis. Part I. - M .: Nauka, 1981.

First mean theorem

Wording

Proof

Note

Summary

Proof

Literature

More articles: