Teorema del valor medio

On the existence of a tangent to an arc parallel to the line through its endpoints

Para conocer el teorema de la teoría de la función armónica, consulte Función armónica § La propiedad del valor medio .

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v t mi

Para cualquier función que sea continua en y diferenciable en , existe algo en el intervalo tal que la secante que une los puntos finales del intervalo es paralela a la tangente en .${\displaystyle [a,b]}$${\displaystyle (a,b)}$${\displaystyle c}$${\displaystyle (a,b)}$${\displaystyle [a,b]}$${\displaystyle c}$

En matemáticas , el teorema del valor medio establece, aproximadamente, que para un arco plano dado entre dos puntos finales, hay al menos un punto en el que la tangente al arco es paralela a la secante a través de sus puntos finales. Es uno de los resultados más importantes del análisis real . Este teorema se utiliza para probar afirmaciones sobre una función en un intervalo a partir de hipótesis locales sobre derivadas en puntos del intervalo.

Más precisamente, el teorema establece que si es una función continua en el intervalo cerrado y diferenciable en el intervalo abierto , entonces existe un punto en el que la tangente en es paralela a la recta secante que pasa por los puntos finales y , es decir,${\displaystyle f}$ ${\displaystyle [a,b]}$ ${\displaystyle (a,b)}$${\displaystyle c}$${\displaystyle (a,b)}$${\displaystyle c}$${\displaystyle (a,f(a))}$${\displaystyle (b,f(b))}$

${\displaystyle f'(c)={\frac {f(b)-f(a)}{b-a}}.}$

Historia

Un caso especial de este teorema fue descrito por primera vez por Parameshvara (1370-1460), de la Escuela de Astronomía y Matemáticas de Kerala en India , en sus comentarios sobre Govindasvāmi y Bhāskara II . ^[1] Michel Rolle demostró una forma restringida del teorema en 1691; el resultado fue lo que ahora se conoce como teorema de Rolle , y se demostró solo para polinomios, sin las técnicas de cálculo. El teorema del valor medio en su forma moderna fue establecido y probado por Augustin Louis Cauchy en 1823. ^[2] Desde entonces se han demostrado muchas variaciones de este teorema.^{[3] [4]}

Declaración formal

La función obtiene la pendiente de la secante entre y como derivada en el punto .${\displaystyle f}$${\displaystyle a}$${\displaystyle b}$${\displaystyle \xi \in (a,b)}$

También es posible que haya múltiples tangentes paralelas a la secante.

Sea una función continua en el intervalo cerrado y diferenciable en el intervalo abierto , donde . Entonces existe algo en tal que${\displaystyle f:[a,b]\to \mathbb {R} }$ ${\displaystyle [a,b]}$${\displaystyle (a,b)}$${\displaystyle a<b}$${\displaystyle c}$${\displaystyle (a,b)}$

${\displaystyle f'(c)={\frac {f(b)-f(a)}{b-a}}.}$

El teorema del valor medio es una generalización del teorema de Rolle , que asume que el lado derecho de arriba es cero.${\displaystyle f(a)=f(b)}$

El teorema del valor medio sigue siendo válido en un contexto un poco más general. Sólo hay que asumir que es continua en , y que por cada en el límite${\displaystyle f:[a,b]\to \mathbb {R} }$${\displaystyle [a,b]}$${\displaystyle x}$${\displaystyle (a,b)}$

${\displaystyle \lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}}$

existe como un número finito o igual a o . Si es finito, ese límite es igual . Un ejemplo en el que se aplica esta versión del teorema es el mapeo de la función de raíz cúbica de valor real , cuya derivada tiende a infinito en el origen.${\displaystyle \infty }$${\displaystyle -\infty }$${\displaystyle f'(x)}$${\displaystyle x\mapsto x^{1/3}}$

Tenga en cuenta que el teorema, como se indicó, es falso si una función diferenciable tiene un valor complejo en lugar de un valor real. Por ejemplo, defina para todo real . Luego${\displaystyle f(x)=e^{xi}}$${\displaystyle x}$

${\displaystyle f(2\pi )-f(0)=0=0(2\pi -0)}$

mientras que para cualquier real .${\displaystyle f'(x)\neq 0}$${\displaystyle x}$

These formal statements are also known as Lagrange's Mean Value Theorem.^[5]

Proof

The expression ${\textstyle {\frac {f(b)-f(a)}{b-a}}}$ gives the slope of the line joining the points ${\displaystyle (a,f(a))}$ and ${\displaystyle (b,f(b))}$, which is a chord of the graph of ${\displaystyle f}$, while ${\displaystyle f'(x)}$ gives the slope of the tangent to the curve at the point ${\displaystyle (x,f(x))}$. Thus the mean value theorem says that given any chord of a smooth curve, we can find a point on the curve lying between the end-points of the chord such that the tangent of the curve at that point is parallel to the chord. The following proof illustrates this idea.

Definir , donde es una constante. Dado que es continuo y diferenciable , lo mismo es válido para . Ahora queremos elegir de modo que satisfaga las condiciones del teorema de Rolle . A saber${\displaystyle g(x)=f(x)-rx}$${\displaystyle r}$${\displaystyle f}$${\displaystyle [a,b]}$${\displaystyle (a,b)}$${\displaystyle g}$${\displaystyle r}$${\displaystyle g}$

${\displaystyle {\begin{aligned}g(a)=g(b)&\iff f(a)-ra=f(b)-rb\\&\iff r(b-a)=f(b)-f(a)\\&\iff r={\frac {f(b)-f(a)}{b-a}}.\end{aligned}}}$

Según el teorema de Rolle , dado que es diferenciable y , hay algunos en los cuales , y de la igualdad se sigue que,${\displaystyle g}$${\displaystyle g(a)=g(b)}$${\displaystyle c}$${\displaystyle (a,b)}$${\displaystyle g'(c)=0}$${\displaystyle g(x)=f(x)-rx}$

${\displaystyle {\begin{aligned}&g'(x)=f'(x)-r\\&g'(c)=0\\&g'(c)=f'(c)-r=0\\&\Rightarrow f'(c)=r={\frac {f(b)-f(a)}{b-a}}\end{aligned}}}$

Implicaciones

Theorem 1: Assume that f is a continuous, real-valued function, defined on an arbitrary interval I of the real line. If the derivative of f at every interior point of the interval I exists and is zero, then f is constant in the interior.

Prueba: Suponga que la derivada de f en cada punto interior del intervalo I existe y es cero. Vamos ( un , b ) Ser un intervalo abierto arbitrario en I . Por el teorema del valor medio, existe un punto c en ( a , b ) tal que

${\displaystyle 0=f'(c)={\frac {f(b)-f(a)}{b-a}}.}$

Esto implica que f ( a ) = f ( b ) . Por tanto, f es constante en el interior de I y, por tanto, es constante en I por continuidad. (Consulte a continuación para obtener una versión multivariable de este resultado).

Observaciones:

• Only continuity of f, not differentiability, is needed at the endpoints of the interval I. No hypothesis of continuity needs to be stated if I is an open interval, since the existence of a derivative at a point implies the continuity at this point. (See the section continuity and differentiability of the article derivative.)
• The differentiability of f can be relaxed to one-sided differentiability, a proof given in the article on semi-differentiability.

Teorema 2: Si f ' ( x ) = g' ( x ) para todo x en un intervalo ( a , b ) del dominio de estas funciones, entonces f - g es constante o f = g + c donde c es una constante en ( a , b ).

Demostración: Sea F = f - g , luego F '= f' - g '= 0 en el intervalo ( a , b ), por lo que el teorema 1 anterior dice que F = f - g es una constante c o f = g + c .

Theorem 3: If F is an antiderivative of f on an interval I, then the most general antiderivative of f on I is F(x) + c where c is an constant.

Proof: It is directly derived from the above theorem 2.

Cauchy's mean value theorem

Cauchy's mean value theorem, also known as the extended mean value theorem,^[6] is a generalization of the mean value theorem. It states: if the functions ${\displaystyle f}$ and ${\displaystyle g}$ are both continuous on the closed interval ${\displaystyle [a,b]}$ and differentiable on the open interval ${\displaystyle (a,b)}$, then there exists some ${\displaystyle c\in (a,b)}$, such that^[5]

Geometrical meaning of Cauchy's theorem

${\displaystyle (f(b)-f(a))g'(c)=(g(b)-g(a))f'(c).}$

Of course, if ${\displaystyle g(a)\neq g(b)}$ and ${\displaystyle g'(c)\neq 0}$, this is equivalent to:

${\displaystyle {\frac {f'(c)}{g'(c)}}={\frac {f(b)-f(a)}{g(b)-g(a)}}.}$

Geometrically, this means that there is some tangent to the graph of the curve^[7]

${\displaystyle {\begin{cases}[a,b]\to \mathbb {R} ^{2}\\t\mapsto (f(t),g(t))\end{cases}}}$

which is parallel to the line defined by the points ${\displaystyle (f(a),g(a))}$ and ${\displaystyle (f(b),g(b))}$. However, Cauchy's theorem does not claim the existence of such a tangent in all cases where ${\displaystyle (f(a),g(a))}$ and ${\displaystyle (f(b),g(b))}$ are distinct points, since it might be satisfied only for some value ${\displaystyle c}$ with ${\displaystyle f'(c)=g'(c)=0}$, in other words a value for which the mentioned curve is stationary; in such points no tangent to the curve is likely to be defined at all. An example of this situation is the curve given by

${\displaystyle t\mapsto \left(t^{3},1-t^{2}\right),}$

which on the interval ${\displaystyle [-1,1]}$ goes from the point ${\displaystyle (-1,0)}$ to ${\displaystyle (1,0)}$, yet never has a horizontal tangent; however it has a stationary point (in fact a cusp) at ${\displaystyle t=0}$.

Cauchy's mean value theorem can be used to prove L'Hôpital's rule. The mean value theorem is the special case of Cauchy's mean value theorem when ${\displaystyle g(t)=t}$.

Proof of Cauchy's mean value theorem

The proof of Cauchy's mean value theorem is based on the same idea as the proof of the mean value theorem.

• Suppose ${\displaystyle g(a)\neq g(b)}$. Define ${\displaystyle h(x)=f(x)-rg(x)}$, where ${\displaystyle r}$ is fixed in such a way that ${\displaystyle h(a)=h(b)}$, namely

  ${\displaystyle {\begin{aligned}h(a)=h(b)&\iff f(a)-rg(a)=f(b)-rg(b)\\&\iff r(g(b)-g(a))=f(b)-f(a)\\&\iff r={\frac {f(b)-f(a)}{g(b)-g(a)}}.\end{aligned}}}$

  Since ${\displaystyle f}$ and ${\displaystyle g}$ are continuous on ${\displaystyle [a,b]}$ and differentiable on ${\displaystyle (a,b)}$, the same is true for ${\displaystyle h}$. All in all, ${\displaystyle h}$ satisfies the conditions of Rolle's theorem: consequently, there is some ${\displaystyle c}$ in ${\displaystyle (a,b)}$ for which ${\displaystyle h'(c)=0}$. Now using the definition of ${\displaystyle h}$ we have:

  ${\displaystyle 0=h'(c)=f'(c)-rg'(c)=f'(c)-\left({\frac {f(b)-f(a)}{g(b)-g(a)}}\right)g'(c).}$

  Therefore:

  ${\displaystyle f'(c)={\frac {f(b)-f(a)}{g(b)-g(a)}}g'(c),}$

  which implies the result.^[5]
• If ${\displaystyle g(a)=g(b)}$, then, applying Rolle's theorem to ${\displaystyle g}$, it follows that there exists ${\displaystyle c}$ in ${\displaystyle (a,b)}$ for which ${\displaystyle g'(c)=0}$. Using this choice of ${\displaystyle c}$, Cauchy's mean value theorem (trivially) holds.

Generalization for determinants

Assume that ${\displaystyle f,g,}$ and ${\displaystyle h}$ are differentiable functions on ${\displaystyle (a,b)}$ that are continuous on ${\displaystyle [a,b]}$. Define

${\displaystyle D(x)={\begin{vmatrix}f(x)&g(x)&h(x)\\f(a)&g(a)&h(a)\\f(b)&g(b)&h(b)\end{vmatrix}}}$

There exists ${\displaystyle c\in (a,b)}$ such that ${\displaystyle D'(c)=0}$.

Notice that

${\displaystyle D'(x)={\begin{vmatrix}f'(x)&g'(x)&h'(x)\\f(a)&g(a)&h(a)\\f(b)&g(b)&h(b)\end{vmatrix}}}$

and if we place ${\displaystyle h(x)=1}$, we get Cauchy's mean value theorem. If we place ${\displaystyle h(x)=1}$ and ${\displaystyle g(x)=x}$ we get Lagrange's mean value theorem.

The proof of the generalization is quite simple: each of ${\displaystyle D(a)}$ and ${\displaystyle D(b)}$ are determinants with two identical rows, hence ${\displaystyle D(a)=D(b)=0}$. The Rolle's theorem implies that there exists ${\displaystyle c\in (a,b)}$ such that ${\displaystyle D'(c)=0}$.

Mean value theorem in several variables

The mean value theorem generalizes to real functions of multiple variables. The trick is to use parametrization to create a real function of one variable, and then apply the one-variable theorem.

Let ${\displaystyle G}$ be an open convex subset of ${\displaystyle \mathbb {R} ^{n}}$, and let ${\displaystyle f:G\to \mathbb {R} }$ be a differentiable function. Fix points ${\displaystyle x,y\in G}$ , and define ${\displaystyle g(t)=f{\big (}(1-t)x+ty{\big )}}$. Since ${\displaystyle g}$ is a differentiable function in one variable, the mean value theorem gives:

${\displaystyle g(1)-g(0)=g'(c)}$

for some ${\displaystyle c}$ between 0 and 1. But since ${\displaystyle g(1)=f(y)}$ and ${\displaystyle g(0)=f(x)}$, computing ${\displaystyle g'(c)}$ explicitly we have:

${\displaystyle f(y)-f(x)=\nabla f{\big (}(1-c)x+cy{\big )}\cdot (y-x)}$

where ${\displaystyle \nabla }$ denotes a gradient and ${\displaystyle \cdot }$ a dot product. Note that this is an exact analog of the theorem in one variable (in the case ${\displaystyle n=1}$ this is the theorem in one variable). By the Cauchy–Schwarz inequality, the equation gives the estimate:

${\displaystyle {\Bigl |}f(y)-f(x){\Bigr |}\leq {\Bigl |}\nabla f{\big (}(1-c)x+cy{\big )}{\Bigr |}\ {\Bigl |}y-x{\Bigr |}.}$

In particular, when the partial derivatives of ${\displaystyle f}$ are bounded, ${\displaystyle f}$ is Lipschitz continuous (and therefore uniformly continuous).

As an application of the above, we prove that ${\displaystyle f}$ is constant if ${\displaystyle G}$ is open and connected and every partial derivative of ${\displaystyle f}$ is 0. Pick some point ${\displaystyle x_{0}\in G}$, and let ${\displaystyle g(x)=f(x)-f(x_{0})}$. We want to show ${\displaystyle g(x)=0}$ for every ${\displaystyle x\in G}$. For that, let ${\displaystyle E=\{x\in G:g(x)=0\}}$. Then E is closed and nonempty. It is open too: for every ${\displaystyle x\in E}$ ,

${\displaystyle {\Big |}g(y){\Big |}={\Big |}g(y)-g(x){\Big |}\leq (0){\Big |}y-x{\Big |}=0}$

for every ${\displaystyle y}$ in some neighborhood of ${\displaystyle x}$. (Here, it is crucial that ${\displaystyle x}$ and ${\displaystyle y}$ are sufficiently close to each other.) Since ${\displaystyle G}$ is connected, we conclude ${\displaystyle E=G}$.

The above arguments are made in a coordinate-free manner; hence, they generalize to the case when ${\displaystyle G}$ is a subset of a Banach space.

Mean value theorem for vector-valued functions

There is no exact analog of the mean value theorem for vector-valued functions.

In Principles of Mathematical Analysis, Rudin gives an inequality which can be applied to many of the same situations to which the mean value theorem is applicable in the one dimensional case:^[8]

Theorem. For a continuous vector-valued function ${\displaystyle \mathbf {f} :[a,b]\to \mathbb {R} ^{k}}$ differentiable on ${\displaystyle (a,b)}$, there exists ${\displaystyle x\in (a,b)}$ such that ${\displaystyle \left|\mathbf {f} '(x)\right|\geq {\frac {1}{b-a}}|\mathbf {f} (b)-\mathbf {f} (a)|}$.

Jean Dieudonné in his classic treatise Foundations of Modern Analysis discards the mean value theorem and replaces it by mean inequality as the proof is not constructive and one cannot find the mean value and in applications one only needs mean inequality. Serge Lang in Analysis I uses the mean value theorem, in integral form, as an instant reflex but this use requires the continuity of the derivative. If one uses the Henstock–Kurzweil integral one can have the mean value theorem in integral form without the additional assumption that derivative should be continuous as every derivative is Henstock–Kurzweil integrable. The problem is roughly speaking the following: If f : U → R^m is a differentiable function (where U ⊂ Rⁿ is open) and if x + th, x, h ∈ Rⁿ, t ∈ [0, 1] is the line segment in question (lying inside U), then one can apply the above parametrization procedure to each of the component functions f_i (i = 1, …, m) of f (in the above notation set y = x + h). In doing so one finds points x + t_ih on the line segment satisfying

${\displaystyle f_{i}(x+h)-f_{i}(x)=\nabla f_{i}(x+t_{i}h)\cdot h.}$

But generally there will not be a single point x + t*h on the line segment satisfying

${\displaystyle f_{i}(x+h)-f_{i}(x)=\nabla f_{i}(x+t^{*}h)\cdot h.}$

for all i simultaneously. For example, define:

${\displaystyle {\begin{cases}f:[0,2\pi ]\to \mathbb {R} ^{2}\\f(x)=(\cos(x),\sin(x))\end{cases}}}$

Then ${\displaystyle f(2\pi )-f(0)=\mathbf {0} \in \mathbb {R} ^{2}}$, but ${\displaystyle f_{1}'(x)=-\sin(x)}$ and ${\displaystyle f_{2}'(x)=\cos(x)}$ are never simultaneously zero as ${\displaystyle x}$ ranges over ${\displaystyle \left[0,2\pi \right]}$.

However a certain type of generalization of the mean value theorem to vector-valued functions is obtained as follows: Let f be a continuously differentiable real-valued function defined on an open interval I, and let x as well as x + h be points of I. The mean value theorem in one variable tells us that there exists some t* between 0 and 1 such that

${\displaystyle f(x+h)-f(x)=f'(x+t^{*}h)\cdot h.}$

On the other hand, we have, by the fundamental theorem of calculus followed by a change of variables,

${\displaystyle f(x+h)-f(x)=\int _{x}^{x+h}f'(u)\,du=\left(\int _{0}^{1}f'(x+th)\,dt\right)\cdot h.}$

Thus, the value f′(x + t*h) at the particular point t* has been replaced by the mean value

${\displaystyle \int _{0}^{1}f'(x+th)\,dt.}$

This last version can be generalized to vector valued functions:

Lemma 1. Let U ⊂ Rⁿ be open, f : U → R^m continuously differentiable, and x ∈ U, h ∈ Rⁿ vectors such that the line segment x + th, 0 ≤ t ≤ 1 remains in U. Then we have:

${\displaystyle f(x+h)-f(x)=\left(\int _{0}^{1}Df(x+th)\,dt\right)\cdot h,}$

where Df denotes the Jacobian matrix of f and the integral of a matrix is to be understood componentwise.

Proof. Let f₁, …, f_m denote the components of f and define:

${\displaystyle {\begin{cases}g_{i}:[0,1]\to \mathbb {R} \\g_{i}(t)=f_{i}(x+th)\end{cases}}}$

Then we have

${\displaystyle {\begin{aligned}&f_{i}(x+h)-f_{i}(x)=g_{i}(1)-g_{i}(0)=\int _{0}^{1}g_{i}'(t)\,dt\\={}&\int _{0}^{1}\left(\sum _{j=1}^{n}{\frac {\partial f_{i}}{\partial x_{j}}}(x+th)h_{j}\right)\,dt=\sum _{j=1}^{n}\left(\int _{0}^{1}{\frac {\partial f_{i}}{\partial x_{j}}}(x+th)\,dt\right)h_{j}.\end{aligned}}}$

The claim follows since Df is the matrix consisting of the components ${\displaystyle {\tfrac {\partial f_{i}}{\partial x_{j}}}}$.

Lemma 2. Let v : [a, b] → R^m be a continuous function defined on the interval [a, b] ⊂ R. Then we have

${\displaystyle \left\|\int _{a}^{b}v(t)\,dt\right\|\leq \int _{a}^{b}\|v(t)\|\,dt.}$

Proof. Let u in R^m denote the value of the integral

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

Now we have (using the Cauchy–Schwarz inequality):

${\displaystyle \|u\|^{2}=\langle u,u\rangle =\left\langle \int _{a}^{b}v(t)\,dt,u\right\rangle =\int _{a}^{b}\langle v(t),u\rangle \,dt\leqslant \int _{a}^{b}\|v(t)\|\cdot \|u\|\,dt=\|u\|\int _{a}^{b}\|v(t)\|\,dt}$

Now cancelling the norm of u from both ends gives us the desired inequality.

Mean Value Inequality. If the norm of Df(x + th) is bounded by some constant M for t in [0, 1], then

${\displaystyle \|f(x+h)-f(x)\|\leq M\|h\|.}$

Proof. From Lemma 1 and 2 it follows that

${\displaystyle \|f(x+h)-f(x)\|=\left\|\int _{0}^{1}(Df(x+th)\cdot h)\,dt\right\|\leq \int _{0}^{1}\|Df(x+th)\|\cdot \|h\|\,dt\leq M\|h\|.}$

Mean value theorems for definite integrals

First mean value theorem for definite integrals

Geometrically: interpreting f(c) as the height of a rectangle and b–a as the width, this rectangle has the same area as the region below the curve from a to b^[9]

Let f : [a, b] → R be a continuous function. Then there exists c in [a, b] such that

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

Since the mean value of f on [a, b] is defined as

${\displaystyle {\frac {1}{b-a}}\int _{a}^{b}f(x)\,dx,}$

we can interpret the conclusion as f achieves its mean value at some c in (a, b).^[10]

In general, if f : [a, b] → R is continuous and g is an integrable function that does not change sign on [a, b], then there exists c in (a, b) such that

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

Proof of the first mean value theorem for definite integrals

Suppose f : [a, b] → R is continuous and g is a nonnegative integrable function on [a, b]. By the extreme value theorem, there exists m and M such that for each x in [a, b], ${\displaystyle m\leq f(x)\leq M}$ and ${\displaystyle f[a,b]=[m,M]}$. Since g is nonnegative,

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

Now let

${\displaystyle I=\int _{a}^{b}g(x)\,dx.}$

If ${\displaystyle I=0}$, we're done since

${\displaystyle 0\leq \int _{a}^{b}f(x)g(x)\,dx\leq 0}$

means

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

so for any c in (a, b),

${\displaystyle \int _{a}^{b}f(x)g(x)\,dx=f(c)I=0.}$

If I ≠ 0, then

${\displaystyle m\leq {\frac {1}{I}}\int _{a}^{b}f(x)g(x)\,dx\leq M.}$

By the intermediate value theorem, f attains every value of the interval [m, M], so for some c in [a, b]

${\displaystyle f(c)={\frac {1}{I}}\int _{a}^{b}f(x)g(x)\,dx,}$

that is,

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

Finally, if g is negative on [a, b], then

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

and we still get the same result as above.

QED

Second mean value theorem for definite integrals

There are various slightly different theorems called the second mean value theorem for definite integrals. A commonly found version is as follows:

If G : [a, b] → R is a positive monotonically decreasing function and φ : [a, b] → R is an integrable function, then there exists a number x in (a, b] such that

${\displaystyle \int _{a}^{b}G(t)\varphi (t)\,dt=G(a^{+})\int _{a}^{x}\varphi (t)\,dt.}$

Here ${\displaystyle G(a^{+})}$ stands for ${\textstyle {\lim _{x\to a^{+}}G(x)}}$, the existence of which follows from the conditions. Note that it is essential that the interval (a, b] contains b. A variant not having this requirement is:^[11]

If G : [a, b] → R is a monotonic (not necessarily decreasing and positive) function and φ : [a, b] → R is an integrable function, then there exists a number x in (a, b) such that

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

Mean value theorem for integration fails for vector-valued functions

If the function ${\displaystyle G}$ returns a multi-dimensional vector, then the MVT for integration is not true, even if the domain of ${\displaystyle G}$ is also multi-dimensional.

For example, consider the following 2-dimensional function defined on an ${\displaystyle n}$-dimensional cube:

${\displaystyle {\begin{cases}G:[0,2\pi ]^{n}\to \mathbb {R} ^{2}\\G(x_{1},\dots ,x_{n})=\left(\sin(x_{1}+\cdots +x_{n}),\cos(x_{1}+\cdots +x_{n})\right)\end{cases}}}$

Then, by symmetry it is easy to see that the mean value of ${\displaystyle G}$ over its domain is (0,0):

${\displaystyle \int _{[0,2\pi ]^{n}}G(x_{1},\dots ,x_{n})dx_{1}\cdots dx_{n}=(0,0)}$

However, there is no point in which ${\displaystyle G=(0,0)}$, because ${\displaystyle |G|=1}$ everywhere.

A probabilistic analogue of the mean value theorem

Let X and Y be non-negative random variables such that E[X] < E[Y] < ∞ and ${\displaystyle X\leq _{st}Y}$ (i.e. X is smaller than Y in the usual stochastic order). Then there exists an absolutely continuous non-negative random variable Z having probability density function

${\displaystyle f_{Z}(x)={\Pr(Y>x)-\Pr(X>x) \over {\rm {E}}[Y]-{\rm {E}}[X]}\,,\qquad x\geqslant 0.}$

Let g be a measurable and differentiable function such that E[g(X)], E[g(Y)] < ∞, and let its derivative g′ be measurable and Riemann-integrable on the interval [x, y] for all y ≥ x ≥ 0. Then, E[g′(Z)] is finite and^[12]

${\displaystyle {\rm {E}}[g(Y)]-{\rm {E}}[g(X)]={\rm {E}}[g'(Z)]\,[{\rm {E}}(Y)-{\rm {E}}(X)].}$

Generalization in complex analysis

As noted above, the theorem does not hold for differentiable complex-valued functions. Instead, a generalization of the theorem is stated such:^[13]

Let f : Ω → C be a holomorphic function on the open convex set Ω, and let a and b be distinct points in Ω. Then there exist points u, v on L_ab (the line segment from a to b) such that

${\displaystyle \operatorname {Re} (f'(u))=\operatorname {Re} \left({\frac {f(b)-f(a)}{b-a}}\right),}$

${\displaystyle \operatorname {Im} (f'(v))=\operatorname {Im} \left({\frac {f(b)-f(a)}{b-a}}\right).}$

Where Re() is the real part and Im() is the imaginary part of a complex-valued function.

See also

• Newmark-beta method
• Mean value theorem (divided differences)
• Racetrack principle
• Stolarsky mean

Notes

1. J. J. O'Connor and E. F. Robertson (2000). Paramesvara, MacTutor History of Mathematics archive.
2. Ádám Besenyei. "Historical development of the mean value theorem" (PDF).
3. Lozada-Cruz, German (2020-10-02). "Some variants of Cauchy's mean value theorem". International Journal of Mathematical Education in Science and Technology. 51 (7): 1155–1163. doi:10.1080/0020739X.2019.1703150. ISSN 0020-739X.
4. Sahoo, Prasanna. (1998). Mean value theorems and functional equations. Riedel, T. (Thomas), 1962-. Singapore: World Scientific. ISBN 981-02-3544-5. OCLC 40951137.
5. Kirshna's Real Analysis: (General). Krishna Prakashan Media.
6. W., Weisstein, Eric. "Extended Mean-Value Theorem". mathworld.wolfram.com. Retrieved 2018-10-08.
7. "Cauchy's Mean Value Theorem". Math24. Retrieved 2018-10-08.
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9. "Mathwords: Mean Value Theorem for Integrals". www.mathwords.com.
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12. Di Crescenzo, A. (1999). "A Probabilistic Analogue of the Mean Value Theorem and Its Applications to Reliability Theory". J. Appl. Probab. 36 (3): 706–719. doi:10.1239/jap/1032374628. JSTOR 3215435.
13. "Complex Mean-Value Theorem". PlanetMath. PlanetMath.

External links

• "Cauchy theorem", Encyclopedia of Mathematics, EMS Press, 2001 [1994]
• PlanetMath: Mean-Value Theorem
• Weisstein, Eric W. "Mean value theorem". MathWorld.
• Weisstein, Eric W. "Cauchy's Mean-Value Theorem". MathWorld.
• "Mean Value Theorem: Intuition behind the Mean Value Theorem" at the Khan Academy