View source for Mean value theorem

{{Short description|Theorem in mathematics}}
{{For|the theorem in harmonic function theory|Harmonic function#The mean value property}}{{Calculus}}
In [[mathematics]], the '''mean value theorem''' (or '''Lagrange's mean value theorem''') states, roughly, that for a given planar [[arc (geometry)|arc]] between two endpoints, there is at least one point at which the [[tangent]] to the arc is parallel to the [[secant line|secant]] through its endpoints. It is one of the most important results in [[real analysis]]. This theorem is used to prove statements about a function on an [[interval (mathematics)|interval]] starting from local hypotheses about derivatives at points of the interval.

{{TOC limit|2}}

== History ==

A special case of this [[theorem]] for inverse interpolation of the sine was first described by [[Parameshvara]] (1380–1460), from the [[Kerala School of Astronomy and Mathematics]] in [[India]], in his commentaries on [[Govindasvāmi]] and [[Bhāskara II]].<ref>J. J. O'Connor and E. F. Robertson (2000). [https://mathshistory.st-andrews.ac.uk/Biographies/Paramesvara/ Paramesvara], ''[[MacTutor History of Mathematics archive]]''.</ref> A restricted form of the theorem was proved by [[Michel Rolle]] in 1691; the result was what is now known as [[Rolle's theorem]], and was proved only for polynomials, without the techniques of calculus. The mean value theorem in its modern form was stated and proved by [[Augustin Louis Cauchy]] in 1823.<ref>{{cite web|author=Ádám Besenyei|title=Historical development of the mean value theorem| url=http://abesenyei.web.elte.hu/publications/meanvalue.pdf}}</ref> Many variations of this theorem have been proved since then.<ref>{{Cite journal|last=Lozada-Cruz|first=German|date=2020-10-02|title=Some variants of Cauchy's mean value theorem|url=https://www.tandfonline.com/doi/full/10.1080/0020739X.2019.1703150|journal=International Journal of Mathematical Education in Science and Technology|language=en|volume=51|issue=7|pages=1155–1163|doi=10.1080/0020739X.2019.1703150|bibcode=2020IJMES..51.1155L|s2cid=213335491|issn=0020-739X|url-access=subscription}}</ref><ref>{{Cite book|last=Sahoo, Prasanna.|title=Mean value theorems and functional equations|date=1998|publisher=World Scientific|others=Riedel, T. (Thomas), 1962-|isbn=981-02-3544-5|location=Singapore|oclc=40951137}}</ref>

== Statement ==
[[File:Mittelwertsatz3 c version.svg|thumb|The function <math>f</math> attains the slope of the secant between <math>a</math> and <math>b</math> as the derivative at the point <math>c\in(a,b)</math>.]]
[[File:Mittelwertsatz6.svg|thumb|It is also possible that there are multiple tangents parallel to the secant.]]
Let <math>f:[a,b]\to\R</math> be a [[continuous function]] on the [[closed interval]] {{nowrap|<math>[a,b]</math>,}} and [[differentiable function|differentiable]] on the [[open interval]] {{nowrap|<math>(a,b)</math>,}} where {{nowrap|<math>a < b</math>.}} Then there exists some <math>c</math> in <math>(a,b)</math> such that:{{sfn|Rudin|1976|p=108}}

:<math>f'(c)=\frac{f(b)-f(a)}{b-a}.</math>

The mean value theorem is a generalization of [[Rolle's theorem]], which assumes <math>f(a)=f(b)</math>, so that the right-hand side above is zero.

The mean value theorem is still valid in a slightly more general setting. One only needs to assume that <math>f:[a,b]\to\R</math> is [[continuous function|continuous]] on <math>[a,b]</math>, and that for every <math>x</math> in <math>(a,b)</math> the [[limit of a function|limit]]

:<math>\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}</math>

exists as a finite number or equals <math>\infty</math> or <math>-\infty</math>. If finite, that limit equals <math>f'(x)</math>. An example where this version of the theorem applies is given by the real-valued [[cube root]] function mapping <math>x \mapsto x^{1/3}</math>, whose [[derivative]] tends to infinity at the origin.

===Proof===
The expression <math display="inline">\frac{f(b)-f(a)}{b-a}</math> gives the [[slope]] of the line joining the points <math>(a,f(a))</math> and <math>(b,f(b))</math>, which is a [[Chord (geometry)|chord]] of the graph of <math>f</math>, while <math>f'(x)</math> gives the slope of the tangent to the curve at the point <math>(x,f(x))</math>. 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.

Define <math>g(x)=f(x)-rx</math>, where <math>r</math> is a constant. Since <math>f</math> is continuous on <math>[a,b]</math> and differentiable on <math>(a,b)</math>, the same is true for <math>g</math>. We now want to choose <math>r</math> so that <math>g</math> satisfies the conditions of [[Rolle's theorem]]. Namely

:<math>\begin{align}
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{align}</math>

By [[Rolle's theorem]], since <math>g</math> is differentiable and <math>g(a)=g(b)</math>, there is some <math>c</math> in <math>(a,b)</math> for which <math>g'(c)=0</math> , and it follows from the equality <math>g(x)=f(x)-rx</math> that,

:<math>\begin{align}
&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{align}</math>

==Implications==
'''Theorem 1:''' Assume that <math>f</math> is a continuous, real-valued function, defined on an arbitrary interval <math>I</math> of the real line. If the derivative of <math>f</math> at every [[interior (topology)|interior point]] of the interval <math>I</math> exists and is zero, then <math>f</math> is [[constant function|constant]] on <math>I</math>.

'''Proof:''' Assume the derivative of <math>f</math> at every [[interior (topology)|interior point]] of the interval <math>I</math> exists and is zero. Let <math>(a, b)</math> be an arbitrary open interval in <math>I</math>. By the mean value theorem, there exists a point <math>c</math> in <math>(a, b)</math> such that

:<math>0=f'(c)=\frac{f(b)-f(a)}{b-a}.</math>

This implies that {{nowrap|1=<math>f(a) = f(b)</math>}}. Thus, <math>f</math> is constant on the interior of <math>I</math> and thus is constant on <math>I</math> by continuity. (See below for a multivariable version of this result.)

'''Remarks:'''
* Only continuity of <math>f</math>, not differentiability, is needed at the endpoints of the interval <math>I</math>.  No hypothesis of continuity needs to be stated if <math>I</math> is an [[open interval]], since the existence of a derivative at a point implies the continuity at this point.  (See the section [[Derivative#Continuity and differentiability|continuity and differentiability]] of the article [[derivative]].)
* The differentiability of <math>f</math> can be relaxed to [[one-sided derivatives|one-sided differentiability]], a proof is given in the article on [[semi-differentiability]].
'''Theorem 2:''' If <math>f'(x) = g'(x)</math> for all <math>x</math> in an interval <math>(a, b)</math> of the domain of these functions, then <math>f - g</math> is constant, i.e. <math>f = g + c</math> where <math>c</math> is a constant on <math>(a, b)</math>.

'''Proof:''' Let <math>F(x) = f(x) - g(x)</math>, then <math>F'(x)=f'(x)-g'(x)=0</math> on the interval <math>(a, b)</math>, so the above theorem 1 tells that <math>F(x) = f(x) - g(x)</math> is a constant <math>c</math> or <math>f = g + c</math>.

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

'''Proof:''' It directly follows from the theorem 2 above.

==Cauchy's mean value theorem==
<!-- This section is linked from [[Taylor's theorem]] -->
'''Cauchy's mean value theorem''', also known as the '''extended mean value theorem''', is a generalization of the mean value theorem.<ref>{{MathWorld|ExtendedMean-ValueTheorem|Extended Mean-Value Theorem}}</ref>{{sfn|Rudin|1976|pp=107-108}} It states: if the functions <math>f</math> and <math>g</math> are both continuous on the closed interval <math>[a,b]</math> and differentiable on the open interval <math>(a,b)</math>, then there exists some <math>c \in (a,b)</math>, such that
[[File:Cauchy.svg|right|thumb|260px|Geometrical meaning of Cauchy's theorem]]

:<math>(f(b)-f(a))g'(c)=(g(b)-g(a))f'(c).</math>

Of course, if <math>g(a) \neq g(b)</math> and <math>g'(c) \neq 0</math>, this is equivalent to:

:<math>\frac{f'(c)}{g'(c)}=\frac{f(b)-f(a)}{g(b)-g(a)}.</math>

Geometrically, this means that there is some [[tangent]] to the graph of the [[curve]]<ref>{{Cite news|url=https://www.math24.net/cauchys-mean-value-theorem/|title=Cauchy's Mean Value Theorem|work=Math24|access-date=2018-10-08|language=en-US}}</ref>

:<math>\begin{cases}[a,b] \to \R^2\\t\mapsto (f(t),g(t))\end{cases}</math>

which is [[Parallel (geometry)|parallel]] to the line defined by the points <math>(f(a), g(a))</math> and <math>(f(b), g(b))</math>. However, Cauchy's theorem does not claim the existence of such a tangent in all cases where <math>(f(a), g(a))</math> and <math>(f(b), g(b))</math> are distinct points, since it might be satisfied only for some value <math>c</math> with <math>f'(c) = g'(c) = 0</math>, in other words a value for which the mentioned curve is [[Stationary point|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

:<math>t \mapsto \left(t^3,1-t^2\right),</math>

which on the interval <math>[-1,1]</math> goes from the point <math>(-1, 0)</math> to <math>(1, 0)</math>, yet never has a horizontal tangent; however it has a stationary point (in fact a [[cusp (singularity)|cusp]]) at <math>t = 0</math>.

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 <math>g(t) = t</math>.

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

Define <math>h(x) = (g(b)-g(a))f(x) - (f(b)-f(a))g(x)</math>, then we easily see <math>h(a)=h(b)=f(a)g(b)-f(b)g(a)</math>.

Since <math>f</math> and <math>g</math> are continuous on <math>[a,b]</math> and differentiable on <math>(a,b)</math>, the same is true for <math>h</math>. All in all, <math>h</math> satisfies the conditions of [[Rolle's theorem]]. Consequently, there is some <math>c</math> in <math>(a,b)</math> for which <math>h'(c) = 0</math>. Now using the definition of <math>h</math> we have:
:<math>0=h'(c)=(g(b)-g(a))f'(c)-(f(b)-f(a))g'(c)</math>
The result easily follows.

== 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 <math>G</math> be an open subset of <math>\R^n</math>, and let <math>f:G\to\R</math> be a differentiable function. Fix points <math>x,y\in G</math> such that the line segment between <math>x, y</math> lies in <math>G</math>, and define <math>g(t)=f\big((1-t)x+ty\big)</math>. Since <math>g</math> is a differentiable function in one variable, the mean value theorem gives:

:<math>g(1)-g(0)=g'(c)</math>

for some <math>c</math> between 0 and 1. But since <math>g(1)=f(y)</math> and <math>g(0)=f(x)</math>, computing <math>g'(c)</math> explicitly we have:

:<math>f(y)-f(x)=\nabla f\big((1-c)x+cy\big)\cdot (y-x)</math>

where <math>\nabla</math> denotes a [[gradient]] and <math>\cdot</math> a [[dot product]]. This is an exact analog of the theorem in one variable (in the case <math>n=1</math> this ''is'' the theorem in one variable). By the [[Cauchy–Schwarz inequality]], the equation gives the estimate:

:<math>\Bigl|f(y)-f(x)\Bigr| \le \Bigl|\nabla f\big((1-c)x+cy\big)\Bigr|\ \Bigl|y - x\Bigr|.</math>

In particular, when <math>G</math> is convex and the partial derivatives of <math>f</math> are bounded, <math>f</math> is [[Lipschitz continuity|Lipschitz continuous]] (and therefore [[Uniform continuity|uniformly continuous]]).

As an application of the above, we prove that <math>f</math> is constant if the open subset <math>G</math> is connected and every partial derivative of <math>f</math> is 0. Pick some point <math>x_0\in G</math>, and let <math>g(x)=f(x)-f(x_0)</math>. We want to show <math>g(x)=0</math> for every <math>x\in G</math>. For that, let <math>E=\{x\in G:g(x)=0\}</math>. Then <math>E</math> is closed in <math>G</math> and nonempty. It is open too: for every <math>x\in E</math> ,

:<math>\Big|g(y)\Big|=\Big|g(y)-g(x)\Big|\le (0)\Big|y-x\Big|=0</math>

for every <math>y</math> in open ball centered at <math>x</math> and contained in <math>G</math>. Since <math>G</math> is connected, we conclude <math>E=G</math>.

The above arguments are made in a coordinate-free manner; hence, they generalize to the case when <math>G</math> 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 (see below). However, there is 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:{{sfn|Rudin|1976|p=113}}

{{math theorem|''For a continuous vector-valued function <math>\mathbf{f}:[a,b]\to\mathbb{R}^k</math> differentiable on <math>(a,b)</math>, there exists a number <math>c\in(a,b)</math> such that''
:<math>|\mathbf{f}(b)-\mathbf{f}(a)| \le (b-a)\left|\mathbf{f}'(c)\right|</math>.}}

{{math proof| Take <math>\varphi(t) = (\textbf{f}(b) - \textbf{f}(a)) \cdot \textbf{f}(t)</math>. Then <math>\varphi</math> is real-valued and thus, by the mean value theorem,
:<math>\varphi(b) - \varphi(a) = \varphi'(c)(b-a)</math>
for some <math>c \in (a, b)</math>. Now,
<math display="block">\varphi(b) - \varphi(a) = |\textbf{f}(b) - \textbf{f}(a)|^2</math>
and
<math display="block">\varphi'(c) = (\textbf{f}(b) - \textbf{f}(a)) \cdot \textbf{f}'(c).</math>
Hence, using the [[Cauchy–Schwarz inequality]], from the above equation, we get:
:<math>|\textbf{f}(b) - \textbf{f}(a)|^2 \le |\textbf{f}(b) - \textbf{f}(a)| |\textbf{f}'(c) |(b-a).</math>
If <math>\textbf{f}(b) = \textbf{f}(a)</math>, the theorem holds trivially. Otherwise, dividing both sides by <math> |\textbf{f}(b) - \textbf{f}(a)|</math> yields the theorem.}}

===Mean value inequality===
{{see also|Calculus on Euclidean space#Derivative of a map and chain rule}}
[[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 reason why there is no analog of mean value equality is the following: If {{math|''f'' : ''U'' → '''R'''<sup>''m''</sup>}} is a differentiable function (where {{math|''U'' ⊂ '''R'''<sup>''n''</sup>}} is open) and if {{math|''x'' + ''th''}}, {{math|''x'', ''h'' ∈ '''R'''<sup>''n''</sup>, ''t'' ∈ [0, 1]}} is the line segment in question (lying inside {{mvar|U}}), then one can apply the above parametrization procedure to each of the component functions {{math|1=''f<sub>i</sub>'' (''i'' = 1, …, ''m'')}} of ''f'' (in the above notation set {{math|1=''y'' = ''x'' + ''h''}}). In doing so one finds points {{math|''x'' + ''t<sub>i</sub>h''}} on the line segment satisfying

:<math>f_i(x+h) - f_i(x) = \nabla f_i (x + t_ih) \cdot h.</math>

But generally there will not be a ''single'' point {{math|''x'' + ''t''*''h''}} on the line segment satisfying

:<math>f_i(x+h) - f_i(x) = \nabla f_i (x + t^* h) \cdot h.</math>

for all {{mvar|i}} ''simultaneously''. For example, define:

:<math>\begin{cases}
f : [0, 2 \pi] \to \R^2 \\
f(x) = (\cos(x), \sin(x))
\end{cases}</math>

Then <math>f(2\pi) - f(0) = \mathbf{0} \in \R^2</math>, but <math>f_1'(x) = -\sin (x)</math> and <math>f_2'(x) = \cos (x)</math> are never simultaneously zero as <math>x</math> ranges over <math>\left[0, 2 \pi\right]</math>.

The above theorem implies the following:
{{math_theorem
| name = Mean value inequality<ref>{{harvnb|Hörmander|2015|loc=Theorem 1.1.1. and remark following it.}}</ref>
| math_statement = For a continuous function <math>\textbf{f} : [a, b] \to \mathbb{R}^k</math>, if <math>\textbf{f}</math> is differentiable on <math>(a, b)</math>, then
:<math>|\textbf{f}(b) - \textbf{f}(a)| \le (b-a)\sup_{(a, b)} |\textbf{f}'|</math>.
}}

In fact, the above statement suffices for many applications and can be proved directly as follows. (We shall write <math>f</math> for <math>\textbf{f}</math> for readability.)
{{math proof|First assume <math>f</math> is differentiable at <math>a</math> too. If <math>f'</math> is unbounded on <math>(a, b)</math>, there is nothing to prove. Thus, assume <math>\sup_{(a, b)} |f'| < \infty</math>. Let <math>M > \sup_{(a, b)} |f'|</math> be some real number. Let
<math display="block">E = \{ 0 \le t \le 1 \mid |f(a + t(b-a)) - f(a)| \le Mt(b-a) \}.</math>
We want to show <math>1 \in E</math>. By continuity of <math>f</math>, the set <math>E</math> is closed. It is also nonempty as <math>0</math> is in it. Hence, the set <math>E</math> has the largest element <math>s</math>. If <math>s = 1</math>, then <math>1 \in E</math> and we are done. Thus suppose otherwise. For <math>1 > t > s</math>,
:<math>\begin{align}
&|f(a + t(b-a)) - f(a)| \\
&\le |f(a + t(b-a)) - f(a+s(b - a)) - f'(a + s(b-a))(t-s)(b-a)| + |f'(a+s(b-a))|(t-s)(b-a) \\
&+|f(a + s(b-a)) - f(a)|.
\end{align}
</math>

Let <math>\epsilon > 0</math> be such that <math>M - \epsilon > \sup_{(a, b)} |f'|</math>. By the differentiability of <math>f</math> at <math>a + s(b-a)</math> (note <math>s</math> may be 0), if <math>t</math> is sufficiently close to <math>s</math>, the first term is <math>\le \epsilon (t-s)(b-a)</math>. The second term is <math>\le (M - \epsilon) (t-s)(b-a)</math>. The third term is <math>\le Ms(b-a)</math>. Hence, summing the estimates up, we get: <math>|f(a + t(b-a)) - f(a)| \le tM|b-a|</math>, a contradiction to the maximality of <math>s</math>. Hence, <math>1 = s \in M</math> and that means:
:<math>|f(b) - f(a)| \le M(b-a).</math>
Since <math>M</math> is arbitrary, this then implies the assertion. Finally, if <math>f</math> is not differentiable at <math>a</math>, let <math>a' \in (a, b)</math> and apply the first case to <math>f</math> restricted on <math>[a', b]</math>, giving us:
:<math>|f(b) - f(a')| \le (b-a')\sup_{(a, b)} |f'|</math>
since <math>(a', b) \subset (a, b)</math>. Letting <math>a' \to a</math> finishes the proof.}}

== Cases where the theorem cannot be applied ==
All conditions for the mean value theorem are necessary:

# '''<math>\boldsymbol{f(x)} </math> is differentiable on <math>\boldsymbol{(a,b)} </math>'''
# '''<math>\boldsymbol{f(x)} </math> is continuous on <math>\boldsymbol{[a,b]} </math>'''
# '''<math>\boldsymbol{f(x)} </math> is real-valued'''

When one of the above conditions is not satisfied, the mean value theorem is not valid in general, and so it cannot be applied.

[[File:Absolute Value.svg|thumb]]
The necessity of the first condition can be seen by the counterexample where the function <math>f(x)=|x| </math> on [-1,1] is not differentiable.

The necessity of the second condition can be seen by the counterexample where the function
<math display="block">f(x) = \begin{cases} 1, & \text{at }x=0 \\ 0, & \text{if }x\in( 0,1] \end{cases} </math>
satisfies criteria 1 since <math>f'(x)=0  </math> on <math>(0,1) </math> but not criteria 2 since
<math display="block">\frac{f(1)-f(0)}{1-0}=-1 </math> and <math>-1\neq 0 =f'(x)  </math> for all <math>x\in (0,1) </math> so no such <math>c </math> exists.

The theorem is false if a differentiable function is complex-valued instead of real-valued. For example, if <math>f(x) = e^{xi}</math> for all real <math>x</math>, then
<math display="block">f(2\pi)-f(0)=0=0(2\pi-0)</math>
while <math>f'(x)\ne 0</math> for any real <math>x</math>.

== Mean value theorems for definite integrals ==

===First mean value theorem for definite integrals===
[[File:积分中值定理.jpg|thumb|right|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''<ref>{{cite web|url=http://www.mathwords.com/m/mean_value_theorem_integrals.htm|title=Mathwords: Mean Value Theorem for Integrals|website=www.mathwords.com}}</ref>]]
Let ''f'' : [''a'', ''b''] → '''R''' be a continuous function. Then there exists ''c'' in (''a'', ''b'') such that

:<math>\int_a^b f(x) \, dx = f(c)(b - a).</math>

This follows at once from the [[fundamental theorem of calculus]], together with the mean value theorem for derivatives.  Since the mean value of ''f'' on [''a'', ''b''] is defined as

:<math>\frac{1}{b-a} \int_a^b f(x) \, dx,</math>

we can interpret the conclusion as ''f'' achieves its mean value at some ''c'' in (''a'', ''b'').<ref name="Comenetz2002">{{cite book|author=Michael Comenetz|title=Calculus: The Elements| year=2002| publisher=World Scientific|isbn=978-981-02-4904-5|page=159}}</ref>

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

:<math>\int_a^b f(x) g(x) \, dx = f(c) \int_a^b  g(x) \, dx.</math>

===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 <math>G : [a,b]\to \mathbb{R}</math>  is a positive [[monotone function|monotonically decreasing]] function and <math>\varphi : [a,b]\to \mathbb{R}</math> is an integrable function, then there exists a number ''x'' in (''a'', ''b''] such that
::<math> \int_a^b G(t)\varphi(t)\,dt = G(a^+) \int_a^x \varphi(t)\,dt. </math>

Here <math>G(a^+)</math> stands for <math display="inline">{\lim_{x\to a^+}G(x)}</math>, 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:<ref>{{cite journal |first=E. W. |last=Hobson |year=1909 |title=On the Second Mean-Value Theorem of the Integral Calculus |journal=[[Proceedings London Mathematical Society|Proc. London Math. Soc.]] |volume=S2–7 |issue=1 |pages=14–23 |mr=1575669 |doi=10.1112/plms/s2-7.1.14 |bibcode=1909PLMS...27...14H |url=https://zenodo.org/record/1447800 }}</ref>

:If <math>G : [a,b]\to \mathbb{R}</math>  is a [[monotone function|monotonic]] (not necessarily decreasing and positive) function and <math>\varphi : [a,b]\to \mathbb{R}</math>  is an integrable function, then there exists a number ''x'' in (''a'', ''b'') such that
::<math> \int_a^b G(t)\varphi(t)\,dt = G(a^+) \int_a^x \varphi(t)\,dt + G(b^-) \int_x^b \varphi(t)\,dt. </math>

If the function <math>G</math> returns a multi-dimensional vector, then the MVT for integration is not true, even if the domain of <math>G</math> is also multi-dimensional.

For example, consider the following 2-dimensional function defined on an <math>n</math>-dimensional cube:

:<math>\begin{cases}
G: [0,2\pi]^n \to \R^2 \\
G(x_1, \dots, x_n) = \left(\sin(x_1 + \cdots + x_n), \cos(x_1 + \cdots + x_n) \right)
\end{cases} </math>

Then, by symmetry it is easy to see that the mean value of <math>G</math> over its domain is (0,0):

:<math>\int_{[0,2\pi]^n} G(x_1,\dots,x_n) dx_1 \cdots dx_n = (0,0)</math>

However, there is no point in which <math>G=(0,0)</math>, because <math>|G|=1</math> everywhere.

==Generalizations==

===Linear algebra===
Assume that <math>f, g,</math> and <math>h</math> are differentiable functions on <math>(a,b)</math> that are continuous on <math>[a,b]</math>. Define

:<math> D(x) = \begin{vmatrix}
f(x) & g(x) & h(x)\\
f(a) & g(a) & h(a)\\
f(b) & g(b) & h(b)
\end{vmatrix}</math>

There exists <math>c\in(a,b)</math> such that <math> D'(c)=0</math>.

Notice that
:<math> D'(x) = \begin{vmatrix}
f'(x) & g'(x)& h'(x)\\
f(a) & g(a) & h(a)\\
f(b) & g(b)& h(b)
\end{vmatrix}</math>
and if we place <math>h(x)=1</math>, we get '''Cauchy's mean value theorem'''. If we place <math>h(x)=1</math> and <math>g(x)=x</math> we get '''Lagrange's mean value theorem'''.

The proof of the generalization is quite simple: each of <math>D(a)</math> and <math>D(b)</math> are [[determinant]]s with two identical rows, hence <math>D(a)=D(b)=0</math>. The Rolle's theorem implies that there exists <math>c\in (a,b)</math> such that <math>D'(c)=0</math>.

=== Probability theory===
Let ''X'' and ''Y'' be non-negative [[random variable]]s such that E[''X''] < E[''Y''] < ∞ and <math>X \leq_{st} Y</math> (i.e. ''X'' is smaller than ''Y'' in the [[Stochastic ordering|usual stochastic order]]). Then there exists an absolutely continuous non-negative random variable ''Z'' having [[probability density function]]

:<math> f_Z(x)={\Pr(Y>x)-\Pr(X>x)\over {\rm E}[Y]-{\rm E}[X]}\,, \qquad x\geqslant 0.</math>

Let ''g'' be a [[Measurable function|measurable]] and [[differentiable function]] such that E[''g''(''X'')], E[''g''(''Y'')] < ∞, and let its derivative ''g′''  be measurable and [[Riemann integral|Riemann-integrable]] on the interval [''x'', ''y''] for all ''y'' ≥ ''x'' ≥ 0. Then, E[''g′''(''Z'')] is finite and<ref>{{cite journal |first=A. |last=Di Crescenzo |year=1999 |title=A Probabilistic Analogue of the Mean Value Theorem and Its Applications to Reliability Theory |journal=[[J. Appl. Probab.]] |volume=36 |issue=3 |pages=706–719 |jstor=3215435 |doi=10.1239/jap/1032374628 |s2cid=250351233 }}</ref>

:<math> {\rm E}[g(Y)]-{\rm E}[g(X)]={\rm E}[g'(Z)]\,[{\rm E}(Y)-{\rm E}(X)].</math>

=== Complex analysis ===
{{See also|Voorhoeve index|Mean value problem}}
As noted above, the theorem does not hold for differentiable complex-valued functions. Instead, a generalization of the theorem is stated such:<ref>1 J.-Cl. Evard, F. Jafari, A Complex Rolle’s Theorem, American Mathematical Monthly, Vol. 99, Issue 9, (Nov. 1992), pp. 858-861.</ref>

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 the interior of the line segment from ''a'' to ''b'' such that
:<math>\operatorname{Re}(f'(u)) = \operatorname{Re}\left ( \frac{f(b)-f(a)}{b-a} \right),</math>
:<math>\operatorname{Im}(f'(v)) = \operatorname{Im}\left ( \frac{f(b)-f(a)}{b-a} \right).</math>

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

== See also ==
* [[Intermediate value theorem]]
* [[Mean value problem]]
* [[Mean value theorem (divided differences)]]
* [[Newmark-beta method]]
* [[Racetrack principle]]
* [[Stolarsky mean]]

== Notes ==
{{Reflist|50em}}

== References ==
* {{cite book | last=Rudin | first=Walter | title=Principles of Mathematical Analysis | publisher=McGraw-Hill Publishing Company | publication-place=Auckland | date=1976 | isbn=978-0-07-085613-4}}
*{{citation|title=The Analysis of Linear Partial Differential Operators I: Distribution Theory and Fourier Analysis|series=Classics in Mathematics|first=Lars|last= Hörmander|authorlink=Lars Hörmander|publisher=Springer|year= 2015|edition=2nd|
isbn= 9783642614972}}

== External links ==
* {{SpringerEOM|title=Cauchy theorem|id=p/c020990}}
* {{PlanetMath|MeanValueTheorem}}
* {{MathWorld|Mean-ValueTheorem|Mean-Value Theorem}}
* {{MathWorld|CauchysMean-ValueTheorem|Cauchy's Mean-Value Theorem}}
* [https://www.khanacademy.org/math/old-differential-calculus/derivative-applications-dc/mean-value-theorem-dc/v/mean-value-theorem-1 Differential Calculus (2017 edition) • Unit 11: Derivative applications • Mean Value Theorem] at the [[Khan Academy]]


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{{Joseph-Louis Lagrange}}
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