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{{For|the theorem in harmonic function theory|Harmonic function#The mean value property}}
This is a preview for the new '''MathML rendering mode''' (with SVG fallback), which is availble in production for registered users.
{{distinguish2|the [[Intermediate value theorem]]}}
[[Image:Mvt2.svg|thumb|300 px|right|For any function that is continuous on [''a'', ''b''] and differentiable on (''a'', ''b'') there exists some ''c'' in the interval (''a'', ''b'') such that the secant joining the endpoints of the interval [''a'', ''b''] is parallel to the tangent at ''c''.]]
{{Calculus}}


In [[mathematics]], the '''mean value theorem''' states, roughly:  that given a 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.  
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The theorem is used to prove global statements about a function on an interval starting from local hypotheses about derivatives at points of the interval.
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More precisely, if a function ''f'' is [[continuous function|continuous]] on the [[closed interval]] [''a'', ''b''], where ''a'' < ''b'', and differentiable on the [[open interval]] (''a'', ''b''), then there exists a point ''c'' in (''a'', ''b'') such that
'''MathML'''
:<math> f'(c) = \frac{f(b) - f(a)}{b-a} \, .</math><ref name=mathworld>{{cite web|title=Mean-Value Theorem|last=Weisstein|first=Eric|url=http://mathworld.wolfram.com/Mean-ValueTheorem.html|work=[[MathWorld]]|publisher=[[Wolfram Research]]|accessdate=24 March 2011}}</ref>
:<math forcemathmode="mathml">E=mc^2</math>


A special case of this [[theorem]] was first described by [[Parameshvara]] (1370–1460) from the [[Kerala school of astronomy and mathematics]] in his commentaries on [[Govindasvāmi]] and [[Bhaskara II]].<ref>J. J. O'Connor and E. F. Robertson (2000). [http://www-groups.dcs.st-and.ac.uk/~history/Biographies/Paramesvara.html Paramesvara], ''[[MacTutor History of Mathematics archive]]''.</ref> The mean value theorem in its modern form was later stated by [[Augustin Louis Cauchy]] (1789–1857). It is one of the most important results in [[Derivative|differential]] [[calculus]], as well as one of the most important theorems in [[mathematical analysis]], and is useful in proving the [[fundamental theorem of calculus]]. The mean value theorem follows from the more specific statement of [[Rolle's theorem]], and can be used to prove the more general statement of [[Taylor's theorem]] (with Lagrange form of the remainder term).
<!--'''PNG''' (currently default in production)
:<math forcemathmode="png">E=mc^2</math>


== Formal statement ==
'''source'''
Let ''f'' : [''a'', ''b''] → '''R''' be a [[continuous function]] on the closed [[interval (mathematics)|interval]] [''a'', ''b''], and [[differentiable_function|differentiable]] on the open interval (''a'', ''b''), where {{nowrap|''a'' < ''b''.}} Then there exists some ''c'' in (''a'', ''b'') such that
:<math forcemathmode="source">E=mc^2</math> -->


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


The mean value theorem is still valid in a slightly more general setting. One only needs to assume that ''f'' : [''a'', ''b''] → '''R''' is [[continuous function|continuous]] on [''a'', ''b''], and that for every ''x'' in (''a'', ''b'') the [[limit of a function|limit]]  
Here are some [https://commons.wikimedia.org/w/index.php?title=Special:ListFiles/Frederic.wang demos]:


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


exists as a finite number or equals +∞ or −∞. If finite, that limit equals ''f′''(''x''). An example where this version of the theorem applies is given by the real-valued [[cube root]] function mapping ''x'' to ''x''<sup>1/3</sup>, whose [[derivative]] tends to infinity at the origin.
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Note that the theorem, as stated, is false if a differentiable function is complex-valued instead of real-valued. For example, define {{nowrap|''f''(''x'') {{=}} ''e<sup>ix</sup>''}} for all real ''x''. Then
==Test pages ==
:''f''(2π) − ''f''(0) = 0 = 0(2π − 0)


while ''f′''(''x'') ≠ 0 for any real ''x''.
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==Proof==
*[[Inputtypes|Inputtypes (private Wikis only)]]
The expression {{nowrap|(''f''(''b'') − ''f''(''a'')) / (''b'' − ''a'')}} gives the [[slope]] of the line joining the points (''a'', ''f''(''a'')) and (''b'', ''f''(''b'')), which is a [[Chord (geometry)|chord]] of the graph of ''f'', while ''f '''(''x'') gives the slope of the tangent to the curve at the point (''x'', ''f''(''x'')). Thus the Mean value theorem says that given any chord of a smooth curve, we can find a point lying between the end-points of the chord such that the tangent at that point is parallel to the chord. The following proof illustrates this idea.
*[[Url2Image|Url2Image (private Wikis only)]]
 
==Bug reporting==
Define ''g''(''x'') = ''f''(''x'') − ''rx'', where ''r'' is a constant. Since ''f'' is continuous on [''a'', ''b''] and differentiable on (''a'', ''b''), the same is true for ''g''. We now want to choose ''r'' so that ''g'' satisfies the conditions of [[Rolle's theorem]]. Namely
If you find any bugs, please report them at [https://bugzilla.wikimedia.org/enter_bug.cgi?product=MediaWiki%20extensions&component=Math&version=master&short_desc=Math-preview%20rendering%20problem Bugzilla], or write an email to math_bugs (at) ckurs (dot) de .
:<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}\cdot\end{align}</math>
 
By Rolle's theorem, since ''g'' is differentiable and ''g''(''a'') = ''g''(''b''), there is some ''c'' in (''a'', ''b'') for which ''g′''(''c'') = 0, and it follows from the equality ''g''(''x'') = ''f''(''x'') − ''rx'' that,
:<math>f '(c)=g '(c)+r=0+r=\frac{f(b)-f(a)}{b-a}</math>
as required.
 
== A simple application ==
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 (topology)|interior point]] of the interval ''I'' exists and is zero, then ''f'' is [[constant function|constant]].
 
'''Proof:''' Assume the derivative of ''f'' at every [[interior (topology)|interior point]] of the interval ''I'' exists and is zero. Let (''a'', ''b'') be an arbitrary open interval in ''I''. By the mean value theorem, there exists a point ''c'' in (''a'',''b'') such that
 
:<math>0 = f'(c) = \frac{f(b)-f(a)}{b-a}.</math>
 
This implies that ''f''(''a'') = ''f''(''b''). Thus, ''f'' is constant on the interior of ''I'' and thus is constant on ''I'' by continuity. (See below for a multivariable version of this result.)
 
'''Remarks:'''
* 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 [[Derivative#Continuity and differentiability|continuity and differentiability]] of the article [[derivative]].)
* The differentiability of ''f'' can be relaxed to [[one-sided derivatives|one-sided differentiability]], a proof given in the article on [[semi-differentiability]].
 
==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. It states: If functions ''f'' and ''g'' are both continuous on the closed interval [''a'',''b''], and differentiable on the open interval (''a'','' ''b), then there exists some ''c'' ∈ (''a'',''b''), such that
[[Image:Cauchy svg.svg|right|thumb|306px|Geometrical meaning of Cauchy's theorem]]
 
:<math>(f(b)-f(a))g\,'(c)=(g(b)-g(a))f\,'(c).\,</math>
 
Of course, if {{nowrap|''g''(''a'') ≠ ''g''(''b'')}} and if {{nowrap|''g''&prime;(''c'') ≠ 0}}, this is equivalent to:
 
:<math>\frac{f'(c)}{g'(c)}=\frac{f(b)-f(a)}{g(b)-g(a)}\cdot</math>
 
Geometrically, this means that there is some [[tangent]] to the graph of the [[curve]]
 
:<math>\begin{array}{ccc}[a,b]&\longrightarrow&\mathbb{R}^2\\t&\mapsto&\bigl(f(t),g(t)\bigr),\end{array}</math>
 
which is [[Parallel (geometry)|parallel]] to the line defined by the points (''f''(''a''),''g''(''a'')) and (''f''(''b''),''g''(''b'')). However Cauchy's theorem does not claim the existence of such a tangent in all cases where (''f''(''a''),''g''(''a'')) and (''f''(''b''),''g''(''b'')) are distinct points, since it might be satisfied only for some value ''c'' with {{nowrap|''f''&prime;(''c'') {{=}} ''g''&prime;(''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
 
:<math>t\mapsto(t^3,1-t^2),</math>
 
which on the interval [−1,1] goes from the point (−1,0) to (1,0), yet never has a horizontal tangent; however it has a stationary point (in fact a [[cusp (singularity)|cusp]]) at {{nowrap|''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 {{nowrap|''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 that ''g''(''a'') ≠''g''(''b''). Define ''h''(''x'') = ''f''(''x'') − ''rg(x)'', where ''r'' is fixed in such a way that ''h''(''a'') = ''h''(''b''), namely
:<math>\begin{align}h(a)=h(b)&\iff f(a)-r\,g(a)=f(b)-r\,g(b)\\ &\iff r\,(g(b)-g(a))=f(b)-f(a)\\ &\iff r=\frac{f(b)-f(a)}{g(b)-g(a)}.\end{align}</math>
:Since ''f'' and ''g'' are continuous on [''a'', ''b''] and differentiable on (''a'', ''b''), the same is true for ''h''. All in all, ''h'' satisfies the conditions of [[Rolle's theorem]]: consequently, there is some ''c'' in (''a'', ''b'') for which ''h′''(''c'') = 0.
:From the equality ''h''(''x'') = ''f''(''x'') − ''rg(x)'' it follows that,
::<math>0=h'(c)=f'(c)-r\, g'(c) \Rightarrow(g(b)-g(a))\,f'(c)=(g(b)-g(a))\,r\,g'(c)=(f(b)-f(a))\,g'(c)</math>
:as required.
 
* If instead ''g''(''a'') = ''g''(''b''), then, applying [[Rolle's theorem]] to ''g'', it follows that there exists ''c'' in (''a'', ''b'') for which ''g′''(''c'') = 0. Using this choice of ''c'', Cauchy's mean value theorem (trivially) holds.
 
==Generalization for determinants==
Assume that <math>f</math>, <math>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)=\left|\begin{array}{ccc}f(x) & g(x)& h(x)\\ f(a) & g(a) & h(a)\\ f(b) & g(b)& h(b)\end{array}\right|</math>
 
There exists <math>c\in(a,b)</math> such that <math> D'(c)=0</math>.
 
Notice that
:<math> D'(x)=\left|\begin{array}{ccc}f'(x) & g'(x)& h'(x)\\ f(a) & g(a) & h(a)\\ f(b) & g(b)& h(b)\end{array}\right|</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 determinants 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>.
 
== Mean value theorem in several variables ==
The mean value theorem in one variable generalizes to several variables by applying the theorem in one variable via parametrization.  Let ''G'' be an open connected subset of '''R'''<sup>''n''</sup>, and let {{nowrap|''f'' : ''G'' → '''R'''}} be a differentiable function. Fix points {{nowrap| ''x'', ''y'' ∈ ''G''}} such that the interval {{nowrap|''x'' ''y''}} lies in ''G'', and define {{nowrap|''g''(''t'') {{=}} ''f''((1 − ''t'')''x'' + ''ty'')}}. Since ''g'' is a differentiable function in one variable, the mean value theorem gives:
 
:<math>g(1) - g(0) = g'(c) \!</math>
 
for some ''c'' between 0 and 1. But since {{nowrap|''g''(1) {{=}} ''f''(''y'')}} and {{nowrap|''g''(0) {{=}} ''f''(''x'')}}, computing ''g&prime;''(''c'') explicitly we have:
 
:<math>f(y) - f(x) = \nabla f ((1- c)x + cy) \cdot (y - x)</math>
 
where ∇ denotes a gradient and · a [[dot product]]. Note that this is an exact analog of the theorem in one variable (in the case {{nowrap|''n'' {{=}} 1}} this ''is'' the theorem in one variable). By the [[Schwarz inequality]], the equation gives the estimate:
 
:<math>|f(y) - f(x)| \le |\nabla f ((1- c)x + cy)| \, |y - x|.</math>
 
In particular, when the partial derivatives of ''f'' are bounded, ''f'' is [[Lipschitz continuous]] (and therefore [[uniformly continuous]]). Note that ''f'' is not assumed to be continuously differentiable nor continuous on the closure of ''G''. However, in the above, we used the chain rule so the existence of ∇''f'' would not be sufficient.
 
As an application of the above, we prove that ''f'' is constant if ''G'' is open and connected and every partial derivative of ''f'' is 0.  Pick some point {{nowrap|''x''<sub>0</sub> &isin; ''G''}}, and let {{nowrap|''g''(''x'') {{=}} ''f''(''x'') − ''f''(''x''<sub>0</sub>)}}. We want to show {{nowrap|''g''(''x'') {{=}} 0}} for every {{nowrap|''x'' ∈ ''G''}}. For that, let {{nowrap|''E'' {{=}} {''x'' ∈ ''G'' : ''g''(''x'') {{=}} 0} }}.  Then ''E'' is closed and nonempty.  It is open too: for every {{nowrap|''x'' ∈ ''E''}},
 
:<math>|g(y)| = |g(y) - g(x)| \le (0) |y - x| = 0</math>
 
for every ''y'' in some neighborhood of ''x''. (Here, it is crucial that ''x'' and ''y'' are sufficiently close to each other.) Since ''G'' is connected, we conclude {{nowrap|''E'' {{=}} ''G''}}.
 
Remark that all arguments in the above are made in a coordinate-free manner; hence, they actually generalize to the case when ''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. [[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 by no way one can find the mean value. 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'''<sup>''m''</sup> is a differentiable function (where ''U'' ⊂ '''R'''<sup>''n''</sup> is open) and if ''x'' + ''th'', ''x, h'' ∈ '''R'''<sup>''n''</sup>, ''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<sub>i</sub>'' (''i'' = 1, ..., ''m'') of ''f'' (in the above notation set ''y'' = ''x'' + ''h''). In doing so one finds points ''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 ''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 ''i'' ''simultaneously''.
(As a counterexample one could take ''f'' : [0, 2π] → '''R'''<sup>2</sup> defined via the component functions ''f''<sub>1</sub>(''x'') = cos(''x''), ''f''<sub>2</sub>(''x'') = sin(''x''). Then ''f''(2π) − f(0) = '''0''' ∈ '''R'''<sup>2</sup>, but <math>\,f_1'(x)=-\sin (x)</math> and <math>\,f_2'(x)=\cos (x)</math> are never simultaneously zero as ''x'' ranges over [0, 2π].)
 
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
 
:<math>f(x+h)-f(x) = f'(x+t^*h)\cdot h. \,</math>
 
On the other hand we have, by the [[Fundamental Theorem of Calculus|fundamental theorem of calculus]] followed by a change of variables,
 
:<math> f(x+h)-f(x) = \int_x^{x+h} f'(u)du = \left(\int_0^1 f'(x+th)\,dt\right)\cdot h.</math>
 
Thus, the value ''f′''(''x'' + ''t*h'') at the particular point ''t*'' has been replaced by the mean value
 
:<math>\int_0^1 f'(x+th)\,dt.</math>
 
This last version can be generalized to vector valued functions:
 
Let ''U'' ⊂ '''R'''<sup>''n''</sup> be open, ''f'' : ''U'' → '''R'''<sup>''m''</sup> continuously differentiable, and ''x'' ∈ ''U'', ''h'' ∈ '''R'''<sup>''n''</sup>  vectors such that the whole line segment ''x'' + ''th'',  0 ≤ ''t'' ≤ 1 remains in ''U''. Then we have:
 
:<math>\text{(*)} \qquad f(x+h)-f(x) = \left(\int_0^1 Df(x+th)\,dt\right)\cdot h,</math>
 
where the integral of a matrix is to be understood componentwise. (''Df'' denotes the [[Jacobian matrix]] of ''f''.)
 
From this one can further deduce that if ||''Df''(''x'' + ''th'')|| is bounded for ''t'' between 0 and 1 by some constant ''M'', then
 
:<math> \text{(**)} \qquad \|f(x+h)-f(x)\| \leq M\|h\|.</math>
 
<br />
'''Proof of (*).''' Write ''f<sub>i</sub>'' (''i'' = 1, ..., ''m'') for the real valued components of ''f''. Define the functions ''g<sub>i</sub>'': [0, 1] → '''R''' by ''g<sub>i</sub>''(''t'') := ''f<sub>i</sub>''(''x'' + ''th'').
 
Then we have
 
:<math>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.</math>
 
The claim follows since ''Df'' is the matrix consisting of the components <math>\frac{\partial f_i}{\partial x_j}</math>, q.e.d.
 
'''Proof of (**).''' From (*) it follows that
 
:<math>\|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\|.</math>
 
Here we have used the following
 
'''Lemma.''' Let ''v'' : [''a'', ''b''] → '''R'''<sup>''m''</sup> be a continuous function defined on the interval [''a'', ''b''] ⊂ '''R'''. Then we have
 
<math>\text{(***)}\qquad \left\|\int_a^b v(t)\,dt\right\|\leq \int_a^b \|v(t)\|\,dt.</math>
 
'''Proof of (***).''' Let ''u'' in '''R'''<sup>''m''</sup> denote the value of the integral
 
:<math>u:=\int_a^b v(t)\,dt.</math>
 
Now
 
:<math>\|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  \leq \int_a^b \| v(t) \|\cdot \|u \|\,dt = \|u\| \int_a^b \|v(t)\|\,dt,</math>
 
thus <math>\| u\| \leq \int_a^b \|v(t)\|\,dt</math> as desired. (Note the use of the [[Cauchy&ndash;Schwarz inequality]].) This shows (***) and thereby finishes the proof of (**).
 
== Mean value theorems for integration ==
{{unreferenced section|date=April 2013}}
===First mean value theorem for integration===
The '''first mean value theorem for integration''' states 
 
:If ''G'' : [''a'', ''b''] → '''R''' is a continuous function and <math> \varphi </math> is an integrable function that does not change sign on the interval (''a'', ''b''), then there exists a number ''x'' in [''a'', ''b''] such that
 
::<math>\int_a^b G(t)\varphi (t) \, dt=G(x) \int_a^b \varphi (t) \, dt.</math>
 
In particular, if φ(''t'') = 1 for all ''t'' in [''a'', ''b''], then there exists ''x'' in (''a'', ''b'') such that
 
:<math>\int_a^b G(t) \, dt=\ G(x)(b - a).\,</math>
 
When presented in the equivalent form
:<math>\frac{1}{b-a} \int_a^b G(t) \, dt=\ G(x),</math>
the theorem's conclusion says that the ''mean value'' of ''G''(''t'') on [''a'', ''b'']
(which is defined by the left side) is achieved as the ''point value'' ''G''(''x'')
for some ''x'' in (''a'', ''b'').
 
===Proof of the first mean value theorem for integration===
Without loss of generality assume the one-signed function <math>\varphi(t)\ge 0</math> for all ''t'' (the negative case just changes direction of some inequalities).
It follows from the [[extreme value theorem]] that the continuous function ''G'' has a finite infimum ''m'' and a finite supremum ''M'' on the interval [''a'', ''b'']. From the monotonicity of the integral and the fact that ''m'' ≤ ''G''(''t'') ≤ ''M'', it follows from the non-negativity of <math>\varphi(t)</math> that
 
:<math>m I= \int_a^b m\varphi(t)\,dt \le \int^b_aG(t)\varphi(t) \, dt \le \int_a^b M\varphi(t)\,dt = M I,</math>
 
where
 
:<math>I:=\int^b_a\varphi(t) \, dt</math>
 
denotes the integral of <math>\varphi(t)</math>. Hence, if ''I'' = 0, then the claimed equality holds for every ''x'' in [''a'', ''b'']. Therefore, we may assume ''I'' > 0 in the following. Dividing through by ''I'' we have that
 
:<math>m \le \frac1I\int^b_aG(t)\varphi(t) \, dt\le M.</math>
 
The [[extreme value theorem]] tells us more than just that the infimum and supremum of ''G'' on [''a'', ''b''] are finite; it tells us that both are actually attained.  Thus we can apply the [[intermediate value theorem]], and conclude that the continuous function ''G'' attains every value of the interval [''m'', ''M''], in particular there exists ''x'' in [''a'', ''b''] such that
 
:<math>G(x) = \frac1I\int^b_aG(t)\varphi(t) \, dt.</math>
 
This completes the proof.
 
===Second mean value theorem for integration===
There are various slightly different theorems called the '''second mean value theorem for integration'''.  A commonly found version is as follows:
 
:If ''G'' : [''a'', ''b''] → '''R''' is a positive [[monotone function|monotonically decreasing]] function and φ : [''a'', ''b''] → '''R''' 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+0) \int_a^x \varphi(t)\,dt. </math>
 
Here ''G''(''a'' + 0) stands for <math>{\underset{a_+}{\lim}G}</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>E. W. Hobson (1909). On the Second Mean-Value Theorem of the Integral Calculus. ''Proc. London Math Soc.'' S2-7, no. 1, pp.14-23. [http://www.ams.org/mathscinet-getitem?mr=1575669 MR1575669]. Available online: [http://plms.oxfordjournals.org/content/s2-7/1/14].</ref>
 
:If ''G'' : [''a'', ''b''] → '''R''' is a [[monotone function|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
 
::<math> \int_a^b G(t)\varphi(t)\,dt = G(a+0) \int_a^x \varphi(t)\,dt + G(b-0) \int_x^b \varphi(t)\,dt. </math>
 
==A probabilistic analogue of the mean value theorem==
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\geq 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>A. Di Crescenzo (1999). A probabilistic analogue of the mean value theorem and its applications to reliability theory. [[Applied Probability Trust|J. Appl. Prob.]] 36, 706-719.</ref> 
 
::<math> {\rm E}[g(Y)]-{\rm E}[g(X)]={\rm E}[g'(Z)]\,[{\rm E}(Y)-{\rm E}(X)].</math>
 
== 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:<ref name=PlanetMath>{{cite web|title=Complex Mean-Value Theorem|url=http://planetmath.org/ComplexMeanValueTheorem.html|work=[[PlanetMath]]|publisher=[[PlanetMath]]}}</ref>
 
{{calculus}}
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<sub>ab</sub>'' (the line segment from ''a'' to ''b'') such that
::<math>\mathrm{Re}(f'(u)) = \mathrm{Re}\left( \frac{f(b)-f(a)}{b-a} \right),</math>
::<math>\mathrm{Im}(f'(v)) = \mathrm{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 ==
* [[Newmark-beta method]]
* [[Mean value theorem (divided differences)]]
 
== Notes ==
{{reflist}}
 
== External links ==
* {{springer|title=Cauchy theorem|id=p/c020990}}
* [http://planetmath.org/encyclopedia/MeanValueTheorem.html PlanetMath: Mean-Value Theorem]
* {{MathWorld|Mean-ValueTheorem|Mean value theorem}}
* {{MathWorld|CauchysMean-ValueTheorem|Cauchy's Mean-Value Theorem}}
* [http://www.khanacademy.org/video/mean-value-theorem "Mean Value Theorem: Intuition behind the Mean Value Theorem"] at the [[Khan Academy]]
 
{{DEFAULTSORT:Mean Value Theorem}}
[[Category:Articles containing proofs]]
[[Category:Theorems in calculus]]
[[Category:Theorems in real analysis]]

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