Editing Mean value theorem (section)

== 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.