Mean Value Theorem: Difference between revisions
mNo edit summary |
mNo edit summary |
||
| Line 1: | Line 1: | ||
The '''Mean Value Theorem''' states that if <math>a < b</math> are [[real number]]s and the [[function]] <math>f:[a,b] \to \mathbb{R}</math> is [[ | The '''Mean Value Theorem''' states that if <math>a < b</math> are [[real number]]s and the [[function]] <math>f:[a,b] \to \mathbb{R}</math> is [[differentiable]] on the [[interval]] <math>(a,b)</math>, then there exists a value <math>c</math> in <math>(a,b)</math> such that | ||
<cmath>f(c)=\dfrac{1}{b-a}\int_{a}^{b}f(x)dx.</cmath> | <cmath>f(c)=\dfrac{1}{b-a}\int_{a}^{b}f(x)dx.</cmath> | ||
Revision as of 11:03, 19 November 2016
The Mean Value Theorem states that if 
are real numbers and the function ![$f:[a,b] \to \mathbb{R}$](http://latex.artofproblemsolving.com/f/d/b/fdb437e0031d3b146dc3c59922d4a94e15a7ec29.png)
is differentiable on the interval 
, then there exists a value 
in such that
![\[f(c)=\dfrac{1}{b-a}\int_{a}^{b}f(x)dx.\]](http://latex.artofproblemsolving.com/1/a/8/1a8347c593c9df7eca6e55210e2b2e5a448bdff9.png)
In words, there is a number in such that 
equals the average value of the function in the interval ![$[a,b]$](http://latex.artofproblemsolving.com/8/e/c/8ecbd1ba3da8f2adef66a63f2ab32c47e63fa734.png)
.
This article is a stub. Help us out by expanding it.
