Art of Problem Solving

Mean Value Theorem: Difference between revisions

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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 [[continuous]] on the [[interval]] <math>[a,b]</math>, then there exists a value <math>c</math> in <math>[a,b]</math> such that
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 [[continuous]] 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>


In words, there is a number <math>c</math> in <math>[a,b]</math> such that <math>f(c)</math> equals the average value of the function in the interval <math>[a,b]</math>.
In words, there is a number <math>c</math> in <math>(a,b)</math> such that <math>f(c)</math> equals the average value of the function in the interval <math>[a,b]</math>.


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Revision as of 19:56, 25 November 2013

The Mean Value Theorem states that if $a < b$ are real numbers and the function $f:[a,b] \to \mathbb{R}$ is continuous on the interval $[a,b]$, then there exists a value $c$ in $(a,b)$ such that

\[f(c)=\dfrac{1}{b-a}\int_{a}^{b}f(x)dx.\]

In words, there is a number $c$ in $(a,b)$ such that $f(c)$ equals the average value of the function in the interval $[a,b]$.

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