The Mean Value Theorem for Integrals

Learning Outcomes

  • Describe the meaning of the Mean Value Theorem for Integrals

The Mean Value Theorem for Integrals states that a continuous function on a closed interval takes on its average value at the same point in that interval. The theorem guarantees that if [latex]f(x)[/latex] is continuous, a point [latex]c[/latex] exists in an interval [latex]\left[a,b\right][/latex] such that the value of the function at [latex]c[/latex] is equal to the average value of [latex]f(x)[/latex] over [latex]\left[a,b\right].[/latex] We state this theorem mathematically with the help of the formula for the average value of a function that we presented at the end of the preceding section.

The Mean Value Theorem for Integrals

If [latex]f(x)[/latex] is continuous over an interval [latex]\left[a,b\right],[/latex] then there is at least one point [latex]c\in \left[a,b\right][/latex] such that

[latex]f(c)=\frac{1}{b-a}{\displaystyle\int }_{a}^{b}f(x)dx.[/latex]

 

This formula can also be stated as

[latex]{\displaystyle\int }_{a}^{b}f(x)dx=f(c)(b-a).[/latex]

 

Proof

Since [latex]f(x)[/latex] is continuous on [latex]\left[a,b\right],[/latex] by the extreme value theorem (see Maxima and Minima), it assumes minimum and maximum values—[latex]m[/latex] and M, respectively—on [latex]\left[a,b\right].[/latex] Then, for all [latex]x[/latex] in [latex]\left[a,b\right],[/latex] we have [latex]m\le f(x)\le M.[/latex] Therefore, by the comparison theorem (see The Definite Integral), we have

[latex]m(b-a)\le {\displaystyle\int }_{a}^{b}f(x)dx\le M(b-a).[/latex]

 

Dividing by [latex]b-a[/latex] gives us

[latex]m\le \frac{1}{b-a}{\displaystyle\int }_{a}^{b}f(x)dx\le M.[/latex]

 

Since [latex]\frac{1}{b-a}{\displaystyle\int }_{a}^{b}f(x)dx[/latex] is a number between [latex]m[/latex] and M, and since [latex]f(x)[/latex] is continuous and assumes the values [latex]m[/latex] and M over [latex]\left[a,b\right],[/latex] by the Intermediate Value Theorem (see Continuity), there is a number [latex]c[/latex] over [latex]\left[a,b\right][/latex] such that

[latex]f(c)=\dfrac{1}{b-a}{\displaystyle\int}_{a}^{b}f(x)dx,[/latex]

 

and the proof is complete.

[latex]_\blacksquare[/latex]

Example: Finding the Average Value of a Function

Find the average value of the function [latex]f(x)=8-2x[/latex] over the interval [latex]\left[0,4\right][/latex] and find [latex]c[/latex] such that [latex]f(c)[/latex] equals the average value of the function over [latex]\left[0,4\right].[/latex]

Watch the following video to see the worked solution to Example: Finding the Average Value of a Function.

Try It

Find the average value of the function [latex]f(x)=\dfrac{x}{2}[/latex] over the interval [latex]\left[0,6\right][/latex] and find [latex]c[/latex] such that [latex]f(c)[/latex] equals the average value of the function over [latex]\left[0,6\right].[/latex]

Hint

Use the procedures from the last example to solve the problem

example: FINDING THE POINT WHERE A FUNCTION TAKES ON ITS AVERAGE VALUE

Given [latex]{\displaystyle\int }_{0}^{3}{x}^{2}dx=9,[/latex] find [latex]c[/latex] such that [latex]f(c)[/latex] equals the average value of [latex]f(x)={x}^{2}[/latex] over [latex]\left[0,3\right].[/latex]

Watch the following video to see the worked solution to Example: Finding the Point Where a Function Takes on its Average Value.

Try It

Given [latex]{\displaystyle\int }_{0}^{3}(2{x}^{2}-1)dx=15,[/latex] find [latex]c[/latex] such that [latex]f(c)[/latex] equals the average value of [latex]f(x)=2{x}^{2}-1[/latex] over [latex]\left[0,3\right].[/latex]