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 $f(x)$ is continuous, a point $c$ exists in an interval $\left[a,b\right]$ such that the value of the function at $c$ is equal to the average value of $f(x)$ over $\left[a,b\right].$ 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 $f(x)$ is continuous over an interval $\left[a,b\right],$ then there is at least one point $c\in \left[a,b\right]$ such that

f (c) = \frac{1}{b - a} \int_{a}^{b} f (x) d x .

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

This formula can also be stated as

\int_{a}^{b} f (x) d x = f (c) (b - a) .

${\displaystyle\int }_{a}^{b}f(x)dx=f(c)(b-a).$

Proof

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

m (b - a) \leq \int_{a}^{b} f (x) d x \leq M (b - a) .

$m(b-a)\le {\displaystyle\int }_{a}^{b}f(x)dx\le M(b-a).$

Dividing by $b-a$ gives us

m \leq \frac{1}{b - a} \int_{a}^{b} f (x) d x \leq M .

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

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

f (c) = \frac{1}{b - a} \int_{a}^{b} f (x) d x,

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

and the proof is complete.

$_\blacksquare$

Example: Finding the Average Value of a Function

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

Show Solution

The formula states the mean value of

f (x)

$f(x)$ is given by

\frac{1}{4 - 0} \int_{0}^{4} (8 - 2 x) d x .

$\frac{1}{4-0}{\displaystyle\int }_{0}^{4}(8-2x)dx.$

We can see in Figure 1 that the function represents a straight line and forms a right triangle bounded by the $x$ – and $y$ -axes. The area of the triangle is $A=\frac{1}{2}(\text{base})(\text{height}).$ We have

A = \frac{1}{2} (4) (8) = 16.

$A=\frac{1}{2}(4)(8)=16.$

The average value is found by multiplying the area by $\frac{1}{(4-0)}.$ Thus, the average value of the function is

\frac{1}{4} (16) = 4.

$\frac{1}{4}(16)=4.$

Set the average value equal to $f(c)$ and solve for $c$ .

\begin{matrix} 8 - 2 c & = & 4 c & = & 2 \end{matrix}

$\begin{array}{ccc}8-2c\hfill & =\hfill & 4\hfill \\ \hfill c& =\hfill & 2\hfill \end{array}$

At $c=2,f(2)=4.$

The graph of a decreasing line f(x) = 8 – 2x over [-1,4.5]. The line y=4 is drawn over [0,4], which intersects with the line at (2,4). A line is drawn down from (2,4) to the x-axis and from (4,4) to the y axis. The area under y=4 is shaded.

Figure 1. By the Mean Value Theorem, the continuous function $f(x)$ takes on its average value at c at least once over a closed interval.

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

Closed Captioning and Transcript Information for Video

Try It

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

Hint

Show Solution

Average value

= 1.5

$=1.5$ ;

c = 3

$c=3$

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

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

Show Solution

We are looking for the value of $c$ such that

$f(c)=\frac{1}{3-0}{\displaystyle\int }_{0}^{3}{x}^{2}dx=\frac{1}{3}(9)=3.$

Replacing $f(c)$ with $c$ ², we have

$\begin{array}{ccc}{c}^{2}\hfill & =\hfill & 3\hfill \\ c\hfill & =\hfill & \text{±}\sqrt{3}.\hfill \end{array}$

Since $\text{−}\sqrt{3}$ is outside the interval, take only the positive value. Thus, $c=\sqrt{3}$ (Figure 2).

A graph of the parabola f(x) = x^2 over [-2, 3]. The area under the curve and above the x-axis is shaded, and the point (sqrt(3), 3) is marked.

Figure 2. Over the interval $\left[0,3\right],$ the function $f(x)={x}^{2}$ takes on its average value at $c=\sqrt{3}.$

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

Closed Captioning and Transcript Information for Video

Try It

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

Hint

Show Solution

$c=\sqrt{3}$

Module 1: Integration