# The Mean Value Theorem for Integrals

Calculating average value of function over interval | AP Calculus AB | Khan Academy
Calculating average value of function over interval | AP Calculus AB | Khan Academy

### 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

This formula can also be stated as

### 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

Dividing by $b-a$ gives us

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

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 $\left[0,4\right].$

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

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

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

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

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.

You can view the transcript for this segmented clip of “5.3 The Fundamental Theorem of Calculus” here (opens in new window).

### 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 $\left[0,6\right].$

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 $\left[0,3\right].$

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

Replacing $f(c)$ with $c$2, we have

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

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.

You can view the transcript for this segmented clip of “5.3 The Fundamental Theorem of Calculus” here (opens in new window).

### 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 $\left[0,3\right].$

$c=\sqrt{3}$

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