# Applications of Integration – Mean Value Theorem for Integrals – Average Value of a Function

In arithmetic, we learned how to find the average value of a dataset of points. We simply added all of the numbers then divided that sum by the number of addends:

However, let’s say that we wanted to find the average value of a function over a specific interval. That would be difficult, because a function has an infinite amount of data points.

However, we can use our knowledge of integration to find the average value of a function.

In previous sections, we learned that the integral of a function, or the area under a curve, is the accumulation, or the sum, of all the values over a specified interval. Therefore, for a function, the sum of the values inside an interval is the integral.

The length of the interval will be equal to the number of values available. In conclusion, for a function, the average value can be written as:

Let’s take a look at the formal definition of the mean value theorem.

 Definition: Average Value of a Function If there is a function F(x), which is closed on the interval of [a, b], and that function is continuous and positive, then there must exists a number x = c in that interval such that: Essentially, the average value of a function multiplied by the length of the interval is equal to the integral of the function over that interval. With f(c) being the average value of the function over the interval, the formula can be rewritten as:

Let’s look at a geometrical representation of the mean value theorem for integrals:

For the curve f(x), f(c) is the average value of the function over the interval. c is the x value that gives us the average value of the function. If we draw a rectangle of a height f(c), with a base length that extends from a to b, the area of that rectangle is equivalent to the integral of the function on the interval [a,b].

For a positive function, we can think of the mean value theorem as saying area/width = average height:

Notice how the mean value theorem for integration is similar to the mean value theorem for differentiation:

 Guidelines for applying Mean Value Theorem to a Function: Find the definite integral of the function over the specified interval: Find the length of the interval: b – a Apply the formula to obtain f(c) To find the value(s) of c, set the value f(c) equal to the original function and solve for the variable. Note: If you obtain multiple values of c, use the value of c that falls inside the interval.

Example 1: Find the average value of the function f(x) = x on the interval [1,2]. Find the x value of the average value.

Solution:

Find the definite integral of the function over the specified interval:

Using the power rule,

Find the length of the interval: b – a

Length of interval: b – a = 2 – 1 = 1

Apply the formula to obtain f(c)

f(c) = 1/1[3/2]

f(c) = 3/2 or 1.5

The average value of definite integral is 3/2.

Set the value f(c) equal to the original function and solve for the variable.

f(x) = f(c)

x = 3/2 or 1.5

Therefore, x =3/2 is the value that gives us the average value.

Let’s look at a geometric representation of our results:

Example 2: Find the average value of the function f(x) = 2x-1 on the interval [0,1]. Find the x value of the average value.

Solution:

Find the definite integral of the function over the specified interval:

Use the power rule,

Find the length of the interval: b – a

Length of interval: b – a = 1 – 0 = 1

Apply the formula to obtain f(c)

f(c) = 1/1[0]

f(c) = 0

The average value of the function is zero.

Set the value f(c) equal to the original function and solve for the variable.

f(x) = f(c)

2x-1 = 0

2x = 1

x = ½ or 0.5

Therefore, x = ½ is the value that gives us the average value.

Let’s look at a geometric representation of our results:

One may look at the graphical representation and wonder if this makes sense.

Remember, the average value of the function f(c) is the height of the rectangle. In this case, the value of f(c), or the average height of the rectangle, is zero. Therefore, the area of the rectangle is zero. The net area under the curve, or the definite integral, is also zero, since the areas below and above the x axis are the same but have opposite signs. In conclusion, the graphical representation still aligns with the mean value theorem.

Example 3: Find the average value of the function f(x) = cos(x) on the interval [0,π/2]. Find the x value of the average value.

Solution:

Set the value f(c) equal to the original function and solve for the variable.

f(x) = f(c)

cos(x) = -2/π

arccos(-2/π) = x

arccos(2/π) = x

0.881 = x

Therefore, x = 0.881 is the value that gives us the average value.

Let’s look at a geometric representation of our results:

Example 4: Find the average value of the function f(x) = x2+5x+16 on the interval [0,2]. Find the x value of the average value.

Solution:

f(c) = 1/2 [44.667]

f(c) = 22.333

The average value of the function is 22.333.

Set the value f(c) equal to the original function and solve for the variable.

f(x) = f(c)

x2+5x+16 = 22.333

x2+5x-6.333 = 0

Using the quadratic formula, let’s solve for x:

x = -6.047 or 1.047

We will use the x value that belongs on the interval of [0,2]. x = 1.047 is the x value that satisfies this.

Therefore, x = 1.047 is the value that gives us the average value of the function.

Let’s look at a geometric representation of our results:

 Summary of Section: The Mean Value Theorem for Integrals states that for every definite integral , a rectangle with the same area and width (w = b-a) exists. The top of the rectangle, which intersects the curve, f(c), is the average value of the function. The average value of the function, the width of the interval, and the definite integral are related to each other by the formula: Which states that the average value of the function is equal to the definite integral divided by the length of the interval.