Mean Value of a Function

The mean value (or average value) of a function over an interval is a powerful concept in Further Mathematics. It connects integration to the idea of averaging, and provides elegant geometric and physical interpretations.

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1 Definition

The mean value of a continuous function f(x) over the interval [a, b] is:

f_mean = (1 / (b - a)) * integral from a to b of f(x) dx

This formula says: divide the total area under the curve by the width of the interval. The result is the height of a rectangle with the same base [a, b] and the same area as the region under the curve.

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2 Geometric Interpretation

Consider the curve y = f(x) between x = a and x = b. The area under this curve is:

A = integral from a to b of f(x) dx

Now imagine a horizontal line y = f_mean such that the rectangle with base (b - a) and height f_mean has exactly the same area A. Then:

(b - a) * f_mean = A

so f_mean = A / (b - a), which matches our formula.

This rectangle is sometimes called the mean value rectangle. Its height equals the average height of the function across the interval.

Visual Idea: The mean value "levels out" all the bumps and dips of the curve into a single flat height that encloses the same total area.

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3 Worked Example 1

Find the mean value of f(x) = x^2 over the interval [0, 3].

Solution:

Step 1: Compute the integral.

integral from 0 to 3 of x^2 dx = [x^3 / 3] from 0 to 3 = 27/3 - 0 = 9

Step 2: Divide by (b - a) = 3 - 0 = 3.

f_mean = 9 / 3 = 3

The mean value of x^2 over [0, 3] is 3.

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4 Worked Example 2

Find the mean value of sin x over [0, pi].

Solution:

integral from 0 to pi of sin x dx = [-cos x] from 0 to pi = -cos(pi) - (-cos(0)) = -(-1) + 1 = 2

f_mean = 2 / (pi - 0) = 2/pi

This is approximately 0.637. Notice that the maximum of sin x on [0, pi] is 1, and the minimum is 0, so the average 2/pi sits between them — as expected.

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5 Worked Example 3

Find the mean value of e^x over [0, 1].

Solution:

integral from 0 to 1 of e^x dx = [e^x] from 0 to 1 = e - 1

f_mean = (e - 1) / (1 - 0) = e - 1

This is approximately 1.718.

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6 The Mean Value Theorem for Integrals

This theorem guarantees that the mean value is actually attained somewhere on the interval:

Theorem: If f is continuous on [a, b], then there exists at least one point c in (a, b) such that:

f(c) = (1 / (b - a)) * integral from a to b of f(x) dx

In other words, the function actually takes its mean value at some point in the interval.

Application

For the example f(x) = x^2 on [0, 3], we found f_mean = 3. We need c such that c^2 = 3, giving c = sqrt(3) (approximately 1.732), which lies in (0, 3). The theorem is satisfied.

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7 Mean Value for Trigonometric Functions

Worked Example 4

Find the mean value of cos x over [0, pi/2].

Solution:

integral from 0 to pi/2 of cos x dx = [sin x] from 0 to pi/2 = 1 - 0 = 1

f_mean = 1 / (pi/2 - 0) = 1 / (pi/2) = 2/pi

Interestingly, the mean value of cos x over [0, pi/2] equals the mean value of sin x over [0, pi]. This is not a coincidence — it follows from the relationship sin x = cos(pi/2 - x).

Worked Example 5

Find the mean value of sin^2 x over [0, 2pi].

Solution:

Use the identity sin^2 x = (1 - cos 2x) / 2.

integral from 0 to 2pi of sin^2 x dx = integral from 0 to 2pi of (1 - cos 2x)/2 dx

= [x/2 - sin(2x)/4] from 0 to 2pi

= (2pi/2 - sin(4pi)/4) - (0 - 0) = pi

f_mean = pi / (2pi - 0) = pi / (2pi) = 1/2