Average Value of a Function Revision Notes for Leaving Cert Mathematics

Average Value of a Function

What does the average value of a function mean?

The average value of a function provides a single representative value of a function over an interval. It is a way to determine the "mean height" of the graph of the function across a specified interval.

For a continuous function $f(x)$ defined on the interval $[a, b]$ , the average value is calculated using the formula:

f_{\text{avg}} = \frac{1}{b-a} \int_a^b f(x) \, dx

Where:

$\int_a^b f(x) \, dx$ is the definite integral of $f(x)$ over $[a, b]$ , representing the total "area under the curve."
$b-a$ is the length of the interval, scaling the total area to compute the average. This concept has applications in physics, engineering, and economics, where it is used to find averages of rates, costs, or quantities that vary over a range.

Interpretation

The average value $f_{\text{avg}}$ is equivalent to the constant height of a horizontal line that would enclose the same area under the curve $y = f(x)$ over $[a, b]$ .

Worked Examples

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Example 1: Average Value of a Linear Function

Problem:

Find the average value of $f(x) = 2x + 3$ on the interval $[1,4]$

Solution:

Step 1: Write the formula for the average value:

f_{\text{avg}} = \frac{1}{b-a} \int_a^b f(x) \, dx

Step 2: Identify the interval and function:

$a = 1, b = 4$ and $f(x) = 2x + 3$

Step 3: Set up the integral:

f_{\text{avg}} = \frac{1}{4-1} \int_1^4 (2x + 3) \, dx

Step 4: Simplify the fraction and integrate:

f_{\text{avg}} = \frac{1}{3} \left[ \int_1^4 2x \, dx + \int_1^4 3 \, dx \right]

Step 5: Solve each integral:

\int_1^4 2x \, dx = \left[x^2 \right]_1^4 = 4^2 - 1^2 = 16 - 1 = 15

\int_1^4 3 \, dx = \left[3x \right]_1^4 = 3(4) - 3(1) = 12 - 3 = 9

Step 6: Add the results and multiply by $\frac{1}{3}$

f_{\text{avg}} = \frac{1}{3} (15 + 9) = \frac{1}{3} \times 24 = 8

Answer: The average value of $f(x) = 2x + 3$ on $[1, 4]$ is $8$

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Example 2: Average Value of a Quadratic Function

Problem:

Find the average value of $f(x) = x^2$ on the interval $[0,3]$

Solution:

Step 1: Write the formula for the average value:

f_{\text{avg}} = \frac{1}{b-a} \int_a^b f(x) \, dx

Step 2: Identify the interval and function:

$a = 0, b = 3$ and $f(x) = x^2$

Step 3: Set up the integral:

f_{\text{avg}} = \frac{1}{3-0} \int_0^3 x^2 \, dx

Step 4: Simplify the fraction and integrate:

f_{\text{avg}} = \frac{1}{3} \int_0^3 x^2 \, dx = \frac{1}{3} \left[\frac{x^3}{3}\right]_0^3

Step 5: Evaluate the integral:

\left[\frac{x^3}{3}\right]_0^3 = \frac{3^3}{3} - \frac{0^3}{3} = \frac{27}{3} - 0 = 9

Step 6: Multiply by $\frac{1}{3}$

f_{\text{avg}} = \frac{1}{3} \times 9 = 3

Answer: The average value of $f(x) = x^2$ on $[0, 3]$ is $3$ .

Summary

The average value of a function $f(x)$ over an interval $[a,b]$ is given by:

f_{\text{avg}} = \frac{1}{b-a} \int_a^b f(x) \, dx

The formula calculates the "mean height" of the function over the interval.
Steps to calculate the average value:
1. Identify the interval $[a, b]$ and the function $f(x)$
2. Set up the formula and the integral.
3. Compute the definite integral:

\int_a^b f(x) \, dx

Divide by the interval length $b - a$

Applications include physics, engineering, and economics to find averages of varying rates or quantities.

Average Value of a Function (Leaving Cert Mathematics): Revision Notes