Understanding the Average Value of a Function

The average value of a function is a fundamental concept in calculus. It calculates the mean value of a function over a specific interval, making it useful in analyzing performance metrics, trends, and cumulative behaviors in various fields, including mathematics, engineering, and physics.

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What is the Average Value of a Function?

The average value of a function over an interval $[a, b]$ represents the mean output of the function across that range. This is calculated using definite integrals and provides a valuable measure of how the function behaves overall within a given interval.

Formula:

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

Components:

• $f(x)$: The function whose average value is being computed.
• $[a, b]$: The interval over which the average value is calculated.
• $\int_a^b f(x) \, dx$: The integral representing the total area under the curve.

This formula divides the total area under the curve by the interval length, yielding the average value.

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How to Interpret the Average Value of a Function

The average value of a function is essentially the height the function would maintain if its area were evenly distributed across the interval. This single value simplifies the analysis of cumulative behavior in continuous systems.

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Steps to Calculate the Average Value

1. Define the Function and Interval:
   Start with the function $f(x)$ and its interval $[a, b]$.

2. Integrate Over the Interval:
   Compute $\int_a^b f(x) \, dx$ using integral calculus.

3. Divide by Interval Length:
   Multiply the integral by $\frac{1}{b - a}$ to obtain the average value.

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Examples

Example 1: $f(x) = x^2$ over $[0, 3]$

1. Compute the Integral: $$\int_0^3 x^2 \, dx = \left[\frac{x^3}{3}\right]_0^3 = \frac{27}{3} - 0 = 9$$
2. Divide by the Interval Length: $$f_{\text{avg}} = \frac{1}{3 - 0} \cdot 9 = 3$$

Result: The average value is $3$.

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Example 2: $f(x) = \sin(x)$ over $[0, \pi]$

1. Compute the Integral: $$\int_0^\pi \sin(x) \, dx = \left[-\cos(x)\right]_0^\pi = -\cos(\pi) + \cos(0) = 2$$
2. Divide by the Interval Length: $$f_{\text{avg}} = \frac{1}{\pi - 0} \cdot 2 = \frac{2}{\pi}$$

Result: The average value is $\frac{2}{\pi}$.

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Practice Problem

Question 1:

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

Answer:

$\boxed{17}$

Explanation:

Step 1: Set Up the Average Value Formula

The average value $f_{\text{avg}}$ of a function $f(x)$ over the interval $[a, b]$ is given by:

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

For $f(x) = 2x^2 + 3$ over $[1, 4]$:

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

Step 2: Compute the Integral

Integrate $f(x)$:

$$\int (2x^2 + 3) \, dx = \frac{2}{3}x^3 + 3x + C$$

Evaluate from 1 to 4:

$$\left[ \frac{2}{3}x^3 + 3x \right]_1^4 = \left( \frac{2}{3}(64) + 12 \right) - \left( \frac{2}{3}(1) + 3 \right) = \frac{128}{3} + 12 - \frac{2}{3} - 3 = \frac{153}{3} = 51$$

Step 3: Calculate the Average Value

$$f_{\text{avg}} = \frac{1}{3} \times 51 = 17$$

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Question 2:

Determine the average value of $f(x) = \sin(x)$ on the interval $[0, \pi]$.

Answer:

$\boxed{\dfrac{2}{\pi} \approx 0.6366}$

Explanation:

Step 1: Set Up the Average Value Formula

$$f_{\text{avg}} = \frac{1}{\pi - 0} \int_{0}^{\pi} \sin(x) \, dx = \frac{1}{\pi} \int_{0}^{\pi} \sin(x) \, dx$$

Step 2: Compute the Integral

Integrate $f(x)$:

$$\int \sin(x) \, dx = -\cos(x) + C$$

Evaluate from 0 to $\pi$:

$$\left[ -\cos(x) \right]_0^{\pi} = -\cos(\pi) - (-\cos(0)) = 1 + 1 = 2$$

Step 3: Calculate the Average Value

$$f_{\text{avg}} = \frac{2}{\pi} \approx 0.6366$$

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Question 3:

Determine the average value of $f(x) = e^{x}$ over the interval $[0, 1]$.

Answer:

$\boxed{e - 1 \approx 1.71828}$

Explanation:

Step 1: Set Up the Average Value Formula

$$f_{\text{avg}} = \frac{1}{1 - 0} \int_{0}^{1} e^{x} \, dx = \int_{0}^{1} e^{x} \, dx$$

Step 2: Compute the Integral

Integrate $f(x)$:

$$\int e^{x} \, dx = e^{x} + C$$

Evaluate from 0 to 1:

$$\left[ e^{x} \right]_0^1 = e^{1} - e^{0} = e - 1 \approx 2.71828 - 1 = 1.71828$$

Step 3: Calculate the Average Value

$$f_{\text{avg}} = e - 1 \approx 1.71828$$

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Question 4:

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

Answer:

$\boxed{1}$

Explanation:

Step 1: Set Up the Average Value Formula

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

Step 2: Compute the Integral

Integrate $f(x)$:

$$\int (x^3 - 4x + 1) \, dx = \frac{1}{4}x^4 - 2x^2 + x + C$$

Evaluate from $-2$ to $2$:

$$\left[ \frac{1}{4}x^4 - 2x^2 + x \right]_{-2}^{2} = \left( \frac{1}{4}(16) - 2(4) + 2 \right) - \left( \frac{1}{4}(16) - 2(4) - 2 \right) = (-2) - (-6) = 4$$

Step 3: Calculate the Average Value

$$f_{\text{avg}} = \frac{1}{4} \times 4 = 1$$

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