5.2.2 Mean Value of a Function

Understanding the Mean Value of a Function

The mean value of a function $f(x)$ over an interval $[a, b]$ is the average value of $f(x)$ across all points in the interval. It is calculated using the formula:

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

This formula computes the integral (the "total area under the curve") and divides it by the length of the interval, $b - a$ . The result represents a constant height that the function would have if it were a rectangle with the same area over the interval.

Key Steps to Evaluate the Mean Value

Set up the formula: Write the formula for the mean value:

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

Evaluate the integral: Compute $\int_a^b f(x) \, dx$ using standard integration techniques.
Divide by the interval length: Divide the result of the integral by $b - a$ .

Worked Examples

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Example 1**:** Find the mean value of $f(x) = x^2$ over the interval $[1, 3]$

Step 1: Set up the formula:

\text{Mean Value} = \frac{1}{3-1} \int_1^3 x^2 \, dx = \frac{1}{2} \int_1^3 x^2 \, dx

Step 2: Evaluate the integral:

The antiderivative of $x^2$ is $\frac{x^3}{3}$ :

\int_1^3 x^2 \, dx = \left[ \frac{x^3}{3} \right]_1^3 = \frac{3^3}{3} - \frac{1^3}{3} = \frac{27}{3} - \frac{1}{3} = \frac{26}{3}

Step 3: Divide by the interval length:

\text{Mean Value} = \frac{1}{2} \times \frac{26}{3} = \frac{26}{6} = \frac{13}{3}

Result:

The mean value of $f(x) = x^2$ over $[1, 3]$ is $\frac{13}{3}$ .

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Example 2: Find the mean value of $f(x) = e^{-x}$ over the interval $[2, 4]$ .

Step 1: Set up the formula:

\text{Mean Value} = \frac{1}{4-2} \int_2^4 e^{-x} \, dx = \frac{1}{2} \int_2^4 e^{-x} \, dx

Step 2: Evaluate the integral:

The antiderivative of $e^{-x}$ is $-e^{-x}$ :

\int_2^4 e^{-x} \, dx = \left[ -e^{-x} \right]_2^4 = -e^{-4} + e^{-2}

Step 3: Simplify:

\int_2^4 e^{-x} \, dx = e^{-2} - e^{-4}

Step 4: Divide by the interval length:

\text{Mean Value} = \frac{1}{2} (e^{-2} - e^{-4})

Result: The mean value of $f(x) = e^{-x}$ over $[2, 4]$ is $\frac{1}{2} (e^{-2} - e^{-4})$

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Example 3**:** Find the mean value of $f(x) = \sin x$ over the interval $[0, \pi]$

Step 1: Set up the formula:

\text{Mean Value} = \frac{1}{\pi-0} \int_0^\pi \sin x \, dx = \frac{1}{\pi} \int_0^\pi \sin x \, dx

Step 2: Evaluate the integral: The antiderivative of $\sin x$ is $\cos x$ :

\int_0^\pi \sin x \, dx = \left[ -\cos x \right]_0^\pi = -\cos(\pi) + \cos(0) = -(-1) + 1 = 2

Step 3: Divide by the interval length:

\text{Mean Value} = \frac{1}{\pi} \times 2 = \frac{2}{\pi}

Result: The mean value of $f(x) = \sin x\ \text{over}\ [0,π][0, \pi]\ \text{is}\ :highlight[$ \frac{2}{\pi}$]

Note Summary

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Common Mistakes:

Forgetting to divide by $b-a$ : The mean value is the integral divided by the interval length, not just the integral.
Incorrect integration bounds: Always use the correct bounds $a$ and $b$ for the interval.
Overlooking simplifications: When integrating functions with clear antiderivatives (e.g., polynomials or trigonometric functions), simplify where possible.
Misapplying the formula: The mean value formula only applies to definite integrals over a specific interval.
Prematurely approximating results: Keep exact forms (e.g., $\pi, e^{-x}$ ) unless explicitly asked for a decimal approximation.

infoNote

Key Formulas:

Mean Value Formula:

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

Area Under a Curve:

A = \int_a^b f(x) \, dx

Common Integrals:

$\int x^n \, dx = \frac{x^{n+1}}{n+1} + C (for ) n \neq -1$
$\int e^x \, dx = e^x + C$
$\int \sin x \, dx = -\cos x + C$

Mean Value of a Function Simplified Revision Notes for A-Level Edexcel Further Maths Core Pure

5.2.2 Mean Value of a Function

Understanding the Mean Value of a Function

Key Steps to Evaluate the Mean Value

Worked Examples

Note Summary

Common Mistakes:

Key Formulas:

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