Mean Value of a Function (Edexcel A-Level Further Mathematics): Revision Notes

📚 Revision Notes
Revision NotesModel AnswersQuizzesFlashcards

5.2.2 Mean Value of a Function

Understanding the Mean Value of a Function

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

Mean Value=1b−a∫abf(x) dx\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−ab - 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

  1. Set up the formula: Write the formula for the mean value:
Mean Value=1b−a∫abf(x) dx\text{Mean Value} = \frac{1}{b-a} \int_a^b f(x) \, dx
  1. Evaluate the integral: Compute ∫abf(x) dx\int_a^b f(x) \, dx using standard integration techniques.
  2. Divide by the interval length: Divide the result of the integral by b−ab - a.

Worked Examples

lightbulbExample

Example 1**:** Find the mean value of f(x)=x2f(x) = x^2 over the interval [1,3][1, 3]

Step 1: Set up the formula:

Mean Value=13−1∫13x2 dx=12∫13x2 dx\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 x2x^2 is x33\frac{x^3}{3}:

∫13x2 dx=[x33]13=333−133=273−13=263\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:

Mean Value=12×263=266=133\text{Mean Value} = \frac{1}{2} \times \frac{26}{3} = \frac{26}{6} = \frac{13}{3}

Result:

The mean value of f(x)=x2f(x) = x^2 over [1,3][1, 3] is 133\frac{13}{3}.

lightbulbExample

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

Step 1: Set up the formula:

Mean Value=14−2∫24e−x dx=12∫24e−x dx\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−xe^{-x} is −e−x-e^{-x}:

∫24e−x dx=[−e−x]24=−e−4+e−2\int_2^4 e^{-x} \, dx = \left[ -e^{-x} \right]_2^4 = -e^{-4} + e^{-2}

Step 3: Simplify:

∫24e−x dx=e−2−e−4\int_2^4 e^{-x} \, dx = e^{-2} - e^{-4}

Step 4: Divide by the interval length:

Mean Value=12(e−2−e−4)\text{Mean Value} = \frac{1}{2} (e^{-2} - e^{-4})

Result: The mean value of f(x)=e−xf(x) = e^{-x} over [2,4][2, 4] is 12(e−2−e−4)\frac{1}{2} (e^{-2} - e^{-4})

lightbulbExample

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

Step 1: Set up the formula:

Mean Value=1π−0∫0πsin⁡x dx=1π∫0πsin⁡x dx\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\sin x is cos⁡x\cos x:

∫0πsin⁡x dx=[−cos⁡x]0π=−cos⁡(π)+cos⁡(0)=−(−1)+1=2\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:

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

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

Note Summary

infoNote

Common Mistakes:

  1. Forgetting to divide by b−ab-a: The mean value is the integral divided by the interval length, not just the integral.

  2. Incorrect integration bounds: Always use the correct bounds aa and bb for the interval.

  3. Overlooking simplifications: When integrating functions with clear antiderivatives (e.g., polynomials or trigonometric functions), simplify where possible.

  4. Misapplying the formula: The mean value formula only applies to definite integrals over a specific interval.

  5. Prematurely approximating results: Keep exact forms (e.g., π,e−x\pi, e^{-x}) unless explicitly asked for a decimal approximation.

infoNote

Key Formulas:

  1. Mean Value Formula:
Mean Value=1b−a∫abf(x) dx\text{Mean Value} = \frac{1}{b-a} \int_a^b f(x) \, dx
  1. Area Under a Curve:
A=∫abf(x) dxA = \int_a^b f(x) \, dx
  1. Common Integrals:
  • ∫xn dx=xn+1n+1+C(for)n≠−1\int x^n \, dx = \frac{x^{n+1}}{n+1} + C (for ) n \neq -1
  • ∫ex dx=ex+C\int e^x \, dx = e^x + C
  • ∫sin⁡x dx=−cos⁡x+C\int \sin x \, dx = -\cos x + C

Explore Edexcel A-Level Further Mathematics Model Answers by Topics

Volumes of Revolution

Methods in Calculus

Explore Edexcel A-Level Further Mathematics Quizzes by Topics

Volumes of Revolution

Methods in Calculus

Explore Edexcel A-Level Further Mathematics Flashcards by Topics

Volumes of Revolution

Methods in Calculus

Join 100,000+ A-Level students studying Revision Notes with us.

Select your subjects, and get access to A+ resources today.