Lesson 4: Average Value of a Function
Estimated time: 30-40 minutes
Learning Objectives
By the end of this lesson, you will be able to:
- Define and compute the average value of a continuous function on an interval
- State and apply the Mean Value Theorem for Integrals
- Interpret the average value geometrically
- Apply average value in real-world contexts
Motivation: Averaging Continuously
The average of n numbers x1, x2, ..., xn is (x1 + x2 + ... + xn)/n. But how do we average a function that takes on infinitely many values over an interval?
The answer comes from recognizing that a Riemann sum of f over [a, b] with n equal subintervals is Δx · ∑ f(xi). Dividing the integral by the interval length (b − a) gives us the continuous analog of a discrete average.
The Average Value Formula
Average Value of a Function
If f is continuous on [a, b], the average value of f on [a, b] is:
favg = (1/(b − a)) ∫ab f(x) dx
Example 1: Average Value of a Quadratic
Find the average value of f(x) = x² on [0, 3].
Solution: favg = (1/(3 − 0)) ∫03 x² dx = (1/3) [x³/3]03 = (1/3)(9) = 3.
Example 2: Average Value of a Trig Function
Find the average value of f(x) = sin x on [0, π].
Solution: favg = (1/π) ∫0π sin x dx = (1/π) [−cos x]0π = (1/π)(1 + 1) = 2/π ≈ 0.637.
Geometric Interpretation
The average value favg is the height of the rectangle with base [a, b] whose area equals the area under the curve:
favg · (b − a) = ∫ab f(x) dx
In other words, a horizontal line at height favg creates a rectangle with exactly the same area as the region under f.
Mean Value Theorem for Integrals
Mean Value Theorem for Integrals (MVT-I)
If f is continuous on [a, b], then there exists a number c in [a, b] such that:
f(c) = (1/(b − a)) ∫ab f(x) dx = favg
That is, a continuous function actually attains its average value at some point.
Example 3: Finding c from MVT-I
For f(x) = x² on [0, 3], we found favg = 3. Find the value c where f(c) = 3.
Solution: c² = 3, so c = √3 ≈ 1.732. (We take the positive root since c must be in [0, 3].) The function attains its average value at x = √3.
Example 4: MVT-I with Exponential
Find the average value of f(x) = ex on [0, 2] and the value c where it is attained.
Solution: favg = (1/2) ∫02 ex dx = (1/2)(e² − 1) ≈ 3.195.
Find c: ec = (e² − 1)/2, so c = ln[(e² − 1)/2] ≈ 1.161.
Applications
Example 5: Average Temperature
The temperature in a city over 12 hours is modeled by T(t) = 60 + 15 sin(πt/12) (in degrees F). Find the average temperature from t = 0 to t = 12.
Solution: Tavg = (1/12) ∫012 [60 + 15 sin(πt/12)] dt
= (1/12) [60t − 15(12/π) cos(πt/12)]012
= (1/12) [(720 − (180/π)(−1)) − (0 − (180/π)(1))]
= (1/12) [720 + 360/π] = 60 + 30/π ≈ 69.55°F.
Example 6: Average Velocity
A particle moves with velocity v(t) = 3t² − 6t on [0, 4]. Find the average velocity.
Solution: vavg = (1/4) ∫04 (3t² − 6t) dt = (1/4)[t³ − 3t²]04 = (1/4)(64 − 48) = 4.
Note: average velocity = displacement / time = (s(4) − s(0)) / 4, consistent with the net change theorem.
Average Value vs. Average Rate of Change
Do not confuse:
- Average value of f on [a, b]: (1/(b − a)) ∫ab f(x) dx. This is the average height of f.
- Average rate of change of f on [a, b]: [f(b) − f(a)] / (b − a). This is the slope of the secant line.
These are conceptually different quantities. However, the average value of f' equals the average rate of change of f (by FTC).
Key Takeaways
- Average value: favg = (1/(b − a)) ∫ab f(x) dx.
- Geometrically, favg is the height of a rectangle with the same area as the region under f.
- MVT for Integrals: a continuous function attains its average value at some point c in [a, b].
- Average value of f' = average rate of change of f (a nice FTC connection).
Check Your Understanding
1. Find the average value of f(x) = 4x − x² on [0, 4].
2. Find the average value of f(x) = cos x on [0, π/2].
3. For f(x) = x² on [1, 3], find the value c guaranteed by MVT-I.
4. True or false: The average value of a non-negative function on [a, b] is always non-negative.