Unit 8 · Topic 8.1 Beta

Average Value of a Function

The average value of f on [a, b] is 1/(b − a) ∫ from a to b of f(x) dx , the height of the equal-area rectangle, in the units of f .

Practice 1: Implementing Mathematical ProcessesPractice 4: Communication and Notation

Question set for this topic

Part 1 · Hook

Why this matters

A weather report says the average temperature yesterday was 19 °C. It was never "19 for the whole day"; it was colder at dawn and warmer at three. The average is the one steady temperature that would have built up the same total. For a function, that total is an integral, and the average value is the integral divided by the length of the interval.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. ∫023x2dx\int_0^2 3x^2\,dx==

  1. 8
  2. 12
  3. 6
  4. 4
Show the answer

x3|02=8x^3\Big|_0^2 = 8.

  • Correct: 8:
  • 12:
  • 6:
  • 4:

2. The average rate of change of ff on [1,3][1, 3] is

  1. f(3)−f(1)3−1\frac{f(3) - f(1)}{3 - 1}
  2. f(3)+f(1)2\frac{f(3) + f(1)}{2}
  3. f′(2)f'(2)
  4. f′(3)−f′(1)2\frac{f'(3) - f'(1)}{2}
Show the answer

Change in output over change in input: the slope of the secant line.

  • Correct: f(3)−f(1)3−1\frac{f(3) - f(1)}{3 - 1}:
  • f(3)+f(1)2\frac{f(3) + f(1)}{2}:
  • f′(2)f'(2):
  • f′(3)−f′(1)2\frac{f'(3) - f'(1)}{2}:

Part 4 · See it

See it first

The parabola f(x) = 6x − x² on 0 ≤ x ≤ 6 with the area under it shaded in crimson, 36 square units. A dashed blue rectangle of width 6 and height 6 has the same area, so the average value of f on [0, 6] is 6. The curve meets the line y = 6 at x = 3 − √3 and x = 3 + √3, marked with blue points.
f(x)=6x−x2f(x) = 6x - x^2 on [0,6][0, 6]: the area under ff is 36, the same as a 6×66 \times 6 rectangle, so the average value is 6. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. ∫abf(x)dx\int_a^b f(x)\,dx is the signed area, or the total accumulatedspread it evenly over the interval to get one steady height
  2. Divide by the length b−ab - afavgf_{\text{avg}}=1b−a∫abf(x)dx= \frac{1}{b - a}\int_a^b f(x)\,dx, in the units of ff
  3. ff is continuous on [a,b][a, b]ff takes its average value at some cc in [a,b][a, b] (between its minimum and maximum, by the Intermediate Value Theorem)
  4. The integrand is f′f' instead of ff1b−a∫abf′(x)dx\frac{1}{b - a}\int_a^b f'(x)\,dx=f(b)−f(a)b−a= \frac{f(b) - f(a)}{b - a}, the average rate of change of ff

Part 6 · Key ideas

Key ideas

  • Average value of ff on [a,b][a, b]: 1b−a∫abf(x)dx\frac{1}{b - a}\int_a^b f(x)\,dx. The integrand is ff itself.
  • Average value is a height, in the units of ff. Average rate of change, f(b)−f(a)b−a\frac{f(b) - f(a)}{b - a}, is a slope, in units of ff per unit of xx.
  • Areas below the axis count as negative, exactly as in the integral.
  • If ff is continuous, some cc in [a,b][a, b] has f(c)=favgf(c) = f_{\text{avg}}.
  • From a graph use geometry for the integral; from a table use a Riemann or trapezoidal sum; from a formula use the Fundamental Theorem or a calculator.

Part 7 · Misconception

A common mistake

The wrong idea: The average value of ff on [0,8][0, 8] is f(0)+f(8)2\frac{f(0) + f(8)}{2}, or f(8)−f(0)8\frac{f(8) - f(0)}{8}.

What actually happens: The first averages two values and ignores the rest; the second is the average rate of change. The average value uses every value: 18∫08f(x)dx\frac{1}{8}\int_0^8 f(x)\,dx.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Graph

The graph of ff

12345678−3−2−11234567xy

1. Find the average value of ff on [0,8][0, 8].

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

∫08f(x)dx\int_0^8 f(x)\,dx=8= 8+12+ 12+6−2+ 6 - 2=24= 24 (trapezoid, rectangle, triangle, triangle below the axis), so the average is 248=3\frac{24}{8} = 3.

  • Answer: 3

2. Find the average value of ff on [2,6][2, 6].

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

∫26f(x)dx\int_2^6 f(x)\,dx=12= 12+6+ 6=18= 18, and 184=92\frac{18}{4} = \frac{9}{2}.

  • Answer: 92\frac{9}{2} ≈ 4.500

3. On [0,8][0, 8], how do the average value of ff and the average rate of change of ff compare?

  1. Both are 3.
  2. The average value is −12-\frac{1}{2}; the average rate of change is 3.
  3. The average value is 3; the average rate of change is −12-\frac{1}{2}.
  4. The average value is 0, because f(8)<0<f(0)f(8) < 0 < f(0).
Show the answer

Average value: area over length, 248=3\frac{24}{8} = 3. Average rate of change: rise over run, −48=−12\frac{-4}{8} = -\frac{1}{2}.

  • Both are 3.: One is a height and the other a slope; they are different numbers here.
  • The average value is −12-\frac{1}{2}; the average rate of change is 3.: The two are swapped.
  • Correct: The average value is 3; the average rate of change is −12-\frac{1}{2}.: 18∫08f\frac{1}{8}\int_0^8 f=3= 3, while f(8)−f(0)8\frac{f(8) - f(0)}{8}=−2−28= \frac{-2 - 2}{8}.
  • The average value is 0, because f(8)<0<f(0)f(8) < 0 < f(0).: The average value comes from the whole area, not the endpoint signs.

4. Find the average value of f(x)=x2f(x) = x^2 on [0,3][0, 3].

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

13∫03x2dx\frac{1}{3}\int_0^3 x^2\,dx=13⋅x33|03= \frac{1}{3} \cdot \frac{x^3}{3}\Big|_0^3=13⋅9= \frac{1}{3} \cdot 9=3= 3.

  • Answer: 3

5. Which expression gives the average value of gg on [−1,5][-1, 5]?

  1. 14∫−15g(x)dx\frac{1}{4}\int_{-1}^{5} g(x)\,dx
  2. 15∫−15g(x)dx\frac{1}{5}\int_{-1}^{5} g(x)\,dx
  3. 16∫−15g(x)dx\frac{1}{6}\int_{-1}^{5} g(x)\,dx
  4. g(5)−g(−1)6\frac{g(5) - g(-1)}{6}
Show the answer

gavgg_{\text{avg}}=15−(−1)∫−15g(x)dx= \frac{1}{5 - (-1)}\int_{-1}^{5} g(x)\,dx.

  • 14∫−15g(x)dx\frac{1}{4}\int_{-1}^{5} g(x)\,dx: 5−1=45 - 1 = 4 ignores the sign: the length is 5−(−1)5 - (-1).
  • 15∫−15g(x)dx\frac{1}{5}\int_{-1}^{5} g(x)\,dx: Divide by the length of the interval, 6, not by the right endpoint.
  • Correct: 16∫−15g(x)dx\frac{1}{6}\int_{-1}^{5} g(x)\,dx: The length of [−1,5][-1, 5] is 5−(−1)=65 - (-1) = 6.
  • g(5)−g(−1)6\frac{g(5) - g(-1)}{6}: That is the average rate of change of g.

6. For a function ff with continuous derivative, the average value of f′f' on [a,b][a, b] equals

  1. f′(b)−f′(a)b−a\frac{f'(b) - f'(a)}{b - a}
  2. f(b)−f(a)b−a\frac{f(b) - f(a)}{b - a}
  3. f(b)−f(a)f(b) - f(a)
  4. 1b−a∫abf(x)dx\frac{1}{b - a}\int_a^b f(x)\,dx
Show the answer

By the Fundamental Theorem, ∫abf′(x)dx\int_a^b f'(x)\,dx=f(b)−f(a)= f(b) - f(a); dividing by b−ab - a gives the average rate of change of ff.

  • f′(b)−f′(a)b−a\frac{f'(b) - f'(a)}{b - a}: That is the average rate of change of f′f', which is the average value of f″f''.
  • Correct: f(b)−f(a)b−a\frac{f(b) - f(a)}{b - a}: 1b−a∫abf′(x)dx\frac{1}{b - a}\int_a^b f'(x)\,dx=f(b)−f(a)b−a= \frac{f(b) - f(a)}{b - a} by the Fundamental Theorem.
  • f(b)−f(a)f(b) - f(a): That is ∫abf′(x)dx\int_a^b f'(x)\,dx, not yet divided by b−ab - a.
  • 1b−a∫abf(x)dx\frac{1}{b - a}\int_a^b f(x)\,dx: That is the average value of ff, not of f′f'.

Part 9 · Summary

Summary

The average value of ff on [a,b][a, b] is 1b−a∫abf(x)dx\frac{1}{b - a}\int_a^b f(x)\,dx, the height of the equal-area rectangle, in the units of ff. A continuous ff reaches it somewhere on the interval. Do not confuse it with the average rate of change f(b)−f(a)b−a\frac{f(b) - f(a)}{b - a}, which is the average value of f′f'.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections