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8.1 Finding the Average Value of a Function on an Interval

Syllabus
2020
Topic
8.1
Level
—

Average Value Is Accumulation per Unit Interval

For a continuous function on [a,b][a,b], the average value is its total signed accumulation divided by the interval length. It is the constant height whose rectangle of width b−ab-a has the same signed area as the integral of ff.

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

  1. Identify the complete interval [a,b][a,b].
  2. Evaluate ∫abf(x) dx\int_a^b f(x)\,dx.
  3. Divide by the positive interval length b−ab-a.
  4. Report the result in the same units as ff.

For f(x)=x2f(x)=x^2 on [0,3][0,3], favg=13−0∫03x2 dx=13[x3/3]03=13(9)=3f_{\mathrm{avg}}=\frac{1}{3-0}\int_0^3x^2\,dx=\frac13[x^3/3]_0^3=\frac13(9)=3. The accumulated value is 99 function-units times input-units; dividing by the interval length 33 leaves the units of ff.

Average value uses 1b−a∫abf(x) dx\frac{1}{b-a}\int_a^b f(x)\,dx; average rate of change uses f(b)−f(a)b−a\frac{f(b)-f(a)}{b-a}. They answer different questions. Because the integral is signed, an average value may be zero or negative even when the graph encloses positive geometric area elsewhere.