Key takeaway: The average value of a function on an interval is its definite integral over that interval divided by the interval’s length, the height of a rectangle with the same area.
Grade 12 – AP Calculus ABAP Calculus AB Unit 8Calculus
Definition of Average Value of a Function
The average value of a function on \([m,n]\) generalizes the ordinary average to infinitely many inputs. Instead of adding a few outputs and dividing by their count, integrate to add all outputs and divide by the length of the interval.
Picture the area under the curve being leveled into a rectangle with the same base. The height of that rectangle is the average value. For a continuous function, the curve reaches this height at least once inside the interval.
Formula and notation
For a continuous function \(w\) on \([m,n]\):
\[ w_{\text{avg}}=\frac{1}{n-m}\int_{m}^{n} w(x)\,dx. \]
The factor \(\frac{1}{n-m}\) is one over the interval length. The input where the curve equals this value is guaranteed by the mean value theorem for integrals.
Key facts
- The result has the same units as the function, not the units of the integral.
- A function that is constant has an average equal to that constant.
- The average is not simply the mean of the endpoint outputs, except for straight lines.
- Total area under the curve equals average value times interval length.
Average Value of a Function example
Find the average value of \(f(x)=x^2+2\) on \([1,4]\), and the input where the function takes that value.
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Integrate: \(\int_{1}^{4}(x^2+2)\,dx=[\frac{x^3}{3}+2x]_{1}^{4}=(\frac{64}{3}+8)-(\frac{1}{3}+2)=27\).
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Divide by the length \(4-1=3\): \(\frac{27}{3}=9\).
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Solve \(x^2+2=9\): \(x=\sqrt{7}\approx 2.65\), which lies in \([1,4]\).
The average value is 9, reached at \(x=\sqrt{7}\approx 2.65\).
Common mistake
Watch out
A student averages the endpoint outputs, \(\frac{f(1)+f(4)}{2}=\frac{3+18}{2}=10.5\), as the average value of a function.
Why it is wrong
Two endpoints do not capture a curved graph. The parabola climbs slowly at first and quickly later, so the middle of the interval is weighted differently. The integral accounts for every input and gives 9, not 10.5.
Related terms and lessons
Browse the glossary: A–Z list · AP Calculus AB vocabulary · Calculus