Average Value of a Function
In short: Average value of a function over [a, b] is the definite integral of the function divided by b minus a, the constant height that would give the same area.
AP Calculus ABCalculusAP Calculus AB Unit 8
What is Average Value of a Function?
The average value of a function answers the question: if the output were smoothed out into one steady height across an interval, how tall would it be? Averaging a list of numbers means adding and dividing by the count. A continuous function has infinitely many values, so the sum becomes an integral and the count becomes the length of the interval.
\[f_{\text{avg}} = \frac{1}{b – a}\int_a^b f(x)\,dx\]
Geometrically, a rectangle of that height over the same interval has exactly the same area as the region under the curve.
How to spot it
- The question asks for the average temperature, speed, cost or population during a time interval.
- A rate is given as a formula, not a table, and a typical value over the period is wanted.
- The wording “average value of a function” appears together with an interval [a, b].
- You are asked to find the input where the function equals its average, as in the mean value theorem for integrals.
The method
- To find the average value of a function, compute the interval length \(b – a\).
- Evaluate the definite integral \(\int_a^b f(x)\,dx\) with an antiderivative.
- Divide the integral by \(b – a\).
- To locate where the function takes that value, solve \(f(c)\) equal to the average for \(c\) in the interval.
Average Value of a Function: solved example
Problem
A coffee shop’s customer count per hour is modeled by \(C(t) = 3t^2 – 2t + 4\) for \(0 \le t \le 3\) hours after opening. Find the average rate over the first 3 hours.
Solution
Integral: \(\int_0^3 (3t^2 – 2t + 4)\,dt = \left[t^3 – t^2 + 4t\right]_0^3 = 27 – 9 + 12 = 30\).
Divide by the interval length 3: \(\frac{30}{3} = \) 10 customers per hour.
Try it
Question 1
Find the average value of \(f(x) = 4x^3\) on \([1, 3]\).
Show the answer
\(\frac{1}{2}\int_1^3 4x^3\,dx = \frac{1}{2}\left[x^4\right]_1^3 = \frac{81 – 1}{2} = 40\).
Question 2
Find the average value of \(\sin x\) on \([0, \pi]\).
Show the answer
\(\frac{1}{\pi}\int_0^{\pi} \sin x\,dx = \frac{2}{\pi} \approx 0.637\).
Average Value of a Function vs. Mean Value Theorem
The average value of a function compares the integral with the interval length and describes outputs. The mean value theorem compares the endpoint values and describes slopes: it finds a point where the derivative equals the average rate of change. They sit on opposite sides of the fundamental theorem.
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