Average Value of a Function (College Board AP® Calculus AB)

Revision Note

Author

Roger B

Expertise

Maths

Average Value of a Function

What is the average value of a function?

If $f$ is a continuous function, then the average value of $f$ over the interval $[a, b]$ is
- average value of $f$ on $[a, b]$ $= \frac{1}{b - a} \int_{a}^{b} f (x) d x$
The average value of a function will be a number $k$
- where $k = f (c)$ for some $c$ in $[a, b]$
  - and such that $k \cdot (b - a) = \int_{a}^{b} f (x) d x$
- This result is referred to as the mean value theorem for integrals
This means that the constant function $g$ defined by $g (x) = k$
- will represent the same accumulation of change as $f$ between $x = a$ and $x = b$
  - Because $\int_{a}^{b} k d x = k {[x]}_{a}^{b} = k (b - a)$
This can also be interpreted geometrically, as seen in the following diagram

Graph showing the Mean Value Theorem for integrals, with a curve y=f(x) and a horizontal line y=k. Two shaded areas are shown with equal areas. Text explains the theorem.

Exam Tip

Remember that you can't talk about the 'average value of a function' in general

The average value is only defined for a particular interval $[a, b]$
The average value will usually be different for different intervals

Worked Example

Let $f$ be the function defined by $f (x) = \sin x$ .

Calculate the average value of $f$ over the interval $[0, π]$ .

Answer:

Use average value of $f$ on $[a, b]$ $= \frac{1}{b - a} \int_{a}^{b} f (x) d x$

$\begin{array}{rcl} average value & = & \frac{1}{π - 0} \int_{0}^{π} \sin x d x \\ = & \frac{1}{π} {[- \cos x]}_{0}^{π} \\ = & \frac{1}{π} (- \cos (π) - (- \cos (0))) \\ = & \frac{1}{π} (- (- 1) - (- 1)) \\ = & \frac{1}{π} (1 + 1) \\ = & \frac{2}{π} \end{array}$

average value $= \frac{2}{π}$

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