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

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 $\left[a,b\right]$

• The average value will usually be different for different intervals

Do not confuse average value (this section) with average rate of change (Unit 2). They are different quantities:

• Average value of $f$ on $[a,b]$ is $\frac{1}{b-a}{\int }_a^{b}f\left(x\right) \text{d}x$(an integral)

• Average rate of change of $f$ on $[a,b]$ is $\frac{f\left(b\right)-f\left(a\right)}{b-a}$ (a difference quotient)

AP FRQs sometimes ask both in the same problem to test that you can tell them apart.

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

Calculate the average value of $f$ over the interval $\left[0,\pi \right]$.

Answer:

Use average value of $f$ on $\left[a,b\right]$$=\frac{1}{b-a}{\int }_a^{b}f\left(x\right)dx$

$\begin{array}{rcl}\mathrm{average}\mathrm{value}& =& \frac{1}{\pi -0}{\int }_0^{\pi }\sin xdx\\ & =& \frac{1}{\pi }{\left[-\cos x\right]}_0^{\pi }\\ & =& \frac{1}{\pi }\left(-\cos \left(\pi \right)-\left(-\cos \left(0\right)\right)\right)\\ & =& \frac{1}{\pi }\left(-\left(-1\right)-\left(-1\right)\right)\\ & =& \frac{1}{\pi }\left(1+1\right)\\ & =& \frac{2}{\pi }\end{array}$

Average value $=\frac{2}{\pi }$

Always write the setup in an FRQ, such as $\begin{array}{rcl}& & \frac{1}{\pi -0}{\int }_0^{\pi }\sin xdx\end{array}$ in the example above.