Question

What is the average value of the function `f(x)=cos(16x)` on the interval `[0,pi/2]`?

Answer 1

This question asks for average value, but is posted under the topic "average rate of change" Here are solutions to both.

Explanation:

Average Rate of Change

The average rate of change of a continuous function `f` on a closed interval `[a,b]` is `(f(b)-f(a))/(b-a)`.

In this case, we have `(cos(16 (pi/2))-cos(0))/(pi/2-0) = (1-1)/(pi/2) =0`

The question asks for Average Value

The average value of an continuous function `f` on interval `[a,b]` is

`1/(b-a) int_a^b f(x) dx`.

Here, we have
`1/(pi/2-0) int_0^(pi/2) cos 16x dx =2/pi [1/16 sin(16x)]_0^(pi/2)`

`= 1/(8pi) [sin(8pi)-sin(0)] = 1/(8pi) [0] = 0`