Question

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

Answer 1

Average value is not the same as average rate of change.

Explanation:

The question is posted under "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 `(sin^2(pi)-sin^2(0))/(pi-0) = (0-0)/pi =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-0) int_0^pi sin^2 x dx = 1/pi (pi/2) = 1/2`