To find the average value of `f(x)` on the interval [a,b], you compute:

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

So in our case, we would do:

`1/(pi-0) int_0^pi "sin" x dx`

`=1/pi [-"cos" x]|_0^pi`

`=1/pi (1 + 1) = 2/pi`

So, the average value of `"sin" x` on `[o,pi]` is `2/pi`...

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To find the average value of `f(x)` on the interval [a,b], you compute:

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

So in our case, we would do:

`1/(pi-0) int_0^pi "sin" x dx`

`=1/pi [-"cos" x]|_0^pi`

`=1/pi (1 + 1) = 2/pi`

So, the average value of `"sin" x` on `[o,pi]` is `2/pi`

The value of x where the average occurs:

We want to find c in `[0,pi]` such that `"sin" c = 2/pi`

Take `"Sin"^(-1) (2/pi) ~~.690107 `

This is different from the mean value theorem.

The mean value theorem states that:

If f(x) is continuous on [a,b] and differentiable on (a,b), then there is some c in (a,b) where:

`f'(c) = (f(b)-f(a))/(b-a)`

That is, there is some place where the derivative is equal to the slope of the secant line that connects the endpoints

For us, sin(pi) = sin(0) = 0

f'(x) = cos (x)

So the mean value theorem says:

There is some c in [0,pi] such that:

`"cos" (c) = (0-0)/(pi-0) = 0`

And sure enough, at c=`pi/2` , cos(c) = 0