Question

How do you find the average value of the function for `f(x)=sinxcosx, 0<=x<=pi/2`?

Answer 1

The average value is `1/pi`.

Explanation:

The average value of a function `f(x)` on a closed interval `[a, b]` is given by

`A = 1/(b - a)int_a^b F(x)`

Where `A` is the average value and `f'(x) = F(x)`. Our expression will therefore be

`A = 1/(pi/2 - 0) int_0^(pi/2) sinxcosx dx`

`A = 2/piint_0^(pi/2) sinxcosxdx`

The trick here is to realize that `2sinxcosx = sin2x`. Therefore, `sinxcosx = 1/2sin2x`.

`A = 2/piint_0^(pi/2) 1/2sin2xdx`

Now let `u = 2x`. Then `du = 2dx` and `dx = (du)/2`. The bounds of integration become `0` to `pi`.

`A = 2/piint_0^(pi) 1/4sinu du`

`A = 1/4(2/pi)int_0^(pi)sinudu`

`A = 1/(2pi)int_0^(pi)sinudu`

`A = 1/(2pi)[-cosu]_0^(pi)`

`A = 1/(2pi)[-cos2x]_0^(pi/2)`

`A = 1/(2pi)(-cospi - (-cos0))`

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

`A = 1/pi`

Hopefully this helps!