Question

How do you find the average value of the function for `f(x)=(3x)/sqrt(1-x^2), -1/2<=x<=1/2`?

Answer 1

The average value is `0`.

Explanation:

The average value of a function `f(x)` continuous and defined on `[a, b]` is given by

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

So our equation will be

`A = 1/(1/2 - (-1/2)) int _(-1/2)^(1/2) (3x)/sqrt(1 - x^2) dx`

`A = int_(-1/2)^(1/2) (3x)/sqrt(1 - x^2) dx`

We can integrate this using the substitution `u = 1 - x^2`. Then `du = -2x dx` and `dx= (du)/(-2x)`.

`A = int_(3/2)^(1/2) (3x)/(sqrt(u) * -2x) du`

`A = -3/2int_(-1/2)^(1/2) 1/sqrt(u) du`

`A = -3/2 int_(-1/2)^(1/2) u^(-1/2) du`

`A = -3/2[2u^(1/2)]_(-1/2)^(1/2)`

But you can't evaluate this just yet. We haven't reverted to the initial variable, and since we didn't change the bounds, we can't evaluate in `u`.

`A = -3/2[2(1 - x^2)^(1/2)]_(-1/2)^(1/2)`

`A = -3/2(2(sqrt(3)/2) - 2(sqrt(3)/2))`

`A = -3/2(0)`

`A = 0`

The average value therefore is `0`. This makes complete sense if you look at the graph--the function is symmetric about the y-axis, and therefore if `|a| = |b|`, then the average value will be `0`.

graph{y = (3x)/sqrt(1 - x^2) [-10, 10, -5, 5]}

Hopefully this helps!