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

##### 1 Answer

The average value is

#### Explanation:

The average value of a function

#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

#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

#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

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

Hopefully this helps!