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

Question

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

Evelyn Agard · Accepted Answer

The average value is #0#.

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!

Samantha Adkison · Answer

undefined To find the average value of the function $ f(x) = \frac{3x}{\sqrt{1-x^2}} $ over the interval $-\frac{1}{2} \leq x \leq \frac{1}{2}$, you need to evaluate the definite integral of $f(x)$ over that interval and then divide it by the length of the interval.

1. First, compute the definite integral of $f(x)$ over the given interval:
$$ \int_{-1/2}^{1/2} \frac{3x}{\sqrt{1-x^2}} \, dx $$

2. Next, evaluate this integral.

3. Then, find the length of the interval, which is $ \frac{1}{2} - (-\frac{1}{2}) = 1 $.

4. Finally, divide the result from step 2 by the length of the interval (step 3) to find the average value of the function over the interval.

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

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