How do you find the average value of #f(x)=4x^(1/2)# as x varies between #[0,3]#?

Answer 1

Aurora Alvarado

The average value of #f# on #[a,b]# is #1/(b-a)int_a^bf(x) dx#

#1/(3-0) int_0^3 4x^(1/2)dx = 4/3 int_0^3 x^(1/2) dx#

# = 4/3[ {: 2/3 x^(3/2)]_0^3]#

# = 4/3 [2/3(3^(3/2) - 0)]#

# = 4/3[2/3 (3sqrt3)]#

# = (8sqrt3)/3#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Adam Adkins

To find the average value of (f(x) = 4x^{1/2}) as (x) varies between ([0,3]), you can use the formula for the average value of a function on a closed interval. It is given by (\frac{1}{b-a} \int_a^b f(x) , dx), where (a) and (b) are the limits of integration. Substitute (a = 0) and (b = 3) into this formula and calculate the definite integral of (f(x)) over the interval ([0,3]). Then divide the result by (b-a), which is (3 - 0 = 3). This will give you the average value of (f(x)) on the interval ([0,3]).

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.