How do you find the average value of the function #f(x) = 12/(sqrt(2x-1))#
on the interval of [1,5]?

Question

Charles Arocha · Accepted Answer

Have a look:

Emily Ammerman · Answer

The idea for positive functions:
The average value is the height of a horizontal line that gives a rectangle on the interval whose area is equal to the area under the curve. Average value of #f(x)# on #[a,b]# is  #1/(b-a) int_a^b f(x) dx# So find
# int_1^5 12/sqrt(2x-1) dx# (By substitution.) #= 12 sqrt(2x-1)]_1^5# #=12(3) - 12(1) = 24# Now divide by the length of the interval: #24/(5-1) = 24/4 = 6# The average value is #6#

Elijah Abram · Answer

undefined To find the average value of the function $ f(x) = \frac{12}{\sqrt{2x-1}} $ on the interval [1,5], follow these steps:

1. Find the definite integral of the function $ f(x) $ over the interval [1,5].
2. Divide the result from step 1 by the length of the interval [1,5], which is $ 5 - 1 = 4 $.

The formula to find the average value $ \bar{f} $ of a function $ f(x) $ on the interval [a,b] is:

$$ \bar{f} = \frac{1}{b - a} \int_{a}^{b} f(x) \, dx $$

Using this formula, integrate $ f(x) $ over the interval [1,5]:

$$ \bar{f} = \frac{1}{4} \int_{1}^{5} \frac{12}{\sqrt{2x-1}} \, dx $$

After integrating, you'll have the average value of the function over the interval [1,5].

How do you find the average value of the function #f(x) = 12/(sqrt(2x-1))# on the interval of [1,5]?