Science

Math

Humanities

... and beyond

Socratic Meta
Featured Answers

Topics

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

1 Answer

Dec 16, 2016

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

Explanation:

#1/(3-2)int_2^3 1/x dx = ln(3/2)#

Related questions

The profit (in dollars) from the sale of x lawn mowers is ...
What is the average value of the function #f(x)=(x-1)^2# on the interval from x=1 to x=5?
What is the average value of the function #u(x) = 10xsin(x^2)# on the interval #[0,sqrt pi]#?
What is the average value of the function #f(x)=cos(x/2) # on the interval #[-4,0]#?
What is the average value of the function #f(x) = x^2# on the interval #[0,3]#?
What is the average value of the function #f(t)=te^(-t^2 )# on the interval #[0,5]#?
What is the average value of the function #f(x) = x - (x^2) # on the interval #[0,2]#?
What is the average value of the function #f(t) = t (sqrt (1 + t^2) )# on the interval #[0,5]#?
What is the average value of the function #f(x) = sec x tan x# on the interval #[0,pi/4]#?
What is the average value of the function #f(x) = 2x^3(1+x^2)^4# on the interval #[0,2]#?

See all questions in The Average Value of a Function

Impact of this question

1336 views around the world

You can reuse this answer
Creative Commons License