What is the average value of the function #f(x) = 3/x# on the interval #[1,2e]#?

Question

1513 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-12T16:40:51

#3(ln2+1)/(2e-1)~~1.1449#

The average value can be given by:

#1/(b-a)int_a^bf(x)dx#

In this question we can use:

#1/(2e-1)int_1^(2e)3/xdx#

Evaluating the integrand we get:

#1/(2e-1)[3ln(x)]_1^(2e)=3/(2e-1)(ln(2e)-ln(1))#

We can now tidy this up a little, using the rules of logarithms to re write:

#=3/(2e-1)(ln2 +lne -ln1)#

#=3(ln2+1)/(2e-1)#

We can leave it at this exact value or approximate the solution by calculator to be:

#~~1.1449#