What is the average value of a function #y=6/x# on the interval #[1,e]#?

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Answer 1 · 2016-07-30T23:30:16

#6/(e-1)approx3.4918602412...#

The average value of the function #f(x)# on the interval #[a,b]# can be expressed as:

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

So, where the function is #f(x)=6/x# and the interval is #[1,e]#, the average value of the function is:

#1/(e-1)int_1^e 6/xdx#

The #6# can be brought out of the integrand:

#=6/(e-1)int_1^e 1/xdx#

Note that since the derivative of #ln(x)# is #1/x#, the integral of #1/x# is #ln(x)#.

#=6/(e-1)[ln(x)]_1^e#

Now evaluating:

#=6/(e-1)[ln(e)-ln(1)]#

#=6/(e-1)(1-0)#

#=6/(e-1)#