What is the average value of the function #f(x)=2/x # on the interval #[1,10]#? Calculus Derivatives Average Rate of Change Over an Interval 1 Answer Eddie Jun 24, 2016 #= 2/9 ln(10)# Explanation: #f(x)=2/x# By definition: #f_{ave} = (\int_a^b \ f(x) \ dx)/(b-a)# here we have #f_{ave} = (\int_1^{10} \ 2/x \ dx)/(10-1)# #= ( [ 2ln(x) ]_1^{10})/9# #= 2/9 ln(10)# Answer link Related questions How do you find the average rate of change of a function from graph? How do you find the average rate of change of a function between two points? How do you find the average rate of change of #f(x) = sec(x)# from #x=0# to #x=pi/4#? How do you find the average rate of change of #f(x) = tan(x)# from #x=0# to #x=pi/4#? How do you find the rate of change of y with respect to x? How do you find the average rate of change of #y=x^3+1# from #x=1# to #x=3#? What is the relationship between the Average rate of change of a fuction and derivatives? What is the difference between Average rate of change and instantaneous rate of change? What does the Average rate of change of a linear function represent? What is the relationship between the Average rate of change of a function and a secant line? See all questions in Average Rate of Change Over an Interval Impact of this question 1817 views around the world You can reuse this answer Creative Commons License