What is the average value of a function #y=x^2-2x+4 # on the interval #[0,8]#? Calculus Applications of Definite Integrals The Average Value of a Function 1 Answer Eddie Jun 28, 2016 # = 52/3# Explanation: By definition #y_{ave} = (int_a^b \ y \ dx)/(b - a)# so here #y_{ave} = (int_0^8 \ x^2 - 2x + 4 \ dx)/(8-0)# # = 1/8 [ (x^3)/3 - x^2 + 4x]_0^8 # # = 52/3# Answer link 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 3192 views around the world You can reuse this answer Creative Commons License