How do you find the average value of #f(x)=-x^3+7x^2-11x+3# as x varies between #[1,5]#? Calculus Applications of Definite Integrals The Average Value of a Function 1 Answer A. S. Adikesavan Nov 11, 2016 #-2# Explanation: Average #=(int f(x) dx)/(5-1)#, between x = 1 and x = 5 #(1/4)[-x^4/4+7x^3/3-11x^2/3+3x],# between the limits #=(1/4)[-(5^4-1)+7(5^3-1)-11(5^2-1)+3(5-1)]# #=(1/4)[-624+868-264+12]# #=(1/4)(-8)# #=-2# 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 1844 views around the world You can reuse this answer Creative Commons License