How do you find the average value of #x^3# as x varies between -1 and 2? Calculus Applications of Definite Integrals The Average Value of a Function 1 Answer Eddie · Steve M Jan 23, 2017 #= 5/4# Explanation: Average value of #f(x)# over interval #[a,b]# is #bar (f) (x) = (int_a^b f(x) dx)/(b-a)# Here that means that: #bar (f) (x) = (int_(-1)^2 x^3 dx)/(2 - (-1))# #= ([ x^4/4]_(-1)^2)/3 = 5/4# 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 1485 views around the world You can reuse this answer Creative Commons License