How do you find the average value of the function for #f(x)=x^2-sqrtx, 0<=x<=3#?

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Answer 1 · 2017-10-05T16:15:30

#3-(2sqrt(3))/3#

Recall the formula for determining the average value of a function #f(x)# on an interval from #[a,b]#:

#"Avg. Value" = 1/(b-a) int_a^b f(x) dx#

For this problem:

#Avg = 1/(3-0) int_0^3 (x^2-sqrt(x)) dx#

#= 1/3 int_0^3 (x^2 - x^(1//2)) dx #

#= 1/3 [x^3/3 - 2/3 x^3/2]_0^3 #

#= 1/3(3^3/3 - 2/3 3^3/2) = 1/3 (27/3 - 2/3*3sqrt(3)) #

#= 1/3 (9-2sqrt(3)) = 3-(2sqrt(3))/3 #