How do you find the average value of #f(x)=(x-3)^2# as x varies between #[2,5]#?

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Answer 1 · 2018-02-14T13:38:36

The average value is #=1#

The average value of a function #f(x)# over an interval #[a,b]# is

#barx=1/(b-a)int_a^bf(x)dx#

Here,

#f(x)=(x-3)^2=x^2-6x+9#

and #[a,b]=[2,5]#

Therefore,

#barx=1/(5-2)int_2^5(x-3)^2dx#

#=1/3int_2^5(x^2-6x+9)dx#

#=1/3[x^3/3-6x^2/2+9x]_2^5#

#=1/3((125/3-75+45)-(8/3-12+18))#

#=1/3(117/3-36)#

#=1/3*3#

#=1#