How do you find the average value of #f(x)=-x^5+3x^3# as x varies between #[0,1]#?

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Answer 1 · 2016-11-25T22:31:46

#7/12#

The average value of the function #f# on #[a,b]# is found through

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

Which here gives us the average value

#1/(1-0)int_0^1(-x^5+3x^3)dx#

Integrating term by term using #intx^ndx=x^(n+1)/(n+1)# gives

#=[-x^6/6+(3x^4)/4]_0^1#

The #0# part of this will still be #0#, leaving us with just

#-1/6+3/4=7/12#