How do you find the average value of #f(x)=4x^(1/2)# as x varies between #[0,3]#?

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Answer 1 · 2016-12-01T19:14:15

The average value of #f# on #[a,b]# is #1/(b-a)int_a^bf(x) dx#

#1/(3-0) int_0^3 4x^(1/2)dx = 4/3 int_0^3 x^(1/2) dx#

# = 4/3[ {: 2/3 x^(3/2)]_0^3]#

# = 4/3 [2/3(3^(3/2) - 0)]#

# = 4/3[2/3 (3sqrt3)]#

# = (8sqrt3)/3#