How do you find the average value of #f(x)=4x^(1/2)# as x varies between #[0,3]#?
1 Answer
Dec 1, 2016
The average value of
Explanation:
# = 4/3[ {: 2/3 x^(3/2)]_0^3]#
# = 4/3 [2/3(3^(3/2) - 0)]#
# = 4/3[2/3 (3sqrt3)]#
# = (8sqrt3)/3#