How do you find the average value of #f(x)=x-2sqrtx# as x varies between #[0,2]#?
1 Answer
Apr 9, 2018
Average value is given by
#A = 1/(b - a) int_a^b f(x)dx#
Therefore
#A = 1/2int_0^2 x - 2sqrt(x)dx#
#A = 1/2[1/2x^2 - 4/3x^(3/2)]_0^2#
#A = 1/2[2 - 4/3sqrt(8)]#
#A = 1 - 2/3(2)sqrt(2)#
#A = 1 - 4/3sqrt(2) ~~-0.89#
Hopefully this helps!
