Question #c319c

Question

1166 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-09T21:48:16

#2sqrt3#

The average value #s# of the function #f# on the interval #[a,b]# is equal to

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

So, for #f(x)=sqrt(3x)# on #[0,9]# we see that the average value #s# is

#s=1/(9-0)int_0^9sqrt(3x)color(white).dx#

Since #sqrt(3x)=sqrt3(sqrtx)=sqrt3(x^(1/2))#:

#s=sqrt3/9int_0^9x^(1/2)color(white).dx#

First, note that #sqrt3/9=3^(1/2)/3^2=1/3^(3/2)=1/(3sqrt3)#.

Second, for the actual integration, recall that #intx^n=x^(n+1)/(n+1)#:

#s=1/(3sqrt3)[x^(3/2)/(3/2)]_0^9=1/(3sqrt3)[2/3x^(3/2)]_0^9#

Evaluating:

#s=1/(3sqrt3)(2/3(9)^(3/2)-2/3(0)^(3/2))#

Note that #9^(3/2)=(3^2)^(3/2)=3^3=27#:

#s=1/(3sqrt3)(2/3(27)-0)=18/(3sqrt3)=6/sqrt3=(2sqrt3sqrt3)/sqrt3=2sqrt3#