Question

Question #c319c

Answer 1

`2sqrt3`

Explanation:

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`