The average value of the function is given by the following expression:

average = `1/(3-1)int_1^3g(t)dt`

= 1/2 `int_1^3g(t)dt`

Lets calculate the given integral first:

Make a substitution, x = 3+t^2, which means dx = 2t dt or dt = (1/2t) dx

thus, g(t) dt = t dt/(3+t^2)^(1/2) = t. (1/2t) dx /x^(1/2) = (1/2) dx/x^(1/2)

and the limits will change to (3+1^2) = 4 and (3+3^2) = 12

Making this substitution, we get

average = (1/2) `int_4^12f(x)dx` = (1/2) .`int_4^12 (1/2)(1/x^(1/2))dx` = (1/4) (2 x^(1/2)) (with limits 4 and 12) =

= `1/2 (sqrt(12)-2)`

Find average value of `t/sqrt(3+t^2)`

over `[1,3]`

`1/(3-1) int_1^3 t/sqrt(3+t^2) dt`

We will solve this integral using u substitution.

Let `u = 3+t^2`

`du = 2t dt`

`(du) / 2 = t dt`

Substitute the u and du into the equation.

`1/(3-1) int (1/2)(du)/sqrt(u)`

`1/4 *2sqrt(u)`

`1/2 sqrt(3+t^2)`

evaluate the limits 3 and 1

`1/2 (sqrt(12) - sqrt(4))`

`1/2 (sqrt(12) - 2)`