Since you are given a function and the interval, you can use the formula:

where:

`f(x) = f(t) = 6t e^(0.05t^(2))`

b = 10

a = 6

Plug in the given in the formula:

`f_(ave) =int_6^10(6te^(0.05t^(2)))dt`

You can move out 6 so it will look like this:

`f_(ave) = (6)int_6^10(e^(0.05t^(2)))tdt`

` `

To get the integral of `te^(0.05t^(2))`

` `

you can use substitution method, where `e^(0.05t^(2))`

is in the form `e^(u)`

so u = `0.05t^(2)`

Get the derivative of u, so du = 2 * (0.05) t = 0.1t dt.

From that you can get `tdt = (du)/(0.1)`

The integral of `e^(u) du = e^(u)`

so `(6) int (e^(u)/(.01)) du = 60 e^(u)`

Now plug-in `0.05t^(2)`

in place of u then evaluate from 6 to 10.

`60 e^(0.05(10)^(2)) - 60 e^(0.05(6)^(2)) = 8541.8106 mm`

`f_(ave) = 1/(10-6)int_6^10(6te^(0.05t^2))`

`f_(ave) = 1/4 * (8541.8106)`

`f_(ave) = 2135.45265`

``The answer is `2135.45265 mm.`