Question

How do you find the average value of the function `f(t)=4te^(-t^2)` on the interval [0, 5]?

Answer 1

It is `1/5(2-2/e^25) ~~ 2/5`

Explanation:

For a function that is non-negative on an interval `[a,b]` -- like this function is on `[0,5]` -- the average value has a nice geometric explanation:
It is the height that a rectangle on the same base (the interval) would need to have is the area of the rectangle is to be equal to the area under the curve on the interval.

Here is a picture of the area under the graph of `f(t)=4te^(-t^2)` on the interval [0, 5]

graph{(y - 4xe^(-x^2)(sqrt(6.25-(x-2.5)^2))/(sqrt(6.25-(x-2.5)^2))) y <=0 [-1.874, 6.893, -1.357, 3.028]}

To find the area under the curve, we integrate:

`int_0^5 4te^(-t^2) dt`

Let `u = -t^2` to get:

`-2int_0^-25 e^u du = -2e^u]_0^-25`

`= 2(1-e^-25)`

`Area = 2(1-e^-25)`

So if a rectangle on base `[0,5]` is to have that area, its height must be

`"height" = "Area"/"base" = "Area"/5 = (2(1-e^-25))/5` Which can be rewritten:

`"Average Value" = 1/5(2-2/e^25)` or `2/5(1-1/e^25)`

The general formula is: The average value of an function, `f` on interval `[a,b]` is:

`"Average Value" = 1/(b-a) int_a^b f(x) dx`