Question

What is the average value of the function `f(t)=te^(-t^2 )` on the interval `[0,5]`?

Answer 1

It is `1/10(1-e^-25)`

Explanation:

`1/(5-0) int_0^5 te^(-t^2)dt = -1/10 int_0^5 e^(-t^2)(-2t) dt`

`= -1/10[e^(-t^2)]_0^5`

`= -1/10(e^-25 - e^0)`

`= 1/10(1-e^-25)`