Question

What is the average value of a function `f(t)= -2te^(-t^2)` on the interval `[0, 8]`?

Answer 1

The average value is `1/(8e^64) - 1/8 ~~ -0.125`

Explanation:

The average value of a function on the continuous interval `[a, b]` is given by

`A = 1/(b - a) int_a^b f(x) dx`

In this case we would have

`A = 1/8 int_0^8 -2te^(-t^2)dt`

This expression can be integrated using the substitution `u = -t^2`. Then `du = -2tdt` and `dt = (du)/(-2t)`. We change the bounds of integration accordingly.

`A = 1/8int_0^-64 e^u du`

`A = 1/8[e^u]_0^-64`

`A = 1/8[e^(-t^2)])_0^8`

`A = 1/8e^(-64) - 1/8`

If you want an approximation, use a calculator to get

`A = -0.125`

Hopefully this helps!