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

##### 1 Answer

Oct 21, 2017

The average value is

#### Explanation:

The average value of a function on the continuous interval

#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

#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!