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!