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

1 Answer
Jim H
Oct 22, 2016

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)

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