Compute the average of f(x) = e^x / 1 + e^2x over [0,1].(Be sure to set up the definite integral first.)

Question

Compute the average of f(x) = ex

1 + e2xover [0,1].(Be sure to set up the definite integral

Answer 1

The average of a function f(x) over the interval [a,b] is equal to the definite integral of f(x) from a to b, divided by b-a.

Here, f(x) = e^x/(1+e^2x), and the interval is [0,1]. Dividing by b-a is equivalent to dividing by 1-0 = 1, which does nothing to affect the value of the definite integral. Thus, the average of this function over this interval is equal to the definite integral:

∫ e^x/(1+e^2x) dx = ∫ 1/(1+u²) du (using u-substitution: u = e^x, du = e^x dx)

∫ 1/(1+u²) du = tan^-1(u) = tan^-1(e^x) from x=0 to x=1. By the Fundamental Theorem of Calculus, this is tan^-1(e) - tan^-1(1), or just tan^-1(e) - π/4.

Answer 2

Substitute u=e^x and you will get an easy trig integral.

Since the length of the interval is 1, the average value will be the value of the integral.