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

Answer 1

This integral can be done closed form. Just a substitution.

u = 1 + e^2x

du = 2e^2x dx

∫(e^2x/(1 + e^2x)) dx = (1/2) ∫ du/u = (1/2) ln u + C = (1/2) ln (1 + e^2x) + C

∫₀¹(e^2x/(1 + e^2x) dx = (1/2)(ln(1+e²)-ln 2) = ln√((1+e²)/2)

Answer 2

Average is the sum of the numbers divided by the number of numbers, right?

The integral provides the sum of the numbers; the difference of the limits of integration provides the "number of numbers".

average value = 1/(1 - 0)•₀∫¹(e^2x/(1 + e^2x) dx = ₀∫¹(e^2x/(1 + e^2x) dx

This is a definite integral that you can just plug into a calculator. My TI-84 says the answer is 0.71689