Question

What is the average value of a function `sec^2x` on the interval `[0, pi/4]`?

Answer 1

`4/pi`

Explanation:

The average value of a function from point `x=a` to point `x=b` is given by:
`1/(b-a)int_a^bf(x)dx`

For `sec^2(x)` on `[0,pi/4]`, this looks like:
`1/(pi/4-0)int_0^(pi/4)sec^2xdx`

The antiderivative of `sec^2x` is `tanx` because `d/dxtanx=sec^2x`; that reduces our problem to:
`1/(pi/4)[tanx]_0^(pi/4)`
`=4/pi(tan(pi/4)-tan(0))`
`=4/pi(1-0)`
`=4/pi=1.273...`