Question

What is the average value of a function `f(x) = 2x sec2 x` on the interval `[0, pi/4]`?

Answer 1

There isn't one

Explanation:

Recall that `secx` has a vertical asymptote at `x=pi/2`.

Thus, `2xsec(2x)` has a vertical asymptote at `x=pi/4` and there will be no average value for the function on the requested interval.

Attempting to find the average value using an integral:

`1/(pi//4-0)int_0^(pi//4)2xsec(2x)dx`

is impossible because the integral diverges.