Question

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

Answer 1

`4/pi`

Explanation:

The average value of a function `y=f(x)` over an interval `[a,b]` is given by `1/(b-a)int_a^b f(x) dx`

So for `y=sec^2 x` we have:

`Avg = 1/(pi/4-0) int_0^(pi/4) sec^2x dx`
`= 1/(pi/4) int_0^(pi/4) sec^2x dx`
`= 4/pi [tanx]_0^(pi/4)`
`= 4/pi {tan (pi/4) - tan0 }`
`= 4/pi {tan (pi/4) - tan0 }`
`= 4/pi {1-0 }`
`= 4/pi`