Question

What is the average value of a function `sec^2x` on the interval `[pi/6, pi/4]`?

Answer 1

`(12-4sqrt3)/pi`

Explanation:

The average value of the function `f(x)` on the interval `[a,b]` is equivalent to

`1/(b-a)int_a^bf(x)dx`

Thus, the average value of `f(x)=sec^2x` on `[pi/6,pi/4]` is

`1/(pi/4-pi/6)int_(pi/6)^(pi/4)sec^2xdx`

Note that `intsec^2xdx=tanx+C`, since the derivative of `tanx` is `sec^2x`.

Also, note that `1/(pi/4-pi/6)=1/(pi/12)=12/pi`.

The average value then equals:

`1/(pi/4-pi/6)int_(pi/6)^(pi/4)sec^2xdx=12/pi[tanx]_(pi/6)^(pi/4)`

`=12/pi[tan(pi/4)-tan(pi/6)]=12/pi[1-sqrt3/3]=12/pi[(3-sqrt3)/3]`

`=4/pi[3-sqrt3]=(12-4sqrt3)/pi`