Question

What is the net area between `f(x)=tanx` in `x in[0,pi/3]` and the x-axis?

Answer 1

`A= int_0^(pi/3) tanx dx = [log(abs(sec(x)))]_0^(pi/3) ~~0.693`

Explanation:

GIven : `f(x)=tanx: ->x in[0,pi/3]`

Required : Area under `f(x) " and the " x " axis"`

Solution: Straight integration of `f(x)` in the interval `[0, pi/3]`

`A= int_0^(pi/3) tanx dx = [log(abs(sec(x)))]_0^(pi/3) ~~0.693`