Question

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

Answer 1

We start by checking for x-intercepts within the given interval.

`f(x) = cosxsinx -> f(x) = 1/2sin2x`

If we set `f(x)` to `0` and solve for `x`.

`0 = 1/2sin2x`

`0 = sin2x`

`2x = 2pi or 2x = pi`

`x = pi or x = pi/2`

Since trig functions repeat, and the period of this function is `pi`, we will have more than two crossings. If we analyze the graph, we realize that in `[0, 3pi]` there are `6` intercepts.

graph{sinxcosx [-4.5, 15.5, -5.16, 4.84]}

This means that there are `3` regions of equal area above the x-axis, and three areas of equal area below the x-axis. Since we're talking about net area, we should subtract the negative area (below the x-axis) from the positive area (above the x-axis).

Our expression for area will be

`A = 3(int_0^(pi/2) 1/2sin2x dx- int_(pi/2)^pi 1/2sin2x dx)`

`A = 3([-cos2x]_ 0^(pi/2) - [-cos2x]_(pi/2)^pi)`

`A = 3(-cospi - cos0 - (-cos2pi - cospi))`

`A = 3(1 - 1 + 1 - 1)`

`A = 0`

A geometric view of this gives sense to this result: whenever a positive area occurs, it is cancelled out by the negative area, giving a final result of `0` (because there are an equal number of positive and negative areas).

Hopefully this helps!