Question

How do you find the area lying above the x axis of `y=sinxcosx` for `-pi<=x<=pi`?

Answer 1

`1`

Explanation:

you are looking for the integral of `y = sin x cos x (= 1/2 sin 2 x)` within certain limits as we want only the area above the x axis `x in [-pi, pi]`.

for these, see the graph below which gives:

`A = 1/2 int_(-pi)^(-pi/2) sin 2x + 1/2 int_(0)^(pi/2) sin 2x`

which because of symmetry can be aggregated as

`A = int_(0)^(pi/2) sin 2x`

`= [ - 1/2 cos 2x ]_(0)^(pi/2)`

`= 1/2 [ cos 2x ]_(pi/2)^(0)`

`= 1/2 ( cos 0 - cos pi ) = 1`

graph{sinx cosx [-3.897, 3.898, -1.95, 1.947]}