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

1 Answer
Aug 29, 2016

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]}