Stuck on pratice problem reguarding integrals

Question

a. Explain why ∫₀^2pi sin(x)dx = 0

 

b. Can you set up another integral, or perhaps multiple integrals, to compute the total (unsigned) area between sin(x) and the x-axis on the interval [0,2pi]

Answer 1

a)

 

at x = 0 , sinx = 0

at x = π , sinx = 0

at x = 2π , sinx = 0

 

So the area between x=0 and x=π is the same as the area between x=π and x=2π.  The first area is positive, and the second area is negative.  So when you add up these areas, the total area is 0.

 

b)

 

^π∫₀ sin(x)dx + ^2π∫_π sin(x)dx