integrate [sinx + cosx] from 0 to 2pie, where [ ] represents greatest integer function .

Answer 1

∫ (sinx + cosx) from x = 0 to 2π
= -cosx + sinx | from 0 to 2π
= -cos(2π) + sin(2π) - (-cos0 + sin0)
= -1+ 0-(-1 + 0)
= 0

ahh, something is not right.
I know.....
when graphed some of the "area" is above and some is below the x-axis, so some of this stuff canceled.

So we have to look at the graph of sinx + cosx

looking at Wolfram's graph
http://www.wolframalpha.com/input/?i=sin%28x%29+%2B+cos%28x%29
shows that we have 2 x-intercepts

sinx + cosx = 0
sinx = -cosx
divide both sides by cos
sinx/cosx = -1
tanx = -1
x = 135° or 315° or (3π/4 or 7π/4)

so we need to do this in 3 parts

∫(sinx + cosx) dx from 0 to 3π/4 + ∫(-sinx - cosx) dx from 3π/4 to 7π/4 + ∫(sinx+cosx) dx from 7π/4 to 2π

guess what, I will let let you do all that beautiful arithmetic.
(Careful when you adjust that second integral)