integrate [sinx + cosx] from 0 to 2pie, where [ ] represents greatest integer function .
To integrate the function [sin(x) + cos(x)] from 0 to 2π, we need to split the integral into two parts: one from 0 to π, and another from π to 2π. This is because the greatest integer function [ ] changes its value at π.
Let's start with the first part, from 0 to π:
∫[sin(x) + cos(x)] dx
First, we integrate sin(x) with respect to x:
∫sin(x) dx = -cos(x)
Next, we integrate cos(x) with respect to x:
∫cos(x) dx = sin(x)
Now, we can evaluate the integral from 0 to π by substituting the upper and lower limits into the antiderivatives:
∫[sin(x) + cos(x)] dx = [-cos(x) + sin(x)] evaluated from 0 to π
= [-(cos(π) + 1) + (sin(π) - sin(0))]
Since cosine of π is -1 and sine of π is 0, the expression simplifies to:
= [-(-1 + 1) -sin(0)]
= [0 - sin(0)]
= -sin(0)
Since sine of 0 is 0, the final result is:
= [0]
Now, let's move on to the second part, from π to 2π:
∫[sin(x) + cos(x)] dx
Using the same antiderivatives as before, we evaluate the integral from π to 2π:
∫[sin(x) + cos(x)] dx = [-cos(x) + sin(x)] evaluated from π to 2π
= [-cos(2π) + sin(2π)] - [-(cos(π) + sin(π))]
Since cosine and sine of 2π and π are both 1 and 0 respectively, the expression simplifies to:
= [-(1) + 0] - [-(-1) + sin(0)]
= -1 + 1 + sin(0)
= sin(0)
Since sine of 0 is 0, the final result is:
= [0]
Therefore, the definite integral of [sin(x) + cos(x)] from 0 to 2π equals 0.
∫ (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)