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Integral involving abs

  1. Sep 10, 2013 #1
    1. The problem statement, all variables and given/known data

    f(x) = |sinx| and g(x)= |cosx|
    I want the area of this between 0 and 2∏


    2. Relevant equations


    3. The attempt at a solution

    So I'm just being careful. I just did the intervals and by doing g(x) - f(x) because it is the upper minus the lower. So, I got sinx + cosx between whatever interval I wanted. (∏/4), (3∏/4)...etc
    But then I was like wait a minute. What is up with the abs.
    So I'm just wondering about this. I feel like my integral is wrong if I don't account for it.

    Thanks,
    J
     
  2. jcsd
  3. Sep 10, 2013 #2

    Mark44

    Staff: Mentor

    You need to be more specific. What does "this" refer to? Do you mean the area of the five regions that lie between the two curves?
     
  4. Sep 10, 2013 #3
  5. Sep 10, 2013 #4
    Your right its 5 sorry.
     
  6. Sep 10, 2013 #5

    Mark44

    Staff: Mentor

    I see five regions between 0 and 2π. The five regions are above the intervals [0, π/4], [π/4, 3π/4], [3π/4, 5π/4], [5π/4, 7π/4], and [7π/4, 2π].

    You can get rid of the absolute values by using the fact that |u| = u if u ≥ 0, and |u| = -u if u < 0. You'll probably need to break up the longer subintervals above so that you can eliminate the absolute values.

    For example, on [0, π], sin(x) ≥ 0, so |sin(x)| = sin(x), but on [π, 2π], sin(x) ≤ 0, so |sin(x)| = -sin(x).
     
  7. Sep 10, 2013 #6
    Also I can just do 2 times the integral between 0 and Pi/4 then take one of the intervals in the middle and do 3 times that integral? Symmetry?
     
  8. Sep 10, 2013 #7

    Mark44

    Staff: Mentor

    Sure, that would work.
     
  9. Sep 10, 2013 #8
    Thanks Mark 44.
     
  10. Sep 10, 2013 #9
    Hi again,
    This is what I did. I kind of took half sections and just multiplied.

    I took the integral from 0 to ∏/4 of |cosx|-|sinx|
    Since both sinx and cosx are positive on this interval I got A=∫(cosx-sinx)dx [0,∏/4]
    which is equal to sinx + cosx] I evaluated this at ∏/4 and 0 I got 2√2-1.
    I then counted the other half sections for the area inbetween curves. I count 7 more of thease half sections between ∏/4 and 2∏. So I for total area I did 7( 2√2-1).
    Yes or no?
    Thanks
     
  11. Sep 10, 2013 #10

    Mark44

    Staff: Mentor

    No. How many of those half sections are there? Hint: not 7.
     
  12. Sep 10, 2013 #11

    Mark44

    Staff: Mentor

    Also, check your work here. It looks like you made a mistake when you evaluated sin(x) + cos(x) (which is the correct antiderivative).
     
  13. Sep 10, 2013 #12
    Well I accounted for the first one. And then I count 2 between pi/4 and 3pi/4, 2 between 3pi/4 and 5pi/4, 2 between 5pi/4 and 7pi/4 and 1 for 7pi/4 and 2pi.
    That is 7.
     
  14. Sep 10, 2013 #13

    Mark44

    Staff: Mentor

    No, that's eight of them altogether, not seven.

    Note also that your value of 2√2 - 1 is incorrect.
     
  15. Sep 10, 2013 #14
    I don't see where. I got 2sqrt2 +2sqrt2 - (0 + 1) = sqrt2 - 1
     
  16. Sep 10, 2013 #15
    But I found the first area between 0 and pi/4 so I don't count that again. I have 7 more of those half sections so that is 7 times.
     
  17. Sep 10, 2013 #16
    No your right about the number never mind there are 8. But I can't find my mistake in my evaluation.
     
  18. Sep 10, 2013 #17
    I just did it on my calculator and with the integral option (ti 83) and I got the same thing that I got when I did it by hand.
     
  19. Sep 10, 2013 #18

    vela

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    What are ##\cos \pi/4## and ##\sin \pi/4## equal to?
     
  20. Sep 10, 2013 #19
    Both are √2/2
    When you add them you get (2√2)/2 = √2
     
  21. Sep 10, 2013 #20

    Mark44

    Staff: Mentor

    Exactly. So where did you get 2√2 that you showed in a previous post?
     
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