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Math Help - An integration that doesn't quite make sense.

  1. #1
    Super Member Showcase_22's Avatar
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    An integration that doesn't quite make sense.



    The question is copied directly from my book but the working is my own.

    The answer isn't 0 and I have a feeling that my limits are wrong. I'm not really sure how else to write the limits in order for me to get an answer than isn't 0.

    help would be greatly appreciated.
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  2. #2
    MHF Contributor

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    To find your error, I looked for the line where you start integrating functions taking negative values. This is the line with \int_0^\pi \cos t\sin t dt. To go from the previous line into this one, you wrote (\cos^2 t)^{1/2}=\cos t, whereas it should have been (\cos^2 t)^{1/2}=|\cos t|.

    Laurent.
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    Super Member Showcase_22's Avatar
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    Okay then. I have absolutely no idea how to integrate moduli.

    The last one I did was easy and it was just finding the area of two triangles and add them together. I don't think I can do the same for this.

    I suppose what i'm asking is "how do you integrate moduli?"
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  4. #4
    Moo
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    Quote Originally Posted by Showcase_22 View Post
    Okay then. I have absolutely no idea how to integrate moduli.

    The last one I did was easy and it was just finding the area of two triangles and add them together. I don't think I can do the same for this.

    I suppose what i'm asking is "how do you integrate moduli?"
    Separate the integral.

    Remember you can write \int_a^b \dots=\int_a^c \dots + \int_c^b \dots

    Find for what values of x between 0 and \pi, |\cos(x)| \ge 0 \implies |\cos(x)|=\cos(x) and for which ones |\cos(x)| \le 0 \implies |\cos(x)|=-\cos(x)
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  5. #5
    Super Member Showcase_22's Avatar
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    My big worry is that i'm not sure if my limits are right.

    I'd be integrating a cos x along the way that will give me a sign x.

    Sin 0=0 and sin pi=0 so it would give me an area of 0.
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  6. #6
    Super Member Showcase_22's Avatar
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    Is this what I need to integrate:



    ?

    I just noticed something else. In the other post you write "mod cos x < (or equal to) 0" (I haven't figured out how to quote properly). Isn't mod cos x always greater than 0? If any of the values were negative then they wouldn't be needed in the question anyway since it wants the area above the x axis.

    ?
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  7. #7
    MHF Contributor

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    Quote Originally Posted by Showcase_22 View Post
    In the other post you write "mod cos x < (or equal to) 0"
    This was of course a typo for: \cos x\leq 0\Leftrightarrow |\cos x|=-\cos x.

    To explicit Moo's hint a little, here is an example of computation of an integral with an absolute value: \int_{-1}^2|x|dx=\int_{-1}^0 |x|dx +\int_0^2 |x|dx=\int_{-1}^0 (-x) dx+\int_0^2 x dx=\cdots. Try to adapt this to your situation.
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  8. #8
    Super Member Showcase_22's Avatar
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    I decided to draw a graph of what I needed:



    ...and it suddenly all made sense! I was supposed to use the symmetry of the graph (integrate the function from 0 to pi/2) and then double it)!!

    \m/ MAD SKILLZ! \m/

    I got it right! thanks a bundle!
    Last edited by Showcase_22; September 12th 2008 at 01:29 AM. Reason: needed to add semi-cool sentence.
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