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Math Help - Evaluate

  1. #1
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    Evaluate

    I have attached an equation that requires an evaluation, I need help working this out, please!
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  2. #2
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    Quote Originally Posted by norivea
    I have attached an equation that requires an evaluation, I need help working this out, please!
    \int_0^{\pi} cos(x) dx.

    Sketch this function over this range - you will see that it is anti-symmetric
    about \pi/2, so the integral is zero.

    RonL
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  3. #3
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    Reply

    So, is the answer:
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  4. #4
    TD!
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    No, you'd get:

    <br />
\int\limits_0^\pi  {\cos xdx}  = \left[ {\sin x} \right]_0^\pi   = \sin \pi  - \sin 0 = 0 - 0 = 0<br />

    You could have found this without integrating with CaptainBlack's argument, remember what an integral represents (signed area!).
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  5. #5
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    Reply

    Ok thank you. I am just uncertain where I went wrong in my working. Could you look over my working to see where I went wrong? I have attached my original working.

    Thanks again
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  6. #6
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    Quote Originally Posted by norivea
    Ok thank you. I am just uncertain where I went wrong in my working. Could you look over my working to see where I went wrong? I have attached my original working.

    Thanks again
    Why did you split the range of integration into 0-\pi/2, and
    \pi/2-\pi?

    The sign of \cos is positive on the first of your sub-ranges and
    negative on the second, so they cancel rather than add (since \cos
    is anti-symmetric about \pi/2).

    RonL
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  7. #7
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    \int^{\pi}_0 \cos x dx
    Thus,
    \left \sin x \right|^{\pi}_0
    Thus,
    \sin \pi -\sin 0=0-0=0
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  8. #8
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    Quote Originally Posted by norivea
    Ok thank you. I am just uncertain where I went wrong in my working. Could you look over my working to see where I went wrong? I have attached my original working.
    Thanks again
    Hello,

    according to your working you didn't evaluate the integral but you calculated the area which is enclosed by the x-axis, the graph of the cosine function and the lines x = 0 and x = pi. So you calculated:
    \int_{0}^{\pi}|{\cos(x)| dx which differs from your orginal integral.

    Greetings

    EB
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