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Math Help - Intergral at a point sint+t t=12&0.

  1. #1
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    Intergral at a point sint+t t=12&0.

    Hi,

    Trying to work this one out,

    \int_0^{12} (\sin{t} + {t}), dx

    The answer shows = (72 - cos12) - (0-1)
    =(73-.84)
    =72.16

    I calculate (72-.98) - (0-1)
    which equals 70.02.

    Anyone willing to have a quick look and try and see where I went wrong?

    Thanks
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  2. #2
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    Its sin(t)+t or sin(x)+x .. ?!!
    Big difference.
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  3. #3
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    Quote Originally Posted by Splint View Post
    Hi,

    Trying to work this one out,

    \int_0^{12} (\sin{t} + {t}), dx

    The answer shows = (72 - cos12) - (0-1)
    =(73-.84)
    =72.16

    I calculate (72-.98) - (0-1)
    which equals 70.02.

    Anyone willing to have a quick look and try and see where I went wrong?

    Thanks
    You are calculating cos (12 degrees). You should be calculating cos (12 radians)
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  4. #4
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    Quote Originally Posted by Debsta View Post
    You are calculating cos (12 degrees). You should be calculating cos (12 radians)
    Thanks Debsta, I can't believe I overlooked that.
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