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Math Help - FTC Part II simple question?

  1. #1
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    FTC Part II simple question?

    Hi again all,

    I'm doing some assignments and one of the questions asked to evaluate the integral:

    \int_{0}^{5}\((6e^x+4cos(x)) dx

    Using FTC P. II I got that the antiderivative was F(x)=6e^x+4sin(x)

    so I got that the integral was: F(5) - F(0) = 884.8

    I would like to know why this is the wrong answer? Thanks
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  2. #2
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    Quote Originally Posted by DannyMath View Post
    the questions asked to evaluate the integral:
    \int_{0}^{5}\((6e^x+4cos(x)) dx
    Using FTC P. II I got that the antiderivative was F(x)=6e^x+4sin(x)
    so I got that the integral was: F(5) - F(0) = 884.8
    I would like to know why this is the wrong answer? Thanks
    An arithmetic error?
    I get F(5)-F(0)=880.643
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  3. #3
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    HAHA I just started Physics this semester and it uses angles in degrees so I forgot that I had switched my calculator from Rad to Deg. THANKS!
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  4. #4
    A Plied Mathematician
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    Quote Originally Posted by DannyMath View Post
    HAHA I just started Physics this semester and it uses angles in degrees so I forgot that I had switched my calculator from Rad to Deg. THANKS!
    Ah, yes. The downfall of many, many calculations in trigonometry.
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  5. #5
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    Geeze, I can't imagine how it would feel to have this happen on your final Trig test :S. Glad I caught it now rather than later! Cheers!
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