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Math Help - area under a curve

  1. #1
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    area under a curve

    could some one please help me and explain how to find the area under the curve Y=sin(x) between x=0 and x=1.7pi

    Thanks
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  2. #2
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    Re: area under a curve

    Since you know that the sine curve has an intercept at \displaystyle \begin{align*} x = \pi \end{align*}, you will need to evaluate

    \displaystyle \begin{align*} A = \left| \int_0^{\pi} \sin{x}\,dx  \right| + \left| \int_{\pi}^{1.7\pi} \sin{x}\,dx \right| \end{align*}
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  3. #3
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    Lightbulb Re: area under a curve

    \int_a^b \text{Sin}[x] \, dx = \text{Cos}[a]-\text{Cos}[b]
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  4. #4
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    Re: area under a curve

    Quote Originally Posted by MaxJasper View Post
    \int_a^b \text{Sin}[x] \, dx = \text{Cos}[a]-\text{Cos}[b]
    MaxJasper, that's incorrect. The OP was asked to find the AREA. Since there is an x intercept, you need to take into account all the "signed areas", and add their absolute values.
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  5. #5
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    Re: area under a curve

    Quote Originally Posted by Prove It View Post
    MaxJasper, that's incorrect. The OP was asked to find the AREA. Since there is an x intercept, you need to take into account all the "signed areas", and add their absolute values.
    It is correct. It's up to him to find proper ranges and signs.
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  6. #6
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    Re: area under a curve

    Quote Originally Posted by MaxJasper View Post
    It is correct. It's up to him to find proper ranges and signs.
    And it's up to us to not be ambiguous, so that our guidance is of some use
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  7. #7
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    Re: area under a curve

    so am i correct in thinking this?

    (-cospi)-(-cos0)=2
    (-cos1.7pi)-(-cospi)=-1.59

    therefore total area is 3.59 units^2
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  8. #8
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    Re: area under a curve

    Looks good to me
    Thanks from rich1979
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