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Math Help - Integration help (Graph Area)

  1. #1
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    Integration help (Graph Area)

    f(x)= e-xcosx with domain [0,3pi/2]

    a) 'A' is the area between the x-axis, y-axis and f(x). 'B' is the area between the x-axis and f(x). Determine the ration of A to B (preferably in terms of 'e')




    I was unable to do this question mainly because I couldn't find a way to integrate f(x), be it by u substitution or by parts. Any help would be greatly appreciated. Thanks in advance.
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    Re: Integration help (Graph Area)

    Quote Originally Posted by PedroB View Post
    f(x)= e-xcosx with domain [0,3pi/2]

    a) 'A' is the area between the x-axis, y-axis and f(x). 'B' is the area between the x-axis and f(x). Determine the ration of A to B (preferably in terms of 'e')




    I was unable to do this question mainly because I couldn't find a way to integrate f(x), be it by u substitution or by parts. Any help would be greatly appreciated. Thanks in advance.
    Use this result
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    Re: Integration help (Graph Area)

    The thing is I got this question in a non-calculator paper, I have to do everything manually, so I was looking for help on doing it like so. Thanks, though.
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  4. #4
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    Re: Integration help (Graph Area)

    Quote Originally Posted by PedroB View Post
    f(x)= e-xcosx with domain [0,3pi/2]

    a) 'A' is the area between the x-axis, y-axis and f(x). 'B' is the area between the x-axis and f(x). Determine the ration of A to B (preferably in terms of 'e')




    I was unable to do this question mainly because I couldn't find a way to integrate f(x), be it by u substitution or by parts. Any help would be greatly appreciated. Thanks in advance.
    Integrate by parts twice, then you'll get a linear algebraic equation with respect to the original integral.

    \begin{aligned} I&= \int {{e^{ - x}}\cos x\,dx}  = \int e^{-x}\,d(\sin x)=  \\ &= {e^{ - x}}\sin x - \int {\sin x\,d({e^{ - x}})}  = e^{-x}\sin x + \int {{e^{ - x}}\sin xdx}  =  \\  &= {e^{ - x}}\sin x - \int {{e^{ - x}}d(\cos x)}  = e^{-x}\sin x - e^{-x}\cos x + \int {\cos x\,d({e^{ - x}})}  =  \\ &= {e^{ - x}}(\sin x - \cos x) - \int {{e^{ - x}}\cos x\,dx}  = {e^{ - x}}(\sin x - \cos x) - I~~ \Leftrightarrow  \\ &\Leftrightarrow~~ I + I = {e^{ - x}}(\sin x - \cos x)~~ \Leftrightarrow~~ I = \frac{e^{-x}}{2}(\sin x - \cos x) \end{aligned}

    We find the intersection point of the graph with the x-axes (y=0)

    \left\{\begin{gathered}{e^{ - x}}\cos x = 0, \hfill \\x \in (0,3\pi /2) \hfill \\ \end{gathered}  \right. \Leftrightarrow \left\{ \begin{gathered}\cos x = 0, \hfill \\x \in (0,3\pi /2) \hfill \\ \end{gathered}  \right. \Leftrightarrow x = \frac{\pi}{2}

    A = \int\limits_0^{\pi /2}e^{-x}\cos x\,dx=  \ldots  = \left. {\frac{e^{ - x}}{2}(\sin x - \cos x)} \right|_0^{\pi/2} =\ldots= \frac{1+e^{-\pi/2}}{2}

    Since the function takes negative values on the interval \left(\frac{\pi}{2},\frac{3\pi}{2}\right)​​, we have

    B = \int\limits_0^{\pi /2} {{e^{ - x}}\cos x\,dx- \int\limits_{\pi /2}^{3\pi /2} {{e^{ - x}}\cos x\,dx}  = A - \left. {\frac{{{e^{ - x}}}}{2}(\sin x - \cos x)} \right|_{\pi /2}^{3\pi /2} =\ldots= \frac{1+2e^{-\pi /2}+ e^{-3\pi /2}}{2}
    Last edited by DeMath; May 3rd 2012 at 04:15 PM.
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  5. #5
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    Re: Integration help (Graph Area)

    Your suppose to use something called "integration by parts". Its derived from the product rule but you just need to know how to use formula. Its

    Integration of uv' = uv - integration of u'v

    Sorry not sure how to plug in formula still new to the site. You use this equation when you cant use u subsitution. In
    Look up examples on youtube for an example like this.
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  6. #6
    Senior Member DeMath's Avatar
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    Re: Integration help (Graph Area)

    This is the same.
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