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Math Help - What did I do wrong? (integration)

  1. #1
    Member integral's Avatar
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    What did I do wrong? (integration)

    I was playing around and did this:
    f(x)g(x)=f_xg_x
    Then:
    \frac{d}{dx}\left [f_xg_x\right ]=\frac{g_xdf_x}{dx}+\frac{f_xdg_x}{dx}
    so
    f_xg_x=\int \frac{g_xdf_x}{dx}+\frac{f_xdg_x}{dx}dx=\int \frac{g_xdf_x}{dx}dx+\int \frac{f_xdg_x}{dx}dx=\int g_xdf_x+\int f_xdg_x
    so
    f_xg_x=g_x \int df_x+f_x \int dg_x
    and
    f_xg_x=f_xg_x+f_xg_x=2f_xg_x

    Any obvious division by zero someone should point out?
    Or do I just not understand the concept of something?
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  2. #2
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    Are you trying to derive the integration by parts formula?

    Let u = u(x) and v = v(x).

    By the product rule:

    \frac{d}{dx}(uv) = u\,\frac{dv}{dx} + v\,\frac{du}{dx}

    uv = \int{\left(u\,\frac{dv}{dx} + v\,\frac{du}{dx}\right)\,dx}

    uv = \int{u\,\frac{dv}{dx}\,dx} + \int{v\,\frac{du}{dx}\,dx}

    uv = \int{u\,dv} + \int{v\,du}

    \int{u\,dv} = uv - \int{v\,du}.
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  3. #3
    Member integral's Avatar
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    But, I thought since in: udv, you where taking the integral of u with respect of dv and thus you could use the scalar multiple rule.
    Samne thing with:
    \int ydx=yx+C
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  4. #4
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    It's not taking the integral of u with respect to v.

    It's just shorthand for \int{u\,\frac{dv}{dx}\,dx}. Do you see how it "looks like" the dxs cancel?


    So you're taking the integral of the product of u and \frac{dv}{dx}, with respect to x.
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