Results 1 to 4 of 4

Thread: What did I do wrong? (integration)

  1. #1
    Member integral's Avatar
    Joined
    Dec 2009
    From
    Arkansas
    Posts
    200

    What did I do wrong? (integration)

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

    Any obvious division by zero someone should point out?
    Or do I just not understand the concept of something?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    Are you trying to derive the integration by parts formula?

    Let $\displaystyle u = u(x)$ and $\displaystyle v = v(x)$.

    By the product rule:

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

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

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

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

    $\displaystyle \int{u\,dv} = uv - \int{v\,du}$.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member integral's Avatar
    Joined
    Dec 2009
    From
    Arkansas
    Posts
    200
    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:
    $\displaystyle \int ydx=yx+C$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    It's not taking the integral of $\displaystyle u$ with respect to $\displaystyle v$.

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


    So you're taking the integral of the product of $\displaystyle u$ and $\displaystyle \frac{dv}{dx}$, with respect to $\displaystyle x$.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Integration goes wrong? ( ln(x)/x )
    Posted in the Calculus Forum
    Replies: 5
    Last Post: Mar 10th 2011, 10:34 AM
  2. integration what am i doing wrong?
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Feb 19th 2011, 03:16 AM
  3. Replies: 2
    Last Post: May 31st 2010, 12:24 AM
  4. What is wrong with my Integration?
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Jan 14th 2010, 01:21 PM
  5. Integration: What is wrong with my method?
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Aug 26th 2009, 12:44 PM

Search Tags


/mathhelpforum @mathhelpforum