What did I do wrong? (integration)

**integral** · August 29th 2010, 01:54 PM

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?

**Prove It** · August 29th 2010, 01:59 PM

Are you trying to derive the integration by parts formula?

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

By the product rule:

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

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

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

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

$\int{u\,dv} = uv - \int{v\,du}$ .

**integral** · August 29th 2010, 02:02 PM

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:
$\int ydx=yx+C$

**Prove It** · August 30th 2010, 05:07 AM

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 $dx$ s cancel?

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

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