Can you use integration by parts?

[MathewsMD]

How would you integrate this:

$\int x df(x)$

In general, how is this solved: $\int 1df(x)$

Can you use integration by parts? I tried, but kept getting 0 since I let $1 = u$ but then $du = 0$ for later purposes. Also, if $df(x) = u$ then I am still stumped on how to take the derivative of $df(x)$ ...is it even possible?

Any help would be great!

 

[jedishrfu]

How would you integrate this:

$\int x df(x)$

In general, how is this solved: $\int 1df(x)$

Can you use integration by parts? I tried, but kept getting 0 since I let $1 = u$ but then $du = 0$ for later purposes. Also, if $df(x) = u$ then I am still stumped on how to take the derivative of $df(x)$ ...is it even possible?

Any help would be great!

This doesn't look right is the df(x) from y=f(x) and so its really a dy? so that you have the integral $\int x dy$

Can you give the whole problem and what book it came from?

if its by parts then $\int udv$ = uv - $\int vdu$

so you dv would be df(x) and v =f(x) and u would be x so we get:

$\int xdf$ = xf - $\int fdx$

 

[brmath]

If you place a problem in this homework section, you are supposed to follow the homework format. Can you please do that.

As jedishrfu says, your problem is not correctly stated. If you are looking at $\int xdx$ that is easily solved. If by df(x) you mean f'(x), you should write it as $\int xf'(x)dx$.

As you've written it, it is a Riemann-Stieltjes integral; in that case we can't proceed without knowing what f(x) is.

Please clarify.

 

[MathewsMD]

This doesn't look right is the df(x) from y=f(x) and so its really a dy? so that you have the integral $\int x dy$

Can you give the whole problem and what book it came from?

if its by parts then $\int udv$ = uv - $\int vdu$

so you dv would be df(x) and v =f(x) and u would be x so we get:

$\int xdf$ = xf - $\int fdx$

I am just wondering what forms you can transform $\int xdy$ into if y is a function of x. I am trying to manipulate another expression, but seem to be stuck once given this problem.

 

[jedishrfu]

Without knowing what y is as a function of x, you can only do the integration by parts rule as mentioned in earlier posts and that's it.

 

[consciousness]

I think you are asking how to integrate functions w.r.t. functions other than x.

That is

$\int g(x)d(f(x))$

The general approach I think is to take

$f(x)=t$

$d(f(x))=dt$

also express x as

$x = f^{-1}(t)$

and finally put this value of x in g(x) to get a simple integral as

$\int g(f^{-1}(t))dt$

which can be solved by usual techniques. After solving, again substitute t=f(x) to get your answer.

 

[CompuChip]

Of course this is provided $f$ is injective so $f^{-1}$ makes sense as a function.

 

[HallsofIvy]

If f is a differentiable function, then [itex]\int x df= \int x \frac{df}{dx}dx[/tex]<br /> <br /> If f is not differentiable but is increasing function then $\int x df$ is not a Riemann integral but is, rather, a Stieljes integral.<br /> <br /> (A Stieljes integral is defined in a way similar to the Riemann integral by dividing the interval into subintervals of length $\Delta x$ (which might vary from subinterval to subinterval), approximating the integral by the sum $\sum g(x_i) \Delta x$ where $x_i$ is a point in the ith subinterval. The definition of the Riemann integral, $\int g(x)dx$ takes $\Delta x$ to be the length of each subinterval, $\alpha_{i+ 1}- \alpha_i$, while the Stieljes integral, $\int g(x) df$, with f an increasing function (not necessarily differentiable or even continuous) $\Delta x= f(\alpha_{ix+1}- f(\alpha_i)$.)[/itex]

 

[jedishrfu]

If f is a differentiable function, then $\int x df= \int x \frac{df}{dx}dx$

If f is not differentiable but is increasing function then $\int x df$
is not a Riemann integral but is, rather, a Stieljes integral.

(A Stieljes integral is defined in a way similar to the Riemann integral by dividing the interval into subintervals of length $\Delta x$ (which might vary from subinterval to subinterval), approximating the integral by the sum $\sum g(x_i) \Delta x$ where $x_i$ is a point in the ith subinterval. The definition of the Riemann integral, $\int g(x)dx$ takes $\Delta x$ to be the length of each subinterval, $\alpha_{i+ 1}- \alpha_i$, while the Stieljes integral, $\int g(x) df$, with f an increasing function (not necessarily differentiable or even continuous) $\Delta x= f(\alpha_{ix+1}- f(\alpha_i)$.)

...fixed the itex tag of HallsOfIvy post...

 

[MathewsMD]

Wow, thank you for the great explanation HoI!