Integration by parts involving partial derivatives

[tjkubo]

Homework Statement

<br /> \int x \frac {\partial f} {\partial x} dx<br /> <br />

where

f=f(x,t)

Homework Equations

<br /> \int u \, dv = uv - \int v \, du<br />

The Attempt at a Solution

<br /> u = x so du = dx

and

<br /> dv = \frac {\partial f} {\partial x} dx so v = \int \frac {\partial f} {\partial x} dx = f?<br />

Are these the correct substitutions? I am especially hesitant about the last term, so can anyone explain why it's right/wrong?

 

[rock.freak667]

you'd get v=f(x,t) + h(t)


EDIT: Sorry, there'd be no h(t)...

 

[n!kofeyn]

The Attempt at a Solution

<br /> u = x so du = dx

and

<br /> dv = \frac {\partial f} {\partial x} dx so v = \int \frac {\partial f} {\partial x} dx = f?<br />

Are these the correct substitutions? I am especially hesitant about the last term, so can anyone explain why it's right/wrong?

The term you are hesitant about is correct. You are integrating, with respect to x, the derivative of f with respect to x, so by the fundamental theorem of calculus that integral is just f.

So, you're right. Just finish the integration by parts and you're done!

 

[foxjwill]

So, you're right. Just finish the integration by parts and you're done!

Just be sure to remember that the "constant" of integration in this case isn't constant--it's an arbitrary function of t.

 

[n!kofeyn]

Just be sure to remember that the "constant" of integration in this case isn't constant--it's an arbitrary function of t.

That's correct. Thanks.

 

[tjkubo]

What if the integral was a definite integral? Would the final result be some function of t only? And would you disregard the arbitrary function of t?

 

[Cyosis]

You wouldn't disregard it nor do you just disregard it for a normal integral. What happens when you evaluate a definite integral is that the constants get subtracted from each other.

<br /> f(x,t)+h(t)]_{x_1}^{x_2}=f(x_1,t)+h(t)-(f(x_2,t)+h(t))=f(x_1,t)-f(x_2,t)<br />