# Proving an integration

1. Jan 24, 2005

### trap

Can someone help me with this question? gladly appreciate any help on this

Suppose f is continuous. Prove that

$$\int_0^{x} f(u)(x-u) du =$$$$\int_0^{x}$$ ( $$\int_0^{u} f(t) dt$$) du.

Last edited: Jan 24, 2005
2. Jan 25, 2005

### matt grime

differentiate both sides wrt x.

3. Jan 25, 2005

### digink

I dont ever understand how you write these out

do you mean $$\int_0^{x} (x-u) dx$$

to me for some reason it always seems like you write in weird form, but maybe its just me.

also for that what I wrote above for the solution wouldn't you just substitute, but is the u susposed to be a constant or what? im confused sorry.

Last edited: Jan 25, 2005
4. Jan 25, 2005

### shmoe

You can also integrate the left hand side by parts.

5. Jan 25, 2005

### digink

oh ok i see what he was asking now... so x is a constant?

6. Jan 25, 2005

### shmoe

x is independant of u, that's what's important here. You can think of both sides as a function of x if you like.

7. Jan 25, 2005

### trap

thx ppl for the help, i'm trying to solve it now

8. Jan 25, 2005

### trap

for the left side, this is how i did,

$$\int_0^{x} f(u)(x-u) du$$
= $$\int_0^{x} f(u)(x) - f(u)u du$$
= $$\int_0^{x} f(u)(x) du -$$ $$\int_0^{x} f(u)(u) du$$
= x $$\int_0^{x} f(u) du -$$ $$\int_0^{x} f(u)(u) du$$

and then derive it..

= x f(x) - xf(0) - f(x)x
=-xf(0)

is what i'm doing right now correct?
if so, can someone help me on how to make this equal to the right side?

Last edited: Jan 25, 2005