How do I prove the integration equation for continuous function f?

[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.

 

[matt grime]

differentiate both sides wrt x.

 

[digink]

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.

I don't 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? I am confused sorry.

 

[shmoe]

You can also integrate the left hand side by parts.

 

[digink]

You can also integrate the left hand side by parts.

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

 

[shmoe]

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

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

 

[trap]

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

 

[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?