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

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

Suppose f is continuous. Prove that

\int_0^{x} f(u)(x-u) du =\int_0^{x} ( \int_0^{u} f(t) dt) du.
 
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differentiate both sides wrt x.
 
trap said:
Can someone help me with this question? gladly appreciate any help on this :smile:

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.
 
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You can also integrate the left hand side by parts.
 
shmoe said:
You can also integrate the left hand side by parts.
oh ok i see what he was asking now... so x is a constant?
 
digink said:
oh ok i see what he was asking now... so x is a constant?

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