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Is there an alternate way to answer this question, if anyone could please help.

The textbook answer it like this:

enter image description here

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up vote 2 down vote accepted

By parts:

$$x-u\to -du$$

$$f(u)\to \int_0^u f(t)dt$$


$$\int_0^x f(u)(x-u)du=\left[(x-u)\int_0^u f(t)dt\right]_0^x+\int_0^x\int_0^u f(t)dt\;du$$

$$\int_0^x f(u)(x-u)du=\int_0^x\int_0^u f(t)dt\;du$$

This of course relies on $f$ being continuous.

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Thanks very much for your help. – Dick Jan 23 '13 at 14:21
Somehow your last comment seems to imply that continuity of $f$ is equivalent to integrability, which is clearly false; continuity is a stronger condition than integrability. You need continuity only to use the fundamental theorem of calculus. – Nils Matthes Jan 23 '13 at 14:26
@NilsMatthes That is what meant, but I realise it was baldy put. Thanks. – L. F. Jan 23 '13 at 14:34
Does this in itself prove that f is continuous, or not, I am confused now. Isn't this answer using the fundamental theorem? – Dick Jan 25 '13 at 22:01
@Dicky The continuity of $f$ is assumed. The use of the FTC is justified by the continuity of $f$ but does not prove it (and we do not need to prove it, since it is a given). – L. F. Jan 25 '13 at 22:09

For given $x>0$ consider the triangle $T:=\{(t,u)\ |\ 0\leq u\leq t\leq x\}$ in the $(t,u)$-plane and on $T$ the function $\phi(t,u):= f(u)$; see the following figure.

enter image description here

By Fubini's theorem we can compute the integral $$J:=\int_T \phi(t,u)\ {\rm d}(t,u)$$ in two ways, namely as $$J=\int_0^x\left(\int_u^x \phi(t,u)\ dt\right)\ du=\int_0^x\left(\int_u^x f(u)\ dt\right)\ du=\int_0^x f(u)(x-u)\ du$$ and as $$J=\int_0^x\left(\int_0^t \phi(t,u)\ du\right)\ dt=\int_0^x\left(\int_0^t f(u)\ du\right)\ dt\ .$$

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Nice, thank you – Dick Jan 23 '13 at 18:51

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