Is there an alternate way to answer this question, if anyone could please help.
The textbook answer it like this:
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By parts: $$x-u\to -du$$ $$f(u)\to \int_0^u f(t)dt$$ Therefore: $$\int_0^x f(u)(x-u)du=\left[(x-u)\int_0^u f(t)dt\right]_0^x+\int_0^x\int_0^u f(t)dt\;du$$ $$\int_0^x f(u)(x-u)du=\int_0^x\int_0^u f(t)dt\;du$$ This of course relies on $f$ being continuous. |
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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.
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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