Integration and The Fundamental Theorem of Calculus Is there an alternate way to answer this question, if anyone could please help.

The textbook answer it like this:


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.
