How to show that $\int_p^x dT \int_{p}^T dt \ f(t) = \int_{p}^x dt\ (x - t) f(t)$

Question

Suppose $f(t)$ is continuous on the interval $[p,q]$ and that $p \leq x \leq q$. Formula (4.631) of Gradshteyn and Ryzhik then states that $$ \int_p^x dT \int_{p}^T dt \ f(t) = \int_{p}^x dt\ (x - t) f(t) \ . $$ How does one prove this result?

Answer 1

(Gradshteyn/Rhyzik 4.631)

$$ \int_p^x dt_{n-1} \int_p^{t_{n-1}} dt_{n-2}\ \dots \dots \int_p^{t_1} \ f(t)\ dt = \frac{1}{(n-1)!} \ \int_p^x\ (t-x)^{n-1} \ f(t)\ dt $$

Using the indicator function $\mathbf{1}_{t \leqslant T}(t,T) = \begin{cases}1, & t \leqslant T\\0, &t > T \end{cases}$ and Fubini's theorem,

$$\int_p^x\ \left(\int_p^Tf(t)\, dt \right)\ dT = \int_p^x \ \left(\int_p^x\ f(t) \ \mathbf{1}_{t \leqslant T}\ dt \right)\ dT=$$ $$\int_p^x \ \left(\int_p^xf(t)\ \mathbf{1}_{t \leqslant T}\ dT\ \right)\ dt = \int_p^x\left(\int_t^xf(t)\, dT \right)\ dt = \int_p^x \ (x-t) \ f(t) \ dt$$