Prove that $\int_{0}^{a}dx\int_{0}^{x}dy\int_{0}^{y}f(x)f(y)f(z)dz=\frac{1}{3!}\left(\int_{0}^{a}f(t)dt\right)^3$. My question is how to prove that $$\int_{0}^{a}dx\int_{0}^{x}dy\int_{0}^{y}f(x)f(y)f(z)dz=\frac{1}{3!}\left(\int_{0}^{a}f(t)dt\right)^3$$. $f$ is a continuous function.
It is an exercise from my calculus book in section: integrate on the bounded sets, so it won't be too difficult.
I have already proved an easier version. $$\int_{0}^{a}dx\int_{0}^{x}f(x)f(y)dy=\frac{1}{2}\left(\int_{0}^{a}f(t)dt\right)^2.$$
proof: Note that $$\int_{0}^{a}dx\int_{0}^{x}f(x)f(y)dy=\int_{0}^{a}dy\int_{0}^{y}f(x)f(y)dx,$$and  the union of the two integral area is $[0,a]^2$, thus $$\int_{0}^{a}dx\int_{0}^{x}f(x)f(y)dy=\frac{1}{2}\int_0^a dx\int_0^a f(x)f(y)dy=\frac{1}{2}\left(\int_{0}^{a}f(t)dt\right)^2.$$
By this, the question is reduced to show
$$\frac{1}{2}\int_0^a \left(\int_0^x f(t)dt\right)^2f(x)dx=\frac{1}{3!}\left(\int_{0}^{a}f(t)dt\right)^3.$$
Any ideas might help!
 A: Hint: you can prove this directly, much like you did for the $2$-dimensional case, by noticing that the cube $[0,a]^3$ splits into six tetrahedra, each one corresponding to a different permutation of the variables of integration $(x_1,x_2,x_3)$. On each one of these simplexes, an inequality of type $0\leqslant x_i \leqslant x_j \leqslant x_k \leqslant a$ holds, where $i,j,k \in \{1,2,3\}$ – which is why there are six: the number of permutations of $3$ symbols is $3!$. Your original integral has only one of those tetrahedra as its domain, and since it doesn't matter over which specific tetrahedron you calculate the integral (variables of integration are mute, so they can be relabeled with impunity), you may integrate over the whole cube and divide by the number of tetrahedra.
Bonus. This easily generalizes to arbitrary $n$, and is the basis for the Dyson series expansion in quantum mechanics! The cool observation is that, if you sum up the repeated integrals in all dimensions $n$, by your calculation you obtain
$$\sum_{n=0}^\infty \int_0^a\int_0^{x_1}\cdots \int_0^{x_{n-1}} f(x_1)\cdots f(x_n)\ dx_n \cdots dx_1 = \sum_{n=0}^\infty \frac{1}{n!}\left(\int_0^a f(t)\ dt \right)^n,  $$
which is just $\exp\left(\int_0^a f(t)\ dt\right)$. The quantum-mechanical version of this argument has a few added subtleties due to the fact that the product of functions in the integrand is not commutative (each $f(x_j)$ is a linear operator, namely the Hamiltonian of the system at time $x_j$), but other than that, this is pretty much it.
A: From the last step of OP:
$$I=\frac{1}{2}\int_0^a \left(\int_0^x f(t)dt\right)^2f(x)dx$$
Let $\int_{0}^{x} f(t) dt =z \implies f(x) dx=dz.$
So $$I=\frac{1}{2}\int_{0}^{a} z^2 dz= \frac{z^3}{6}=\frac{1}{3!}\left(\int_{0}^{a}f(t)dt\right)^3.$$
