Let $f \in L^1\big( (0,1)^2 \big)$. How to prove
$$ \int_{x=0}^1 \int_{t=0}^x f(t,x) \,dt \, dx=\int_{t=0}^1 \int_{x=t}^1 f(t,x) \,dx \, dt.$$
Both integrals cover the triangle bounded by the graph of the function $t(x)=x$ and the axes.
I guess Fubini should be used somewhere, but I am not sure how; Naively, this is not an intergal over a product.
On
$$ \int_0^1 \left( \int_0^x f(t,x) \,dt \right) dx=\int_0^1 \left( \int_t^1 f(t,x) \,dx \right) dt $$
Both of these are equal to $$ \iint\limits_{\{\,(t,x)\,:\, 0 \, \le \, t\, \le\,x\,\le\, 1\,\}} f(t,x) \, d(t,x) $$ Look at the subscript: $$ 0 \le t\le x\le 1. $$ One can say $x$ goes from $0$ to $1,$ and then for each fixed value of $x,$ $t$ goes from $0$ to $x.$
Or one can say $t$ goes from $0$ to $1,$ and then for each fixed value of $t,$ $x$ goes from $t$ to $1.$
Technical note: Fubini's theorem says the three integrals shown above are equal if $$ \iint\limits_{\{\,(t,x)\,:\, 0 \, \le \, t\, \le\,x\,\le\, 1\,\}} |f(t,x)| \, d(t,x) < +\infty. $$ Note the absolute value.
Define $g(t,x) = \displaystyle \left\{ \begin{array}{lc} 1 & 0 \le x \le 1 \text{ and } 0 \le t \le x \\ 0 & \text{otherwise.}\end{array} \right.$
Then $$\int_{x=0}^1 \int_{t=0}^x f(t,x) \,dt \, dx = \int_0^1 \int_0^1 f(t,x)g(t,x) \, dt \, dx.$$
Now apply Fubini's theorem.