Application of Leibniz Rule [closed]

I want to show that if $f:[0,b]\times[0,b]\to\mathbb{R}$ is a continuous function, then$$\int_0^b\int_0^xf(x,y)dydx=\int_0^b\int_y^bf(x,y)dxdy.$$This exercise is in my analysis book in the Leibniz's Rule section. That is, the rule that says we can commute derivatives with integrals. In the same section the change of integration order is proved using Leibniz's Rule. The book is "Curso de Análise vol. 2" by Elon Lages Lima.

Edit: I managed to solve it:First, the statement makes sense because if we consider the set$$\Omega=\{(x,y)\in\mathbb{R}^2: 0\le y\le x\le b\},$$we have$$\int_0^b\int_0^xf(x,y)dydx=\iint_\Omega f(x,y)dxdy$$and$$ \int_0^b\int_y^bf(x,y)dxdy=\iint_\Omega f(x,y)dxdy.$$But, this exercise musb be shown using Leibniz's Rule (of integrals) or inversion of the order of repeated integrals (that is, Fubini's Theorem). For to show it, we can consider the function$\tilde{f}:[0,b]\times[0,b]\to\mathbb{R}$ given by$$\tilde{f}=\begin{cases}f(x,y),\; if\; 0\le y\le x\le b,\\0,\; otherwise.\end{cases}$$Then for a fixed $x\in[0,b]$, we have$$f(x,y)=\tilde{f}(x,y), \forall y\in[0,x].$$Therefore$$\int_0^x f(x,y)dy=\int_0^x \tilde{f}(x,y)dy=\int_0^b \tilde{f}(x,y)dy,$$and then$$\int_0^b\int_0^x f(x,y)dydx=\int_0^b\int_0^b \tilde{f}(x,y)dydx.\tag{$1$}$$In other hands, for $y\in[0,x]$ fixed it's true that$$f(x,y)=\tilde{f}(x,y), \forall x\in[y,b].$$This implies that$$\int_y^b f(x,y)dx=\int_y^b \tilde{f}(x,y)dx=\int_0^b \tilde{f}(x,y)dx.$$It follows that$$\int_0^b\int_y^b f(x,y)dx=\int_0^b\int_0^b \tilde{f}(x,y)dxdy. \tag{$2$}$$By (1) and (2) and by Fubini's Theorem, the statement follows.