Change of variable in the integral Let $f:[0,b]\times[0,b]\rightarrow\mathbb{R}$ be continuous. Prove that $$\int_0^b dx\int_0^x f(x,y) \, dy=\int_0^b dy\int_y^b f(x,y) \, dx$$
La idea es usar los teoremas:
\ Regla de Leibniz, de derivacion bajo el signo de integral o  teorema de inversion del orden de las integrales, el cual dice: 
Para cada $f:[a,b]\times[a,b]\rightarrow\mathbb{R}$ continua, vale $$\int_a^b ds\int_c^d f(s,t) \, dt=\int_c^d dt\int_a^b f(s,t) \, ds$$
 A: Hint:
Let $S = \{ (x,y)\in\mathbb R^2 \mid 0 \le y \le x \le b \}$ and denote the indicator function of $S$ by $\mathbb 1_S$.
Then, we have
\begin{align*}
\int_0^b \left(\int_0^x f(x,y) \; \mathrm dy \right) \mathrm dx = \int_0^b \left( \int_0^b \mathbb 1_S(x,y) f(x,y) \; \mathrm dy \right) \mathrm dx.
\end{align*}
Now apply Fubini's theorem on the right hand side.
A: $$
\int_0^b dx\int_0^x f(x,y) \, dy = \iint_R f(x,y)\,d(x,y)
$$
where $R$ is the set of points $(x,y)$ for which
$$
0 \le y \le x \le b.
$$
And
$$
\int_0^b dy\int_y^b f(x,y) \, dx = \iint_R f(x,y)\,d(x,y)
$$
where $R$ is that same region.
Draw an accurate picture of the region $R$.  You'll see a triangle.  One side is a segment on the $x$-axis; another is a vertical line segment from $(1,0)$ to $(1,1)$; another is a diagonal segment of the line $y=x$.
Notice that $x$ goes from $0$ to $b$, but for any fixed value of $x$ (meaning, in other words, you're drawing a vertical line) $y$ goes from $0$ up to a point on the vertical line, whose $y$-coordinate is $x$.
But also notice that for any fixed value of $y$ (meaning, in other words, you're drawing a horizontal line) $x$ goes from that diagonal line rightward until in reaches the aforementioned vertical segment.
One way of writing this integral considers those vertical lines; the other considers those horizontal lines.
