Given $f(x,y)$ is a continuous function, Do these integrals equal? Given range $\{ 0 \le x \le 1, 0 \le y \le 1\}$
Do these integrals equal?
$\int_0^1(\int_0^y f(x,y)dx)dy = \int_0^1(\int_0^x f(x,y)dy)dx$
Well, the answer is no. It seems like the triangulars are different in LHS, RHS.
I don't understand $ (D_2)$. Here: $D_1 = \{(x,y) | 0 \le x \le 1, 0 \le y \le x\}$,
but why $D_2 = \{(x,y) | 0 \le x \le 1, x \le y \le 1\}$ ?
Especially this part of $D_2: x \le y \le 1$.

"It seems like the rectangles are different in LHS, RHS."
In fact, they are not rectangles!  $\int_0^1\int_0^x f(x, y)dydx$ takes $x$ to be from $x= 0 $to $x= 1$ and, for each $x, y$ from $y= 0$ to $y= x%.  That is a **triangle** with vertices at %(0, 0), (1, 0), and (1, 1).$  $\int_0^1\int_0^y f(x, y)dxdy$ takes $y$ to be from $y= 0$ to $y= 1$ and, for each $y, x$ from $x= 0 $to $x= y$.  That is a triangle with vertices at $(0, 0), (0, 1),$ and $(1, 1).$  The line $y= x$ is a diagonal of the square $0\le x\le 1$, $0\le y\le 1$.  The first integral  covers the triangle below that diagonal, the second the triangle above that diagonal.
