$\textsf{Reverse the order of integration in the following integral and find the value of }$
\begin{align*}\displaystyle
I&=\int_0^3 \int_y^{3} x^2 e^{xy} dxdy
\end{align*}
ok apparently $0 \le x \le 3$ and $y \le y \le 3$
but ???
$\textsf{Reverse the order of integration in the following integral and find the value of }$
\begin{align*}\displaystyle
I&=\int_0^3 \int_y^{3} x^2 e^{xy} dxdy
\end{align*}
ok apparently $0 \le x \le 3$ and $y \le y \le 3$
but ???
The integration area is a triangle with top border being $y=x$, right border being the line x=3, and bottom border being the x-axis
Thus the new integration is
$I = \displaystyle \int_0^3 \int_0^y~x^2 e^{xy} ~dy~dx$
NO. This is $\int_0^3\left(\int_{y}^3 x^2e^{xy}dx\right) dy$.
The "outer integral" is with respect to y so $0\le y\le 3$. The "inner integral" is with respect to x and goes from y to 3: $y\le x\le 3$. (You should have seen that "$y\le y\le 3$" made no sense. Perhaps that was a typo.)
Draw a graph with the horizontal lines y= 0 (x-axis) and y= 3 drawn. Draw the vertical line x= 3 and the line $x= y^3$. The region of integration is the triangle bounded by those four lines. Notice that $x= y$ crosses y= 3 at (3, 3). That is a diagonal that divides the square into two triangles. Since the "inner integral" is from y= x on the left to y= 3 on the right, the region of integration is the lower triangle. To cover that triangle, x must range from 0 to 3 and, for each x, y must go from x to 3. The integral is $\int_0^3\int_x^3 x^2e^{xy}dydx$.