Solving a double integral using change of variables. $$\int ^{1}_{0} \int^{1}_{y}e^{-x^{2}}\,dx\,dy$$
To solve this I know one must use change of variables, but the problem is that I do not know how to approach the actual change.
Just thinking out loud here: Could I do a change of order from $x$ in $[ y , 1 ]$ and $y$ in $[ 0 , 1 ]$ to $y$ in $[x , 1]$ and $x$ in $[0,1]$ ?
 A: Draw out the region of integration.  You should see that it is a right triangle above the line $y=x$.  The integral as stated integrates from this line to the line $y=1$.  To change the order of integration, integrate from the $y$-axis to the line $y=x$, from $y=0$ to $y=1$.  The integral is then
$$\int_0^1 dx \, e^{-x^2/2} \, \int_0^x dy $$
This is easily doable.
A: HINT: Draw out the region of integration. Then change the variables accordingly by fixing a $y$ value first. Then going on and finding the bounds for the $x$ values.
A: The integral is over values of $(x,y)$ such that $0 \le y \le 1$ and $y \le x \le 1$. 
Combining these gives us $0 \le y \le x \le 1$, which can be rewritten as $0 \le x \le 1$ and $0 \le y \le x$. 
So we change the bounds as follows: $\displaystyle\int_{0}^{1}\int_{y}^{1}e^{-x^2}\,dx\,dy = \int_{0}^{1}\int_{0}^{x}e^{-x^2}\,dy\,dx$.
For this particular example, changing the bounds using just inequalities was easy. However, for more complicated examples, it is better to draw a picture of the region of integration and figure out the new bounds of integration from that.
A: If we draw the region first, we get that the region of integration is just a triangle. If we fix some $x$ value we get that $y$ ranges from $0$ to $y = x$. and $x$ goes from $0$ to $1$. So we now have $$\int_0^1\int_0^x e^{-x^2} \mathrm{d}y\;\mathrm{d}x$$
A: Using Iverson Brackets,
$$
\begin{align}
\int_0^1\int_y^1e^{-x^2}\,\mathrm{d}x\,\mathrm{d}y
&=\int_0^1\int_0^1[x\ge y]e^{-x^2}\,\mathrm{d}x\,\mathrm{d}y\\
&=\int_0^1\int_0^1[x\ge y]e^{-x^2}\,\mathrm{d}y\,\mathrm{d}x\\
&=\int_0^1\int_0^xe^{-x^2}\,\mathrm{d}y\,\mathrm{d}x
\end{align}
$$
