I have a function $$\int^1_{y=0}\int^1_{x=y}e^{x^2}dx\ dy$$
Which I want to perform a change in order of integration. I have plotted the graph:
And it seems it's the area bounded by the y-axis and x-axis. The answer I know is $(e-1)/2$ but it doesn't make sense since a quick check can tell the area of the triangle under is $1/2.$
The limits to be changed to is: $$\int^1_{x=0}\int^x_{y=0}e^{x^2}dy\ dx$$ Giving $$ \left[ \frac{1}{2}e^{x^2} \right]^1_0 = \frac{1}{2}(e-1)$$
On
$$ \int^1_0 \left( \int^1_y e^{x^2} \, dx\right) dy = \iint\limits_{\{\,(x,y)\,:\, 0 \,\le\,y\,\le\,x\,\le\,1\,\}} e^{x^2} \, d(x,y) = \int_0^1 \left( \int_0^x e^{x^2} \, dy \right) dx $$
Look carefully at these inequalities: $$ 0\le y\le x \le 1. $$ Here's another way to say what those inequalities say:
$y$ goes from $0$ to $1.$ For any fixed value of $y$ between $0$ and $1,\,\,$ $x$ goes from $y$ to $1.$
$x$ goes from $0$ to $1.$ For any fixed value of $x$ between $0$ and $1,\,\,$ $y$ goes from $0$ to $x.$
You need to think to the integral as
$$\int^1_{y=0}\int^1_{x=y}f(x,y)\,dx\ dy$$
with $f(x,y)=e^{x^2}$, thus we would obtain the value of $\frac12$ for $f(x,y)=1$.