I am having trouble with this integral $$ I = \iint_D e^{x^2} dxdy\\ \text{where $D$ is the triangle with corners $(0,0)$, $(-1,0)$ och $(-1,1)$} $$ I have seen examples of solutions to similar integrals, but I still can't solve this one.
I see that (edit: marked the error in red) $$ \iint_D e^{x^2} dxdy = \int_{-1}^0\int_0^{\color{red}{1-x}}e^{x^2}dydx $$ $$ \begin{cases} t&=1-x\\x&=1-t\\dx&=-dt\end{cases} $$ I try to integrate with respect to $y$ to get some expression similar to $xe^{x^2}$ (instead of $e^{x^2}$) that I can integrate: $$ \int_0^t e^{(t-1)^2} dy = \left[ ye^{(t-x)^2} \right]_0^t = te^{(t-1)^2} $$ Integrate with respect to $t$ by substituting $x=1-t$ and $dx = -dt$:
edit: I marked the mistake pointed out by this answer in red here (should have been $te^{t^2}e^{-2t}e$). The big mistake I made, was pointed out in the accepted answer. $$ \begin{align} I &= \int_{-1}^0 te^{(t-1)^2} dx = \int_2^1 \color{red}{te^{(t-1)^2}}\cdot (-1)~dt \\&=-\int_2^1 \left(\color{red}{te^{t^2} + te^{-2t} + te} \right)dt = -\left[ \int\left(te^{t^2} + te^{-2t} + te \right)dt \right]_2^1 \end{align} $$ But this integral is evaluated (wolfram alpha) to $$ e + e^4/2 - \frac{5e^{-4}}{4} + \frac{3e^{-2}}{4} $$ while the correct answer is supposed to be $$ \frac{e-1}{2} $$ so obviously I'm doing something wrong early. How do you solve this kind of integral?
The region you are integrating over is bounded by the lines $y=0$, $x=-1$ and $y=-x$. So if you set up the right integral, we see that
$$\iint_De^{x^2} dydx=\int_{-1}^0\int_{0}^{-x} e^{x^2} dydx=\int_{-1}^0-xe^{x^2}dx$$ Now substitute $u=x^2$, then $du=2xdx$ and the lower bound becomes $(-1)^2$ so $$\iint_D e^{x^2}dx=-\frac{1}{2}\int_{1}^0e^udu=-\frac{1}{2}\bigg(1-e\bigg)=\frac{e-1}{2}$$