# Evaluate the integral $\iint_De^{-x^2}\,dx\,dy$

Find $$\iint_De^{-x^2}\,dx\,dy$$, where $$D$$ is the triangle formed by points $$O(0,0), A(1,0), B(1,1)$$

I'm totally lost on this one. I only have experience with circle functions, and converting them to polar and finding the bounds in $$π$$ and $$r$$, but here its just points.

Can anyone help me understand this? Do i just bound $$x$$ to $$[0,1]$$ and $$y$$ to $$[0.1]$$?

• Hint: what is $\int_{y_\min(x)}^{y_\max(x)}dy$ for fixed $x\in[0,\,1]$?
– J.G.
Sep 9, 2021 at 20:27
• Sketch the region. This will help you find the correct bounds. Sep 9, 2021 at 20:27
• @J.G. well, we get a right angle triangle, and $y_{min}$ is $0$ and $y_{max}$ is $1$.
– EL02
Sep 9, 2021 at 20:28
• Great. Consider answering your own question as a next step.
– J.G.
Sep 9, 2021 at 22:13
• @J.G. yep, i will take a while because im not too fast with the syntax here. Thanks for the help!
– EL02
Sep 9, 2021 at 22:14

Note that $$D=\{(x,y)\in\mathbb{R}^2:0\leq x,y\leq1\text{ and }x\leq y\}$$. Therefore $$\iint_De^{-x^2}dxdy=\int_0^1\left(e^{-x^2}\int_0^{x}dy\right) \;dx$$ We have that $$\int_0^{x}dy=x$$ and therefore $$\iint_De^{-x^2}\,dx\,dy=\int_0^1xe^{-x^2}\,dx=\frac12\left(1-\frac1e\right)$$

• Plotting the region $D$ will help you to understand. Sep 10, 2021 at 10:57

$$\int_0^x\int_0^1e^{-x^2}\,dx\,dy$$

$$\int_0^1\int_0^xe^{-x^2}\,dy\,dx=\int_0^1(e^{-x^2}y)|_0^x$$ $$\int_0^1xe^{-x^2}\,dx$$

$$u=-x^2$$

$$du=-2x\,dx$$

$$dx=\frac{du}{-2x}$$

$$\int_0^1xe^u\frac{du}{-2x}=-\frac{1}{2}\int_0^{-1}e^u\,du$$

$$-\frac{1}{2}(e^u|_0^{-1})=-\frac{1}{2}(e^{-1}-1)=-\frac{1}{2e}+\frac{1}{2}$$

$$\iint\limits_{x,y\,:\,0\,<\,y\,<\,x\,<\,1} e^{-x^2} \,d(x,y)$$ Here I have written $$d(x,y)$$ rather than $$dx\,dy$$ or $$dy\,dx$$ to indicate this is a double integral rather than an iterated integral and there is no particular order of integration.

The fact that $$e^{-x^2}$$ is everywhere nonnegative is enough to imply the double integral equals this iterated integral, whose evaluation is easy: $$\iint\limits_{x,y\,:\,0\,<\,y\,<\,x\,<\,1} e^{-x^2} \,d(x,y) = \int_0^1 \left( \int_0^x e^{-x^2} \, dy\right) \,dx$$ (The only time a double integral is not equal to an iterated integral in this way is when the integrals of the positive and negative parts of the function are both infinite, in which case the double integral is not well defined and the iterated integral is in some cases still well defined. In that case the two iterated integrals taken in opposite orders ($$dx\,dy$$ versus $$dy\,dx$$ sometimes evaluate to different numbers.)