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Evaluate $$ \int\int_D e^{y^2} dA $$ where D is the triangular region with vertices (0,0), (0,1) and (2,1)

My attempts:

$$ \int^{2}_0 \int^{1}_\frac{x}{2} e^{y^2} dy dx $$ or $$ \int^{0}_1\int^{2y}_2 e^{y^2} dx dy $$

but I couldn't evaluate the integral so I think I must've done something wrong when finding the region D.

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So close on that second one! The left boundary is the $y$-axis--that is, the line $x=0$--so instead of $\int_2^{2y}$, you should have $\int_0^{2y}$ on your inner integral. That goof made your integral problematic to evaluate. –  Cameron Buie Nov 3 '12 at 8:11
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2 Answers

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The integral $$\int_0^2\int_{x/2}^1 e^{y^2} dy dx$$ is hard to evaluate, so you need to get the $dx$ inside. The following will do: $$\int_0^1\int_0^{2y} e^{y^2} dx dy.$$ Evaluation is simple: \begin{align} \int_0^1\int_0^{2y} e^{y^2} dx dy &= \int_0^1 2y e^{y^2} dy \\ &= \left[e^{y^2}\right]_0^1 \\ &= e-1. \end{align}

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Right! It's 3 in the morning so I forget how to evaluate integrals. thank you! –  40Plot Nov 3 '12 at 7:49
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It is a tricky question. The is no elementary-function that is equal to the integral of e to the 'x-squared' power. That integral is very common in calculus books to show this phenomena. (I am not an english-speaking native, I went to a spanish-speaking college and I don't remember the english term)

You can get computed values using a numeric approach.

You could see some discussion about this in http://www.physicsforums.com/showthread.php?t=103752

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