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.
$\endgroup$ 12 Answers
$\begingroup$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}
$\endgroup$ 1 $\begingroup$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
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