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\begin{equation}

\begin{array}{c}{\text { Use the transformation } x=u+(1 / 2) v, y=v \text { to evaluate the }} \\ {\text { integral }} \\ {\int_{0}^{2} \int_{y / 2}^{(y+4) / 2} y^{3}(2 x-y) e^{(2 x-y)^{2}} d x d y} \\ {\text { by first writing it as an integral over a region } G \text { in the uv-plane. }}\end{array}

\end{equation}

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book is so no, we have execute stoop close. Have we y calls to be? We're right at the Geico. Been J if you be equals 21 1/2 zero on one Silly answer is one. And the second step is to determine the region G. So we say we have regional or in the x y plane. And, uh, listen, we transform into the boundaries of tree celeb interest of army retracement transformed to the boundaries of the G. So you are. What we have is two x equals two Why and why Paws four equals to two X and, like zero n y equals two. So that's going to transform to our G by relation here, that seems that happen. So the first equation will be transformed to you. It was zero second ones. You're close to the 3rd 1 This vehicle, Cyril. And the fourth ones, we equals two. Okay, so right now we have the original integrate integral from 0 to 2 from halfway to have a close to Why cubed two x minus Y some see 32 months. So I squared and the x t y. Okay, so we plugging know of this transformation. We're gonna have 0 to 2 0 to 2. Vey cute to you since even before your squared times one which is a coping and the you Devi you everything this integral we have it is extensive, but this one was for Comes way through the for over four from zero to. So if I answer for this, Integral is 8 to 16 minus one.

University of Illinois at Urbana-Champaign