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The question:
Let $H$ be the parallelogram in $\mathbb{R}^2$ whose vertices are $(1,1), (3,2), (4,5), (2,4).$ Find the affine map $T$ which sense $(0,0)$ to $(1,1), (1,0)$ to $(3,2), (0,1)$ to $(2,4)$. Show that $J_T=5$ (the Jacobian). Use $T$ to convert the integral
$$ \alpha = \int_H e^{x-y}dxdy $$
to an integral over $I^2$ (unit square) and thus compute $\alpha$.

So I'm not sure why I'm having some trouble with this. I've done everything up to solving the integral. The affine map I found is $T(x,y) = (1,1) + (2x+y,x+3y)$, and I've checked that it is correct. However, in my mind, it makes sense that in changing the integral to be over the square, I take $T^{-1}(H) = I^2$, that is, obtaining:
$$ \alpha = \int_H e^{x-y}dxdy = \int_{I^2} f(T^{-1}(x,y))|J_{T^{-1}}|dxdy $$ (note I'm kind of abusing the notation of $(x,y)$...should probably be $(u,v)$ or something, but nonetheless)
However, the integral is supposed to be $f(T(x,y))|J_T|$, not of $T^{-1}$. Can someone explain why this is so? I think I'm just forgetting something from lower division calculus, as it's been a while. It made sense back then, but for some reason this problem is screwing me up.

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You are correct about the direction of the mapping. Now reread Theorem 10.9 You are integrating $$\int_H f(y) dy.$$ Observe that $$y=T(x).$$ Then apply Theorem 10.9 to get

$$\int_H f(y)dy = \int_{I^2} f(T(x))\vert J_T(x)\vert dx.$$

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