Question

In the exercise you will use the change of variables $u = 4x + 2y$, $v = 3x + 3y$ to find $\iint_R (4x + 2y)e^{12x^2 + 18xy + 6y^2} dA$ where R is the region bounded by $4x + 2y = -2$, $4x + 2y = 0$, $3x + 3y = 3$ and $3x + 3y = 1$ $\iint_R (4x + 2y)e^{12x^2 + 18xy + 6y^2} dA = \int_A^B \int_C^D F(u, v) dv du$ where $A = -2$, $B = \text{ }$, $C = \text{ }$, $D = 1$, $F(u, v) = \text{ }$ After the change of variables this integral is simple if done the right way. In particular we obtain $\iint_R (4x + 2y)e^{12x^2 + 18xy + 6y^2} dA = $

          In the exercise you will use the change of variables $u = 4x + 2y$, $v = 3x + 3y$ to find
$\iint_R (4x + 2y)e^{12x^2 + 18xy + 6y^2} dA$
where R is the region bounded by $4x + 2y = -2$, $4x + 2y = 0$, $3x + 3y = 3$ and $3x + 3y = 1$
$\iint_R (4x + 2y)e^{12x^2 + 18xy + 6y^2} dA = \int_A^B \int_C^D F(u, v) dv du$
where $A = -2$, $B = \text{ }$, $C = \text{ }$, $D = 1$, $F(u, v) = \text{ }$
After the change of variables this integral is simple if done the right way. In particular we obtain
$\iint_R (4x + 2y)e^{12x^2 + 18xy + 6y^2} dA = $

In the exercise you will use the change of variables u = 4x + 2y, v = 3x + 3y to find
(4x + 2y)e^12x^2 + 18xy + 6y^2 dA
where R is the region bounded by 4x + 2y = -2, 4x + 2y = 0, 3x + 3y = 3 and 3x + 3y = 1
(4x + 2y)e^12x^2 + 18xy + 6y^2 dA = ^B ^D F(u, v) dv du
where A = -2, B =, C =, D = 1, F(u, v) =
After the change of variables this integral is simple if done the right way. In particular we obtain
(4x + 2y)e^12x^2 + 18xy + 6y^2 dA =

Added by Marie S.

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