Consider the following iterated integral $$ \int_{R} \int f(x, y) d A $$ $$ \int_{0}^{1} \int_{0}^{2}(1-4 x+6 y) d y d x $$

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1. Write the iterated integral:
   $$
   \int_{0}^{1} \int_{0}^{2}\left(1 - 4x + 6y\right) \, dy \, dx
   $$

2. Evaluate the inner integral with respect to $$ y $$:
   $$
   \int_{0}^{2}\left(1 - 4x + 6y\right) \, dy
   $$
   Treat $$ x $$ as a constant. Integrate each term:
   $$
   \int_{0}^{2} 1 \, dy = y \Big|_{0}^{2} = 2
   $$
   $$
   \int_{0}^{2} (-4x) \, dy = -4x \cdot y \Big|_{0}^{2} = -8x
   $$
   $$
   \int_{0}^{2} 6y \, dy = 6 \cdot \frac{y^2}{2} \Big|_{0}^{2} = 3y^2 \Big|_{0}^{2} = 3 \cdot 4 = 12
   $$
   Sum the results:
   $$
   2 - 8x + 12 = 14 - 8x
   $$

3. Now, evaluate the outer integral with respect to $$ x $$:
   $$
   \int_{0}^{1} (14 - 8x) \, dx
   $$
   Compute the antiderivative:
   $$
   \int (14 - 8x) \, dx = 14x - 4x^2
   $$
   Evaluate from $$ x = 0 $$ to $$ x = 1 $$:
   $$
   \left[14x - 4x^2 \right]_{0}^{1} = (14 \cdot 1 - 4 \cdot 1^2) - (14 \cdot 0 - 4 \cdot 0^2) = 14 - 4 = 10
   $$

4. The value of the iterated integral is:
   $$
   10
   $$
The value of the iterated integral is 10.

Consider the following iterated integral

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