Double Integrals over Rectangles 1. Evaluate the double integral \[ \iint_{R}(4-2 y) d A, \quad R=[0,1] \times[0,1] \] by first identifying it as the volume of a solid.

We are given the integral \[ \iint_{R}(4-2y)\,dA, \] with the region \[ R = [0,1] \times [0,1]. \] This integral represents the volume under the surface \( z = 4-2y \) above the rectangle \( R \). **Step 1. Express the double integral as an iterated integral.** Since the region \( R \) is rectangular, we write \[ \iint_{R}(4-2y)\,dA = \int_{x=0}^{1} \int_{y=0}^{1}(4-2y)\,dy\,dx. \] **Step 2. Evaluate the inner integral with respect to \( y \).** \[ \int_{y=0}^{1}(4-2y)\,dy. \] Find the antiderivative: \[ \int (4-2y)\,dy = 4y-\frac{2y^2}{2} = 4y-y^2. \] Now, evaluate from \( y=0 \) to \( y=1 \): \[ \left[4y-y^2\right]_{y=0}^{y=1} = (4(1)-1^2) - (4(0)-0^2) = 4-1 = 3. \] **Step 3. Evaluate the outer integral with respect to \( x \).** Since the result of the inner integration is independent of \( x \): \[ \int_{x=0}^{1} 3\,dx = 3 \times (1-0) = 3. \] Thus, the value of the double integral is \[ \boxed{3}. \]