5. Evaluate the double integral \[ \iint_{D}(2 x+y) d A, \quad D=\{(x, y) \mid 1 \leq y \leq 2, \quad y-1 \leq x \leq 1\} \]

To evaluate the double integral \[ \iint_{D}(2 x+y) d A \] where \[ D=\{(x, y) \mid 1 \leq y \leq 2, \quad y-1 \leq x \leq 1\}, \] we first need to understand the region \(D\) defined by the inequalities for \(y\) and \(x\). 1. The bounds for \(y\) are straightforward: \(y\) ranges from \(1\) to \(2\). 2. The bounds for \(x\) depend on \(y\): for a fixed \(y\), \(x\) ranges from \(y - 1\) to \(1\). Now, we set up the double integral with the limits for \(x\) and \(y\): \[ \int_{1}^{2} \int_{y-1}^{1} (2x+y) \, dx \, dy. \] We will first evaluate the inner integral \(\int_{y-1}^{1} (2x+y) \, dx\): \[ \int_{y-1}^{1} (2x+y) \, dx = \int_{y-1}^{1} 2x \, dx + \int_{y-1}^{1} y \, dx. \] Calculating the first part: \[ \int_{y-1}^{1} 2x \, dx = [x^2]_{y-1}^{1} = 1^2 - (y-1)^2 = 1 - (y^2 - 2y + 1) = 2y - y^2. \] Now for the second integral: \[ \int_{y-1}^{1} y \, dx = y \cdot [x]_{y-1}^{1} = y(1 - (y-1)) = y(2 - y). \] Combining these results, we get: \[ \int_{y-1}^{1} (2x+y) \, dx = (2y - y^2) + (y(2 - y)) = 2y - y^2 + 2y - y^2 = 4y - 2y^2. \] Now, we have reduced our double integral to a single integral with respect to \(y\): \[ \int_{1}^{2} (4y - 2y^2) \, dy. \] Next, we will compute the integral: \[ \int (4y - 2y^2) \, dy = [2y^2 - \frac{2}{3}y^3]_{1}^{2}. \] Calculating the definite integral: 1. At \(y = 2\): \[ 2(2^2) - \frac{2}{3}(2^3) = 8 - \frac{16}{3} = \frac{24}{3} - \frac{16}{3} = \frac{8}{3}. \] 2. At \(y = 1\): \[ 2(1^2) - \frac{2}{3}(1^3) = 2 - \frac{2}{3} = \frac{6}{3} - \frac{2}{3} = \frac{4}{3}. \] Now, subtract: \[ \frac{8}{3} - \frac{4}{3} = \frac{4}{3}. \] Thus, the value of the double integral is \[ \boxed{\frac{4}{3}}. \]