2. Calculate the double integral \[ \iint_{R}\left(y+x y^{-2}\right) d A, \quad R=\{(x, y) \mid 0 \leq x \leq 2,1 \leq y \leq 2\} \]

We want to calculate \[ \iint_{R}(y+x y^{-2})\,dA, \] where \[ R=\{(x, y) \mid 0 \leq x \leq 2,\; 1 \leq y \leq 2\}. \] Since \( R \) is a rectangular region, we can write the integral as an iterated integral: \[ \iint_{R}(y+x y^{-2})\,dA = \int_{y=1}^{2} \int_{x=0}^{2} \left( y + x y^{-2} \right) \, dx \, dy. \] ### Step 1. Integrate with respect to \( x \) For a fixed \( y \), the inner integral is: \[ \int_{x=0}^{2} \left( y + x y^{-2} \right) \, dx. \] Since \( y \) and \( y^{-2} \) are constant with respect to \( x \), we can separate the integral: \[ \int_{0}^{2} \left( y + x y^{-2} \right) dx = y \int_{0}^{2} dx + y^{-2} \int_{0}^{2} x \, dx. \] Compute each integral: - \(\displaystyle \int_{0}^{2} dx = 2.\) - \(\displaystyle \int_{0}^{2} x \, dx = \left[\frac{1}{2}x^2\right]_{0}^{2} = \frac{1}{2}(4) = 2.\) Substitute these results: \[ = y\cdot 2 + y^{-2}\cdot 2 = 2y + 2y^{-2}. \] ### Step 2. Integrate with respect to \( y \) Now, substitute the result into the outer integral: \[ \int_{y=1}^{2} \left( 2y + 2y^{-2} \right) dy = 2\int_{1}^{2} y \, dy + 2\int_{1}^{2} y^{-2} \, dy. \] **First integral:** \[ 2\int_{1}^{2} y \, dy = 2\left[\frac{y^2}{2}\right]_{1}^{2} = \left[y^2\right]_{1}^{2} = 2^2 - 1^2 = 4 - 1 = 3. \] **Second integral:** \[ 2\int_{1}^{2} y^{-2} \, dy. \] Recall that \[ \int y^{-2}\,dy = -y^{-1}. \] Thus, \[ 2\int_{1}^{2} y^{-2} \, dy = 2\left[-y^{-1}\right]_{1}^{2} = 2\left(-\frac{1}{2} + 1\right) = 2\left(\frac{1}{2}\right) = 1. \] ### Step 3. Combine the Results Add the two contributions: \[ 3 + 1 = 4. \] Thus, the value of the double integral is: \[ \iint_{R}\left(y+x y^{-2}\right)\,dA = 4. \]