Let's solve each part step-by-step. --- ### **1. Sketch \( S \) and \( P \)** #### **Domain \( S \):** \[ S = \{ (x_1, x_2) \in \mathbb{R}^2 : 2 \leq x_1 \leq 5,\ x_1 - 1 \leq x_2 \leq x_1^2 \} \] - \( x_1 \) ranges from 2 to 5. - For each \( x_1 \), \( x_2 \) ranges from \( x_1 - 1 \) (a line) to \( x_1^2 \) (a parabola). #### **Domain \( P \):** \[ P = \{ (x_1, x_2) \in \mathbb{R}^2 : -1 \leq x_2 \leq 3,\ 10 \leq x_1 \leq e^{x_2} + 10 \} \] - \( x_2 \) ranges from \(-1\) to \(3\). - For each \( x_2 \), \( x_1 \) ranges from 10 to \( e^{x_2} + 10 \). ##### **Sketch:** - For \( S \): Draw vertical strips from \( x_1 = 2 \) to \( x_1 = 5 \), and for each \( x_1 \), draw the region between the line \( x_2 = x_1 - 1 \) and the parabola \( x_2 = x_1^2 \). - For \( P \): Draw horizontal strips from \( x_2 = -1 \) to \( x_2 = 3 \), and for each \( x_2 \), draw the region between \( x_1 = 10 \) and \( x_1 = e^{x_2} + 10 \). --- ### **2. Determine the Domain Type of \( S \) and Transform It** \( S \) is a **Type I (vertical strip) domain** because \( x_1 \) is bounded first and \( x_2 \) is bounded by functions of \( x_1 \): \[ S = \{ (x_1, x_2) : 2 \leq x_1 \leq 5,\ x_1 - 1 \leq x_2 \leq x_1^2 \} \] **Transform to Type II (horizontal strip) domain:** Find the possible range for \( x_2 \): - The minimum for \( x_2 \) is at \( x_1 = 2 \): \( x_2 = 2 - 1 = 1 \) - The maximum for \( x_1 = 5 \): \( x_2 = 25 \) - But for any \( x_2 \), \( x_2 = x_1 - 1 \) and \( x_2 = x_1^2 \) must have \( x_1 \in [2,5] \). - For a fixed \( x_2 \), \( x_1 \) must satisfy \( x_2 = x_1 - 1 \) and \( x_2 = x_1^2 \), so \( x_1 \) runs over the intersection. For a given \( x_2 \), \( x_1 \) ranges between the solutions to: \[ x_1 - 1 \leq x_2 \leq x_1^2 \implies \sqrt{x_2} \geq x_1 \geq x_2 + 1 \] But \( x_1 \) must also satisfy \( 2 \leq x_1 \leq 5 \). So we can write: \[ S = \{ (x_1, x_2) : x_2 \in [1, 25],\ \max(2, x_2 + 1) \leq x_1 \leq \min(5, \sqrt{x_2}) \} \] --- ### **3. Calculate the Riemann Integral over \( S \) for \( f(x_1, x_2) = x_1^2 + 6x_2 \)** Use the Type I setup (as in the original domain): \[ \int_S f(x) \, dx = \int_{x_1=2}^{5} \int_{x_2 = x_1 - 1}^{x_1^2} \left( x_1^2 + 6x_2 \right) dx_2 dx_1 \] **Integrate with respect to \( x_2 \):** \[ \int_{x_2 = x_1 - 1}^{x_1^2} \left( x_1^2 + 6x_2 \right) dx_2 = x_1^2 (x_2) + 3x_2^2 \Big|_{x_2 = x_1 - 1}^{x_2 = x_1^2} \] Plug in limits: \[ = \left[ x_1^2(x_1^2) + 3(x_1^2)^2 \right] - \left[ x_1^2(x_1 - 1) + 3(x_1 - 1)^2 \right] \] \[ = \left[ x_1^4 + 3x_1^4 \right] - \left[ x_1^3 - x_1^2 + 3(x_1^2 - 2x_1 + 1) \right] \] \[ = 4x_1^4 - \left( x_1^3 - x_1^2 + 3x_1^2 - 6x_1 + 3 \right) \] \[ = 4x_1^4 - x_1^3 + x_1^2 + 6x_1 - 3 \] Now, integrate with respect to \( x_1 \) from 2 to 5: \[ \int_{2}^{5} \left( 4x_1^4 - x_1^3 + x_1^2 + 6x_1 - 3 \right) dx_1 \] Integrate term by term: - \( \int 4x_1^4 dx_1 = 4 \frac{x_1^5}{5} \) - \( \int -x_1^3 dx_1 = -\frac{x_1^4}{4} \) - \( \int x_1^2 dx_1 = \frac{x_1^3}{3} \) - \( \int 6x_1 dx_1 = 3x_1^2 \) - \( \int -3 dx_1 = -3x_1 \) So, \[ \left[ \frac{4}{5} x_1^5 - \frac{1}{4} x_1^4 + \frac{1}{3} x_1^3 + 3x_1^2 - 3x_1 \right]_{2}^{5} \] Calculate at \( x_1 = 5 \): - \( \frac{4}{5}(3125) = 2500 \) - \( \frac{1}{4}(625) = 156.25 \) - \( \frac{1}{3}(125) \approx 41.666... \) - \( 3(25) = 75 \) - \( 3(5) = 15 \) So sum for \( x_1 = 5 \): \[ 2500 - 156.25 + 41.666... + 75 - 15 = 2445 - 156.25 + 41.666... = 2445 + 41.666... - 156.25 = 2486.666... - 156.25 = 2330.416... \] Now for \( x_1 = 2 \): - \( \frac{4}{5}(32) = 25.6 \) - \( \frac{1}{4}(16) = 4 \) - \( \frac{1}{3}(8) \approx 2.666... \) - \( 3(4) = 12 \) - \( 3(2) = 6 \) Sum for \( x_1 = 2 \): \[ 25.6 - 4 + 2.666... + 12 - 6 = 21.6 + 2.666... + 12 - 6 = 24.266... + 12 - 6 = 36.266... - 6 = 30.266... \] Final value: \[ 2330.416... - 30.266... = 2300.15 \] --- ## **Final Answer** ### **Step-by-step:** 1. Sketches: See above. 2. Domain type of \( S \): Type I (vertical strip), can be rewritten as Type II. 3. Riemann integral: \[ \int_S (x_1^2 + 6x_2) \, dx = \boxed{2300.15} \] (Rounded to two decimal places. For exact, keep the fractional expressions.)