Let's tackle your exercise step by step: --- ### 1. **Sketch the Domains \( S \) and \( P \)** #### **Domain \( S \):** \[ S = \left\{ (x_1, x_2) \in \mathbb{R}^2 :\ 2 \leq x_1 \leq 5,\quad x_1 - 1 \leq x_2 \leq x_1^2 \right\} \] - \( x_1 \) ranges from \( 2 \) to \( 5 \). - For each \( x_1 \), \( x_2 \) ranges from \( x_1 - 1 \) up to \( x_1^2 \). **Sketch for \( S \):** - Draw vertical lines at \( x_1 = 2 \) and \( x_1 = 5 \). - For each \( x_1 \), draw the lower boundary \( x_2 = x_1 - 1 \) and the upper boundary \( x_2 = x_1^2 \). - The region is the area between these two curves, as \( x_1 \) goes from \( 2 \) to \( 5 \). #### **Domain \( P \):** \[ P = \left\{ (x_1, x_2) \in \mathbb{R}^2 :\ -1 \leq x_2 \leq 3,\quad 10 \leq x_1 \leq e^{x_2} + 10 \right\} \] - \( x_2 \) ranges from \( -1 \) to \( 3 \). - For each \( x_2 \), \( x_1 \) ranges from \( 10 \) up to \( e^{x_2} + 10 \). **Sketch for \( P \):** - Draw horizontal lines at \( x_2 = -1 \) and \( x_2 = 3 \). - For each \( x_2 \), draw the left boundary \( x_1 = 10 \) and the right boundary \( x_1 = e^{x_2} + 10 \). - The region is the area to the right of \( x_1 = 10 \) and to the left of \( x_1 = e^{x_2} + 10 \), as \( x_2 \) goes from \( -1 \) to \( 3 \). --- ### 2. **Domain Type of \( S \) and Transformation** #### **Domain Type:** - **As given:** \( S \) is a **Type I region** (vertical strip): \( x_1 \) is between constants, \( x_2 \) is between functions of \( x_1 \). - **Type II region** (horizontal strip): \( x_2 \) is between constants, \( x_1 \) is between functions of \( x_2 \). #### **Transform \( S \) into Type II:** Given \( 2 \leq x_1 \leq 5 \) and \( x_1 - 1 \leq x_2 \leq x_1^2 \). Find bounds for \( x_2 \): - The **minimum** value of \( x_2 \) is when \( x_1 = 2 \): \( x_2 \geq 2 - 1 = 1 \). - The **maximum** value of \( x_2 \) is when \( x_1 = 5 \): \( x_2 \leq 5^2 = 25 \). But for each \( x_2 \), find the set of \( x_1 \) such that \( 2 \leq x_1 \leq 5 \) **and** \( x_1 - 1 \leq x_2 \leq x_1^2 \). Let’s solve for \( x_1 \) in terms of \( x_2 \): - From \( x_1 - 1 \leq x_2 \implies x_1 \leq x_2 + 1 \) - From \( x_2 \leq x_1^2 \implies x_1 \geq \sqrt{x_2} \) (since \( x_1 \geq 0 \)) Thus, \[ S = \left\{ (x_1, x_2) : 1 \leq x_2 \leq 25,\quad \max(2, \sqrt{x_2}) \leq x_1 \leq \min(5, x_2 + 1) \right\} \] But need to check for which \( x_2 \) the interval is non-empty. - \( x_1 \geq 2 \implies x_2 \geq 1 \) (from \( x_1 - 1 \leq x_2 \) and \( x_1 \geq 2 \)) - \( x_1 \leq 5 \implies x_2 + 1 \geq 2 \implies x_2 \geq 1 \) - \( x_1 \geq \sqrt{x_2} \), but \( x_1 \geq 2 \implies x_2 \leq 25 \) - \( x_2 \geq \sqrt{x_2} \) when \( x_2 \geq 0 \) So, for \( x_2 \) from \( 1 \) to \( 25 \), the region is: \[ S = \left\{ (x_1, x_2) : 1 \leq x_2 \leq 25,\quad \max(2, \sqrt{x_2}) \leq x_1 \leq \min(5, x_2 + 1) \right\} \] --- ### 3. **Calculate the Riemann Integral** Given: \[ f(x_1, x_2) = x_1^2 + 6x_2 \] \[ \iint_S f(x)dx = \int_{x_1=2}^{5} \int_{x_2=x_1-1}^{x_1^2} (x_1^2 + 6x_2) dx_2 dx_1 \] Let’s compute: #### **First, inner integral (in \( x_2 \)):** \[ \int_{x_2=x_1-1}^{x_1^2} (x_1^2 + 6x_2) dx_2 = x_1^2 (x_2)\Big|_{x_1-1}^{x_1^2} + 6 \frac{x_2^2}{2}\Big|_{x_1-1}^{x_1^2} \] \[ = x_1^2 [x_1^2 - (x_1 - 1)] + 3 [ (x_1^2)^2 - (x_1 - 1)^2 ] \] \[ = x_1^2 [x_1^2 - x_1 + 1] + 3 [ x_1^4 - (x_1^2 - 2x_1 + 1) ] \] \[ = x_1^4 - x_1^3 + x_1^2 + 3x_1^4 - 3x_1^2 + 6x_1 - 3 \] \[ = (x_1^4 + 3x_1^4) + (-x_1^3) + (x_1^2 - 3x_1^2) + 6x_1 - 3 \] \[ = 4x_1^4 - x_1^3 - 2x_1^2 + 6x_1 - 3 \] #### **Now, integrate with respect to \( x_1 \) from 2 to 5:** \[ \int_{2}^{5} (4x_1^4 - x_1^3 - 2x_1^2 + 6x_1 - 3) dx_1 \] Integrate term by term: - \( \int 4x_1^4 dx_1 = 4 \frac{x_1^5}{5} \) - \( \int -x_1^3 dx_1 = -\frac{x_1^4}{4} \) - \( \int -2x_1^2 dx_1 = -2 \frac{x_1^3}{3} \) - \( \int 6x_1 dx_1 = 3x_1^2 \) - \( \int -3 dx_1 = -3x_1 \) So, \[ = \left[ \frac{4}{5}x_1^5 - \frac{1}{4}x_1^4 - \frac{2}{3}x_1^3 + 3x_1^2 - 3x_1 \right]_{2}^{5} \] **Now, plug in \( x_1 = 5 \):** - \( x_1^5 = 3125 \) - \( x_1^4 = 625 \) - \( x_1^3 = 125 \) - \( x_1^2 = 25 \) - \( x_1 = 5 \) So, \[ \frac{4}{5} \cdot 3125 = 2500 \\ \frac{1}{4} \cdot 625 = 156.25 \\ \frac{2}{3} \cdot 125 = \frac{250}{3} \approx 83.333 \\ 3 \cdot 25 = 75 \\ 3 \cdot 5 = 15 \] So, \[ \text{At } 5: 2500 - 156.25 - 83.333 + 75 - 15 = (2500 - 156.25) = 2343.75 \\ 2343.75 - 83.333 = 2260.417 \\ 2260.417 + 75 = 2335.417 \\ 2335.417 - 15 = 2320.417 \] **Now, plug in \( x_1 = 2 \):** - \( x_1^5 = 32 \) - \( x_1^4 = 16 \) - \( x_1^3 = 8 \) - \( x_1^2 = 4 \) - \( x_1 = 2 \) So, \[ \frac{4}{5} \cdot 32 = \frac{128}{5} = 25.6 \\ \frac{1}{4} \cdot 16 = 4 \\ \frac{2}{3} \cdot 8 = \frac{16}{3} \approx 5.333 \\ 3 \cdot 4 = 12 \\ 3 \cdot 2 = 6 \] So, \[ 25.6 - 4 = 21.6 \\ 21.6 - 5.333 = 16.267 \\ 16.267 + 12 = 28.267 \\ 28.267 - 6 = 22.267 \] **Final answer:** \[ \boxed{ \int_{S} f(x) dx = 2320.417 - 22.267 = 2298.15 } \] --- ## **Summary of Steps** 1. **Sketch**: See above for region descriptions. 2. **Type**: \( S \) is Type I as given, can be written as Type II as: \[ S = \left\{ (x_1, x_2) : 1 \leq x_2 \leq 25,\ \max(2, \sqrt{x_2}) \leq x_1 \leq \min(5, x_2 + 1) \right\} \] 3. **Integral**: \[ \boxed{ \int_{S} f(x) dx = 2298.15 } \] If you need the sketch or more details, let me know!

Certainly! Here's a description of how to sketch the regions \( S \) and \( P \): --- ### Sketch of Region \( S \): **Domain \( S \):** \[ 2 \leq x_1 \leq 5, \quad x_1 - 1 \leq x_2 \leq x_1^2 \] **Steps to sketch:** 1. Draw the coordinate axes \( x_1 \) (horizontal) and \( x_2 \) (vertical). 2. Mark the vertical lines at \( x_1 = 2 \) and \( x_1 = 5 \). 3. For each \( x_1 \) between 2 and 5: - Plot the lower boundary \( x_2 = x_1 - 1 \). This is a straight line passing through points: - \( (2, 1) \), since \( 2 - 1 = 1 \). - \( (5, 4) \), since \( 5 - 1 = 4 \). - Plot the upper boundary \( x_2 = x_1^2 \). This is a parabola opening upward: - At \( x_1 = 2 \), \( x_2 = 4 \). - At \( x_1 = 3 \), \( x_2 = 9 \). - At \( x_1 = 4 \), \( x_2 = 16 \). - At \( x_1 = 5 \), \( x_2 = 25 \). 4. Shade the region between these two curves for \( x_1 \) from 2 to 5. The region is bounded below by the line \( x_2 = x_1 - 1 \) and above by the parabola \( x_1^2 \). --- ### Sketch of Region \( P \): **Domain \( P \):** \[ -1 \leq x_2 \leq 3, \quad 10 \leq x_1 \leq e^{x_2} + 10 \] **Steps to sketch:** 1. Draw the axes \( x_1 \) (horizontal) and \( x_2 \) (vertical). 2. Draw horizontal lines at \( x_2 = -1 \) and \( x_2 = 3 \). 3. For each \( x_2 \) between \(-1\) and \( 3 \): - Draw the left boundary \( x_1 = 10 \). - Draw the right boundary \( x_1 = e^{x_2} + 10 \). This is an exponential curve: - At \( x_2 = -1 \), \( x_1 \approx e^{-1} + 10 \approx 0.368 + 10 = 10.368 \). - At \( x_2 = 0 \), \( x_1 = e^{0} + 10 = 1 + 10 = 11 \). - At \( x_2 = 3 \), \( x_1 \approx e^{3} + 10 \approx 20.085 + 10 = 30.085 \). 4. Shade the region between \( x_1 = 10 \) and \( x_1 = e^{x_2} + 10 \) for \( x_2 \) from \(-1\) to \( 3 \). The region is bounded on the left by the vertical line \( x_1=10 \) and on the right by the exponential curve. --- ### Summary: - **\( S \)** is a region between a straight line and a parabola, extending from \( x_1=2 \) to \( 5 \). - **\( P \)** is a region between two horizontal bounds in \( x_2 \), with the right boundary increasing exponentially with \( x_2 \). --- If you'd like, I can generate a visual sketch as an image, but this description should guide you well in plotting these regions!