Let's proceed step by step as requested: --- ### **Step 1: Draw the Functions E(t) and S(t)** - **E(t) = 40000 - 2t** (CO₂ emissions, linear decrease) - **S(t) = 50 + 5t** (Carbon sequestration, linear increase) - **Time interval:** t = 0 (2020) to t = 10 (2030) **Explanation:** - E(t) is a straight line starting at 40000 (when t=0) and decreasing by 2 units every year. - S(t) is a straight line starting at 50 (when t=0) and increasing by 5 units every year. - You can sketch both lines on the same axes with t (years since 2020) on the x-axis and emissions/sequestration on the y-axis. --- ### **Step 2: Summing Even Integers from A to B** We need to recall and justify the formula for the sum of even integers from A to B (where both A and B are even). #### **Let A and B be even integers.** - Even integers between A and B (inclusive): A, A+2, ..., B - This forms an arithmetic sequence: - First term: A - Last term: B - Common difference: 2 **Number of terms, n:** \[ n = \frac{B - A}{2} + 1 \] **Sum, S:** \[ S = \frac{n}{2}(A + B) \] Substitute for n: \[ S = \frac{\left( \frac{B-A}{2} + 1 \right)}{2} \cdot (A+B) \] **Explanation:** The sum uses the arithmetic sequence formula. Since only every second integer is included (even numbers), the common difference is 2, and the number of terms is adjusted accordingly. --- **Final Answer for Step 2:** The sum of even integers from A to B (inclusive), where both are even, is: \[ S = \frac{\left( \frac{B-A}{2} + 1 \right)}{2} \cdot (A+B) \] --- **Continue to next steps as needed! Would you like the calculations for the Riemann sum next?**

Let's proceed step by step as requested: --- ### **Step 1: Draw the Functions E(t) and S(t)** - **E(t) = 40000 - 2t** (CO₂ emissions, linear decrease) - **S(t) = 50 + 5t** (Carbon sequestration, linear increase) - **Time interval:** t = 0 (2020) to t = 10 (2030) **Explanation:** - At t=0 (year 2020), emissions are E(0) = 40000. - At t=10 (year 2030), emissions are E(10) = 40000 - 2*10 = 39980. - The graph of E(t) is a straight line decreasing from 40000 to 39980 over 10 years. - For sequestration: S(0) = 50, S(10) = 50 + 5*10 = 100. - The graph of S(t) is a straight line increasing from 50 to 100 over 10 years. You can sketch these two lines on the same axes: t from 0 to 10, and the respective emissions and sequestration values on the y-axis. --- ### **Step 2: Recall and justify the sum of even integers from A to B** **Given:** A and B are even integers, and the sum runs over even integers between A and B (inclusive). **Number of terms, n:** \[ n = \frac{B - A}{2} + 1 \] since the sequence is A, A+2, ..., B. **Sum of these even integers:** \[ S = \frac{n}{2} (A + B) \] because the sum of an arithmetic sequence is the average of the first and last terms times the number of terms. **Explanation:** - The sequence of even integers forms an arithmetic progression with common difference 2. - The total sum is computed using the average of first and last terms multiplied by the total number of terms. --- **Final answer for Step 2:** \[ \boxed{ \text{Sum of even integers from } A \text{ to } B = \frac{\left( \frac{B - A}{2} + 1 \right)}{2} \times (A + B) } \] --- ### **Step 3: Using a left-Riemann sum with 10 rectangles** **Interval:** t from 0 to 10 **Number of rectangles:** 10 **Width of each rectangle:** \(\Delta t = \frac{10 - 0}{10} = 1\) **Left endpoints:** t = 0, 1, 2, ..., 9 --- **Calculate the area under E(t):** \[ \text{Area}_E \approx \sum_{i=0}^{9} E(t_i) \times \Delta t \] where \( t_i = i \). - \( E(t) = 40000 - 2t \) Calculate each \( E(t_i) \): | i | \( t_i \) | \( E(t_i) \) | |---|-----------|--------------| | 0 | 0 | 40000 | | 1 | 1 | 39998 | | 2 | 2 | 39996 | | 3 | 3 | 39994 | | 4 | 4 | 39992 | | 5 | 5 | 39990 | | 6 | 6 | 39988 | | 7 | 7 | 39986 | | 8 | 8 | 39984 | | 9 | 9 | 39982 | Sum: \[ \sum_{i=0}^{9} E(t_i) = 40000 + 39998 + 39996 + \dots + 39982 \] Number of terms: 10 Calculate sum: \[ \text{Sum} = \frac{\text{number of terms}}{2} \times (\text{first} + \text{last}) = \frac{10}{2} \times (40000 + 39982) = 5 \times 79982 = 399,910 \] **Area under E(t):** \[ \text{Area}_E \approx 399,910 \times 1 = 399,910 \] --- **Calculate the area under S(t):** | i | \( t_i \) | \( S(t_i) = 50 + 5 t_i \) | |---|-----------|---------------------------| | 0 | 0 | 50 | | 1 | 1 | 55 | | 2 | 2 | 60 | | 3 | 3 | 65 | | 4 | 4 | 70 | | 5 | 5 | 75 | | 6 | 6 | 80 | | 7 | 7 | 85 | | 8 | 8 | 90 | | 9 | 9 | 95 | Sum: \[ 50 + 55 + 60 + 65 + 70 + 75 + 80 + 85 + 90 + 95 = \] Adding: \[ 50 + 55 = 105 \\ 105 + 60 = 165 \\ 165 + 65 = 230 \\ 230 + 70 = 300 \\ 300 + 75 = 375 \\ 375 + 80 = 455 \\ 455 + 85 = 540 \\ 540 + 90 = 630 \\ 630 + 95 = 725 \] Total sum: 725 **Area under S(t):** \[ \text{Area}_S \approx 725 \times 1 = 725 \] --- ### **Net emission estimate:** \[ \text{Net emission} \approx \text{Area under } E(t) - \text{Area under } S(t) = 399,910 - 725 = \boxed{399,185} \] **Interpretation:** This is an estimate of the total net CO₂ emissions over the decade, considering the average emissions minus sequestration. --- ### **Step 4: More precise estimate using an appropriate linear function** Instead of Riemann sums with left endpoints, we can estimate the total net emissions by integrating the functions over [0,10]: \[ \text{Net total} = \int_0^{10} [E(t) - S(t)] dt \] Calculate the integral: \[ \int_0^{10} (40000 - 2t - (50 + 5t)) dt = \int_0^{10} (39950 - 7t) dt \] Integrate: \[ = \left[ 39950 t - \frac{7}{2} t^2 \right]_0^{10} \] Plug in t=10: \[ 39950 \times 10 - \frac{7}{2} \times 100 = 399,500 - 350 = 399,150 \] --- **Final estimate of net emissions:** \[ \boxed{399,150} \] This is a more precise calculation than the Riemann sum approximation, as it uses the exact integral over the interval. --- ### **Step 5: Observation and interpretation** The Riemann sum estimate (399,185) and the integral calculation (399,150) are very close, indicating the Riemann sum with 10 rectangles provides a good approximation. The slight difference arises because the Riemann sum uses left endpoints, which tend to slightly overestimate or underestimate depending on the function's behavior. **Interpretation:** Both methods show that the country’s net CO₂ emissions over the decade are approximately 399,150 million tons, indicating persistent emissions slightly exceeding sequestration. The close agreement between the numerical approximation and the exact integral confirms the reliability of the methods used. --- **Summary:** - Drawn the functions and justified the sum of even integers. - Estimated total net emissions with Riemann sums. - Calculated a more accurate integral estimate. - Observed minimal difference, confirming the robustness of the analysis.