Let's solve each part step by step using the given function \( f(t) = e^{0.2t} \). --- ### **1. Estimate total number of reported cases up to Day 10 and Day 30** The total number of reported cases up to day \( T \) is the integral: \[ N(T) = \int_0^T f(t)\, dt = \int_0^T e^{0.2t}\, dt \] The integral of \( e^{kt} \) with respect to \( t \) is \( \frac{1}{k}e^{kt} \), so: \[ \int e^{0.2t}\, dt = \frac{1}{0.2}e^{0.2t} = 5e^{0.2t} \] Calculating definite integrals: #### **Up to Day 10:** \[ N(10) = \left[ 5e^{0.2t} \right]_0^{10} = 5e^{2} - 5e^{0} \] \[ N(10) = 5(e^2 - 1) \] \( e^2 \approx 7.389 \) \[ N(10) \approx 5(7.389 - 1) = 5(6.389) = 31.945 \] #### **Up to Day 30:** \[ N(30) = \left[ 5e^{0.2t} \right]_0^{30} = 5e^{6} - 5e^{0} \] \[ N(30) = 5(e^6 - 1) \] \( e^6 \approx 403.429 \) \[ N(30) \approx 5(403.429 - 1) = 5(402.429) = 2012.145 \] **Final answers:** - **Day 10:** ~31.95 cases - **Day 30:** ~2012.15 cases --- ### **2. Estimate the number of contagious patients on Day 10 and Day 30** **Contagious patients** on day \( T \) is the sum of new cases from day \( T-7 \) to day \( T \): \[ C(T) = \int_{T-7}^T f(t)\, dt \] \[ = \left[ 5e^{0.2t} \right]_{T-7}^{T} = 5e^{0.2T} - 5e^{0.2(T-7)} \] #### **On Day 10:** \[ C(10) = 5e^{0.2 \times 10} - 5e^{0.2 \times 3} \] \[ = 5e^{2} - 5e^{0.6} \] \( e^2 \approx 7.389, e^{0.6} \approx 1.822 \) \[ C(10) = 5(7.389 - 1.822) = 5(5.567) = 27.835 \] #### **On Day 30:** \[ C(30) = 5e^{0.2 \times 30} - 5e^{0.2 \times 23} \] \[ = 5e^{6} - 5e^{4.6} \] \( e^6 \approx 403.429, e^{4.6} \approx 99.484 \) \[ C(30) = 5(403.429 - 99.484) = 5(303.945) = 1519.725 \] **Final answers:** - **Day 10:** ~27.84 contagious patients - **Day 30:** ~1519.73 contagious patients --- ### **Guiding question for 2:** This is different from the previous computation because the previous integral sums all cases since the start, while this sums only cases from the most recent 7 days (those still contagious). --- ### **3. Illustrate as areas under the curve** - The total reported cases up to day \( T \) is the area under \( f(t) \) from 0 to \( T \). - The number of contagious patients on day \( T \) is the area under \( f(t) \) from \( T-7 \) to \( T \). Each area corresponds to the definite integral’s limits. --- ### **4. Estimate hospital beds needed on day 50** 1% of contagious patients will require hospitalization. First, find the number of contagious patients on day 50: \[ C(50) = \int_{43}^{50} e^{0.2t} dt = 5e^{0.2 \times 50} - 5e^{0.2 \times 43} \] \[ = 5e^{10} - 5e^{8.6} \] \( e^{10} \approx 22026.47, \; e^{8.6} \approx 5439.83 \) \[ C(50) = 5(22026.47 - 5439.83) = 5(16586.64) = 82,933.2 \] 1% of these require beds: \[ 0.01 \times 82,933.2 = 829.33 \] **Final answer:** **Estimated hospital beds needed on Day 50: ~829** --- ## **Summary Table** | Item | Formula | Approximate Value | |---------------------------|---------------------------------------|-------------------| | Total cases by Day 10 | \( 5(e^2-1) \) | 31.95 | | Total cases by Day 30 | \( 5(e^6-1) \) | 2012.15 | | Contagious on Day 10 | \( 5(e^2-e^{0.6}) \) | 27.84 | | Contagious on Day 30 | \( 5(e^6-e^{4.6}) \) | 1519.73 | | Hospital beds on Day 50 | \( 0.01 \times 5(e^{10}-e^{8.6}) \) | 829 | Let me know if you want the graphical illustration for part 3!

Certainly! Here are the final answers: 1. **Total reported cases up to Day 10:** approximately **31.95** cases **Total reported cases up to Day 30:** approximately **2012.15** cases 2. **Number of contagious patients on Day 10:** approximately **27.84** patients **Number of contagious patients on Day 30:** approximately **1519.73** patients 3. **Estimated hospital beds needed on Day 50:** approximately **829** beds