Let's break down the problem step by step. ### **Step 1: Understand the Given Rate Function** The rate at which the mosquito population is increasing is: \[ n(t) = 2{,}500 \cdot 10e^{0.5t} \] where \( t \) is in weeks. ### **Step 2: Find the Increase in Population between Weeks 5 and 9** To find the total increase in population from week 5 to week 9, integrate the rate function from \( t = 5 \) to \( t = 9 \): \[ \text{Increase} = \int_{5}^{9} 2{,}500 \cdot 10e^{0.5t} \, dt \] \[ = 2{,}500 \cdot 10 \int_{5}^{9} e^{0.5t} \, dt \] \[ = 25{,}000 \int_{5}^{9} e^{0.5t} \, dt \] ### **Step 3: Integrate the Exponential Function** Recall: \[ \int e^{ax} dx = \frac{1}{a}e^{ax} + C \] Here, \( a = 0.5 \): \[ \int e^{0.5t} dt = 2e^{0.5t} \] So: \[ \int_{5}^{9} e^{0.5t} dt = 2e^{0.5 \cdot 9} - 2e^{0.5 \cdot 5} \] \[ = 2e^{4.5} - 2e^{2.5} \] ### **Step 4: Plug Back and Compute** \[ \text{Increase} = 25{,}000 \times [2e^{4.5} - 2e^{2.5}] \] \[ = 50{,}000 (e^{4.5} - e^{2.5}) \] Now, calculate \( e^{4.5} \) and \( e^{2.5} \): - \( e^{4.5} \approx 90.0171 \) - \( e^{2.5} \approx 12.1825 \) So, \[ e^{4.5} - e^{2.5} \approx 90.0171 - 12.1825 = 77.8346 \] \[ \text{Increase} = 50,000 \times 77.8346 \approx 3,891,730 \] ### **Step 5: Round to the Nearest Integer** \[ \boxed{3,891,730} \] --- **Final Answer:** The mosquito population increases by **3,891,730 mosquitoes** between the fifth and ninth weeks of summer (rounded to the nearest integer).