Let's break down the problem step by step: --- ### **Problem Restatement** A yeast culture weighing 2 grams grows at a rate of \( W'(t) = .4e^{.4t} \) grams/hour. **Question:** 1. How much will the weight of the culture increase during the first 10 hours? 2. How much will the weight increase from the end of the 10th hour to the end of the 20th hour? --- ## **Step 1: Weight Increase in the First 10 Hours** We need to find the increase in weight from \( t = \) to \( t = 10 \). \[ \text{Increase} = \int_{}^{10} .4e^{.4t} \, dt \] Find the antiderivative: \[ \int .4e^{.4t} \, dt = e^{.4t} + C \] So, \[ \int_{}^{10} .4e^{.4t} \, dt = \left[ e^{.4t} \right]_{}^{10} = e^{4} - e^{} = e^{4} - 1 \] Now, calculate \( e^4 \approx 54.598 \): \[ e^4 - 1 \approx 54.598 - 1 = 53.598 \] **Explanation:** We integrated the rate of growth to get the total weight increase over the first 10 hours. --- ## **Step 2: Weight Increase from 10th to 20th Hour** Now, find the increase from \( t = 10 \) to \( t = 20 \): \[ \text{Increase} = \int_{10}^{20} .4e^{.4t} \, dt = \left[ e^{.4t} \right]_{10}^{20} = e^{8} - e^{4} \] Calculate \( e^8 \approx 298.958 \): \[ e^8 - e^4 \approx 298.958 - 54.598 = 2926.360 \] **Explanation:** We integrated the same rate over a different interval to get the weight increase from the 10th to the 20th hour. --- ## **Final Answers** - **The weight increase during the first 10 hours is approximately:** \( \boxed{53.598} \) grams - **The weight increase from the end of the 10th hour to the end of the 20th hour is approximately:** \( \boxed{2926.360} \) grams