Worksheet 2.2p2 Integration by Parts Math& 152 10. The concentration of particulate matter (in parts per million) t hours after a factory ceases operation for the day is given by: 20 In(t + 1) C(t) =- (t + 1)² Find the average concentration during the five hours immediately after operation ceases.
Worksheet 2.2p2 Integration by Parts Math& 152 10. The concentration of particulate matter (in parts per million) t hours after a factory ceases operation for the day is given by: 20 In(t + 1) C(t) =- (t + 1)² Find the average concentration during the five hours immediately after operation ceases.
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Transcribed Image Text:**Worksheet: 2.2p2 Integration by Parts – Math& 152**
10. The concentration of particulate matter (in parts per million) \( t \) hours after a factory ceases operation for the day is given by:
\[ C(t) = \frac{20 \ln(t + 1)}{(t + 1)^2} \]
Find the average concentration during the five hours immediately after operation ceases.
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**Explanation:**
This problem deals with determining the average concentration of particulate matter over a specific time interval using a given mathematical function. The function \( C(t) \) represents the concentration in terms of \( t \), where \( \ln \) denotes the natural logarithm.
To find the average concentration over the first five hours, you would need to calculate:
\[
\text{Average concentration} = \frac{1}{5} \int_{0}^{5} C(t) \, dt
\]
This involves integrating the given function from 0 to 5 and dividing by the interval length, which is 5 hours.
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![**Worksheet: 2.2p2 Integration by Parts – Math& 152**
10. The concentration of particulate matter (in parts per million) \( t \) hours after a factory ceases operation for the day is given by:
\[ C(t) = \frac{20 \ln(t + 1)}{(t + 1)^2} \]
Find the average concentration during the five hours immediately after operation ceases.
---
**Explanation:**
This problem deals with determining the average concentration of particulate matter over a specific time interval using a given mathematical function. The function \( C(t) \) represents the concentration in terms of \( t \), where \( \ln \) denotes the natural logarithm.
To find the average concentration over the first five hours, you would need to calculate:
\[
\text{Average concentration} = \frac{1}{5} \int_{0}^{5} C(t) \, dt
\]
This involves integrating the given function from 0 to 5 and dividing by the interval length, which is 5 hours.](https://content.bartleby.com/qna-images/question/58685b70-af35-449c-a900-687c4077d4c2/2f20a43b-65f6-4a4e-89a3-da1f7b2de0d4/f5vms6i_processed.png)
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