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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Chapter1: Functions And Models
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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.
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. --- **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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