# 2011 Calculus AB free response #2 (c & d) | AP Calculus AB | Khan Academy | Summary and Q&A

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September 7, 2011
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2011 Calculus AB free response #2 (c & d) | AP Calculus AB | Khan Academy

## TL;DR

The analysis explains the meaning of evaluating a definite integral and determines the change in temperature for both biscuits and tea.

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### Q: How is the definite integral used to determine the change in temperature?

The definite integral allows for the evaluation of the anti-derivative function, which gives the temperature at different time points. Subtracting the temperature values at the endpoints of integration provides the change in temperature over that time interval.

### Q: What does the negative sign in -23 degrees Celsius signify?

The negative sign indicates that the temperature has decreased by 23 degrees Celsius from 0 to 10 minutes. It represents a decrease in temperature.

### Q: How is the temperature change for the biscuits determined?

By evaluating the definite integral of B prime of (t)dt from 0 to 10, the change in temperature for the biscuits is calculated to be -65.82 degrees Celsius.

### Q: How do the temperatures of the biscuits and tea compare after 10 minutes?

After 10 minutes, the biscuits have a temperature of 34.18 degrees Celsius, while the tea is at 43 degrees Celsius, making the biscuits 8.82 degrees Celsius cooler.

## Summary & Key Takeaways

• Evaluating the definite integral from 0 to 10 of H prime of (t)dt gives the difference in temperature between 0 and 10 minutes, which is a change of -23 degrees Celsius.

• Evaluating the definite integral from 0 to 10 of B prime of (t)dt gives the change in temperature for the biscuits, resulting in a change of -65.82 degrees Celsius.

• Comparing the temperatures, the biscuits are determined to be 8.82 degrees Celsius cooler than the tea after 10 minutes.