AP Calculus BC 8.3 Using Accumulation Functions and Definite Integrals in Applied Contexts Questions

Practice AP Calculus BC questions on interpreting rate integrals as net change and finding final quantities from initial amounts.

Covered syllabus points

CHA-4.D—Interpret the meaning of a definite integral in accumulation problems CHA-4.E—Determine net change using definite integrals in applied contexts

[Maximum number: 2]

The temperature of water in a tub at time t is modeled by a strictly increasing, twice-differentiable function W, where W(t) is measured in degrees Fahrenheit and t is measured in minutes. At time t=0, the temperature of the water is 55°F. The water is heated for 30 minutes, beginning at time t=0. Values of W(t) at selected times t for the first 20 minutes are given in the table above.

Use the data in the table to evaluate $\int_{0}^{20} W^{\prime}(t) d t$ . Using correct units, interpret the meaning of $\int_{0}^{20} W^{\prime}(t) d t$ in the context of this problem.

$\int_{0}^{20} W^{\prime}(t) d t=W(20)-W(0)=71.0-55.0=16$
The water has warmed by 16 °F over the interval from t=0 to t=20 minutes.

[Maximum number: 3]

A student starts reading a book at time t=0 minutes and continues reading for the next 10 minutes. The rate at which the student reads is modeled by the differentiable function R, where R(t) is measured in words per minute. Selected values of R(t) are given in the table shown.

A teacher also starts reading at time t=0 minutes and continues reading for the next 10 minutes. The rate at which the teacher reads is modeled by the function W defined by $W(t)=-\frac{3}{10} t^{2}+8 t+100$ , where W(t) is measured in words per minute. Based on the model, how many words has the teacher read by the end of the 10 minutes? Show the work that leads to your answer.

D A teacher also starts reading at time t=0 minutes and continues reading for the next 10 minutes. The

rate at which the teacher reads is modeled by the function W defined by $W(t)=-\frac{3}{10} t^{2}+8 t+100$ ,

where W(t) is measured in words per minute. Based on the model, how many words has the teacher

read by the end of the 10 minutes? Show the work that leads to your answer.

$\int_{0}^{10} W(t) d t=\int_{0}^{10}\left(-\frac{3}{10} t^{2}+8 t+100\right) d t$	Integrand	Point 7 (P7)
$=\left.\left(-\frac{1}{10} t^{3}+4 t^{2}+100 t\right)\right\|_{0} ^{10}$	Antiderivative	Point 8 (P8)
$\begin{aligned} & =\left(-\frac{1}{10} \cdot 1000+4 \cdot 100+100 \cdot 10\right)-\left(-\frac{1}{10} \cdot 0+4 \cdot 0+100 \cdot 0\right)<br>& =1300 \end{aligned}$ <br>Based on the model, the teacher has read 1300 words by the end of the 10 minutes.	Answer	Point 9 (P9)
Scoring Notes for Part D

- P7 is earned for an indefinite or definite integral with integrand W(t), with or without the

differential d t.

- P8 is earned for the correct antiderivative, with or without the constant of integration.

- To be eligible for P9, a response must have earned P8.

- A response of $\left(-\frac{1}{10} \cdot 1000+4 \cdot 100+100 \cdot 10\right)-\left(-\frac{1}{10} \cdot 0+4 \cdot 0+100 \cdot 0\right)$ or equivalent banks P9

(i.e., subsequent errors in simplification will not be considered in scoring for P9).

Part A (AB or BC): Graphing calculator not allowed

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Covered syllabus points

Question 1

Question 2