AP Calculus AB 8.3: Net Change

Practice AP Calculus questions on using the integral of a rate function to calculate accumulated quantity and net change in context.

[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.

Table for Question CHA-4.E—Determine net change using definite integrals in applied contexts question 1 — AP Calculus AB

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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CHA-4.E—Determine net change using definite integrals in applied contexts question 1