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AP Calculus AB 8.3 Using Accumulation Functions and Definite Integrals in Applied Contexts- FRQs - Exam Style Questions

AP Calculus AB - FRQs and Answer - All Topics

No-Calc Question

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.
\(t\) (minutes)\(0\)\(2\)\(8\)\(10\)
\(R(t)\) (words per minute)\(90\)\(100\)\(150\)\(162\)
A. Approximate \(R'(1)\) using the average rate of change of \(R\) over the interval \(0\le t\le 2\). Show the work that leads to your answer. Indicate units of measure.
B. Must there be a value \(c\), for \(0<c<10\), such that \(R(c)=155\)? Justify your answer.
C. Use a trapezoidal sum with the three subintervals indicated by the data in the table to approximate the value of \(\int_{0}^{10} R(t)\,dt\). 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 \(W(t)=-\frac{3}{10}t^{2}+8t+100\), 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.

Most-appropriate topic codes (CED):

• TOPIC 2.3: Estimating Derivatives of a Function at a Point — part (A)
• TOPIC 1.16: Working with the Intermediate Value Theorem (IVT) — part (B)
• TOPIC 6.2: Approximating Areas with Riemann Sums (trapezoidal) — part (C)
• TOPIC 8.3: Using Accumulation Functions and Definite Integrals in Applied Contexts — part (D)
▶️ Answer/Explanation

(A) Approximate \(R'(1)\)
Use average rate of change on \([0,2]\): \(R'(1)\approx \dfrac{R(2)-R(0)}{2-0}\).
Substitute table values: \(R'(1)\approx \dfrac{100-90}{2}= \dfrac{10}{2}= \boxed{5}\ \text{words per minute per minute}\) \(=\ \text{words}/\text{minute}^{2}\).

(B) Existence of \(c\) with \(R(c)=155\)
\(R\) is differentiable on \([0,10]\Rightarrow R\) is continuous on \([0,10]\).
From the table: \(R(0)=90\), \(R(10)=162\), and \(90<155<162\).
By IVT, there exists \(c\) with \(0<c<10\) such that \(\boxed{R(c)=155}\).

(C) Trapezoidal sum for \(\int_{0}^{10} R(t)\,dt\)
Use subintervals \([0,2]\), \([2,8]\), \([8,10]\).
\(\int_{0}^{10} R(t)\,dt \approx (2-0)\times \dfrac{R(0)+R(2)}{2} \;+\; (8-2)\times \dfrac{R(2)+R(8)}{2} \;+\; (10-8)\times \dfrac{R(8)+R(10)}{2}\).
\(= 2\times \dfrac{90+100}{2} \;+\; 6\times \dfrac{100+150}{2} \;+\; 2\times \dfrac{150+162}{2}\).
\(= 2\times 95 \;+\; 6\times 125 \;+\; 2\times 156\).
\(= 190 \;+\; 750 \;+\; 312 = \boxed{1252\ \text{words}}\).

(D) Teacher’s total words by \(t=10\)
Total words \(=\int_{0}^{10} W(t)\,dt = \int_{0}^{10} \left(-\dfrac{3}{10}t^{2}+8t+100\right) dt\).
Antiderivative: \(-\dfrac{1}{10}t^{3}+4t^{2}+100t\).
Evaluate \(0\to 10\): \(\left[-\dfrac{1}{10}t^{3}+4t^{2}+100t\right]_{0}^{10} = \left(-\dfrac{1}{10}\times 10^{3}\right) + 4\times 10^{2} + 100\times 10 – 0\).
\(= -100 + 400 + 1000 = \boxed{1300\ \text{words}}\).

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