AP Calculus AB 8.3 Using Accumulation Functions and Definite Integrals in Applied Contexts- FRQs - Exam Style Questions
No-Calc Question
\(t\) (minutes) | \(0\) | \(2\) | \(8\) | \(10\) |
\(R(t)\) (words per minute) | \(90\) | \(100\) | \(150\) | \(162\) |
Most-appropriate topic codes (CED):
• 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 \(0\to 10\): \(\left[-\dfrac{1}{10}t^{3}+4t^{2}+100t\right]_{0}^{10} = \left(-\dfrac{1}{10}\times 10^{3}\right) + 4\times 10^{2} + 100\times 10 – 0\).
\(= -100 + 400 + 1000 = \boxed{1300\ \text{words}}\).