AP Calculus BC 8.1 Finding the Average Value of a Function on an Interval - FRQs - Exam Style Questions
Calc-Ok Question
| \(t\) (seconds) | 0 | 60 | 90 | 120 | 135 | 150 |
|---|---|---|---|---|---|---|
| \(f(t)\) (gal/s) | 0 | 0.1 | 0.15 | 0.1 | 0.05 | 0 |
(a) Using correct units, interpret the meaning of \(\displaystyle \int_{60}^{135} f(t)\,dt\) in the context of the problem. Use a right Riemann sum with the three subintervals \([60,90],\,[90,120],\,[120,135]\) to approximate \(\displaystyle \int_{60}^{135} f(t)\,dt\).
(b) Must there exist a value of \(c\), for \(60<c<120\), such that \(f'(c)=0\)? Justify your answer.
(c) The rate of flow of gasoline (gal/s) can also be modeled by \[ g(t)=\frac{t}{500}\cos\!\left(\left(\frac{t}{120}\right)^{\!2}\right), \quad 0\le t\le 150. \] Using this model, find the average rate of flow of gasoline over \(0\le t\le 150\). Show the setup for your calculations.
(d) Using the model \(g\) from part (c), find the value of \(g'(140)\). Interpret the meaning of your answer in the context of the problem.
Most-appropriate topic codes:
TOPIC 6.4: FTC & Accumulation (interpretation of \(\int f\)) — part (a)
TOPIC 5.1: Mean Value Theorem — part (b)
TOPIC 8.1: Average Value of a Function — part (c)
TOPIC 4.1: Interpreting the Derivative in Context — part (d)
▶️ Answer/Explanation
Interpretation with units: \[ \int_{60}^{135} f(t)\,dt \] represents the total number of gallons of gasoline pumped into the tank from \(t=60\) s to \(t=135\) s. (gal/s integrated over seconds \(\Rightarrow\) gallons).
Right Riemann sum (widths \(30, 30, 15\); right endpoints \(90,120,135\)): \[ \int_{60}^{135} f(t)\,dt \approx f(90)(90-60)+f(120)(120-90)+f(135)(135-120). \] \[ = (0.15)(30)+(0.10)(30)+(0.05)(15) = 4.5+3.0+0.75 = \boxed{8.25\ \text{gallons}}. \]
\(f\) is differentiable \(\Rightarrow\) continuous on \([60,120]\).
Compute the average rate of change: \[ \frac{f(120)-f(60)}{120-60}=\frac{0.10-0.10}{60}=0. \] By the Mean Value Theorem, there exists \(c\in(60,120)\) such that \[ f'(c)=0. \] Yes, such a \(c\) must exist.
Average value over \([0,150]\): \[ \frac{1}{150}\int_{0}^{150} g(t)\,dt = \frac{1}{150}\int_{0}^{150}\frac{t}{500}\cos\!\left(\Big(\tfrac{t}{120}\Big)^{2}\right) dt. \] Substitute \(u=\left(\tfrac{t}{120}\right)^{2}\) so \(du=\tfrac{t}{7200}\,dt\) and \(t\,dt=7200\,du\).
Then \[ \int_{0}^{150}\frac{t}{500}\cos(u)\,dt =\frac{7200}{500}\int_{u=0}^{u=(150/120)^{2}}\cos u\,du =14.4\,\sin u\Big|_{0}^{(150/120)^{2}}. \] \[ =14.4\,\sin\!\left(\frac{25}{16}\right). \] Average value: \[ \boxed{\frac{1}{150}\int_{0}^{150} g(t)\,dt = \frac{14.4}{150}\sin\!\left(\frac{25}{16}\right) = \frac{12}{125}\sin\!\left(\frac{25}{16}\right) \approx 0.096\ \text{gal/s}.} \]
\(g(t)=\dfrac{t}{500}\cos\!\left(\left(\tfrac{t}{120}\right)^{2}\right)\).
Let \(u(t)=\left(\tfrac{t}{120}\right)^{2}\) so \(u'(t)=\tfrac{t}{7200}\).
Product/chain rule: \[ g'(t)=\frac{1}{500}\cos u(t)\;-\;\frac{t}{500}\sin u(t)\cdot \frac{t}{7200} =\frac{1}{500}\cos u(t)\;-\;\frac{t^{2}}{3{,}600{,}000}\sin u(t). \] At \(t=140\): \(u(140)=\left(\tfrac{140}{120}\right)^{2}=\tfrac{49}{36}\). \[ g'(140)=\frac{1}{500}\cos\!\left(\frac{49}{36}\right) -\frac{49}{9000}\sin\!\left(\frac{49}{36}\right) \approx \boxed{-0.0049\ \text{(gal/s)}^{2}}. \] Interpretation: At \(t=140\) s, the rate at which gasoline is flowing into the tank is decreasing at about \(0.005\) gallon per second per second.