AP Calculus BC 8.3 Using Accumulation Functions and Definite Integrals- Exam Style Questions - MCQs - New Syllabus
Question
The rate of change of the water level in a lake at time \(t\) (months) is \(r(t)\) feet per month, for \(0\le t\le12\). At what time(s) \(t\) is the water level equal to its level at \(t=0\)?
(A) \(t=5\) only
(B) \(t=5\) and \(t=12\)
(C) \(t=2,\ t=5,\ \text{and}\ t=8\)
(D) \(t=1\ \text{and}\ t=6\)
(B) \(t=5\) and \(t=12\)
(C) \(t=2,\ t=5,\ \text{and}\ t=8\)
(D) \(t=1\ \text{and}\ t=6\)
▶️ Answer/Explanation
Correct answer: (B) \(t=5\) and \(t=12\)
Water level returns to its initial value when the net change is zero:
\(\displaystyle \int_{0}^{t} r(\tau)\,d\tau=0.\)
From the graph, the signed area on \([0,5]\) is \(0\) ⇒ \(t=5\).
The total signed area on \([0,12]\) is also \(0\) ⇒ \(t=12\).
Calc-Ok Question
If \(\displaystyle \frac{dy}{dt}=6e^{-0.08(t-5)^{2}}\), by how much does \(y\) change as \(t\) changes from \(t=1\) to \(t=6\)?
(A) \(3.870\)
(B) \(8.341\)
(C) \(18.017\)
(D) \(22.583\)
(B) \(8.341\)
(C) \(18.017\)
(D) \(22.583\)
▶️ Answer/Explanation
Detailed solution
The change in \(y\) from \(t=1\) to \(t=6\) is \[ \Delta y=y(6)-y(1)=\int_{1}^{6}\frac{dy}{dt}\,dt =\int_{1}^{6}6e^{-0.08(t-5)^{2}}\,dt. \] This integral has no elementary antiderivative (Gaussian form), so evaluate numerically (calculator OK):
\[ \int_{1}^{6}6e^{-0.08(t-5)^{2}}\,dt \approx 22.583. \] ✅ Answer: (D)