Home / AP Calculus BC 8.1 Finding the Average Value of a Function on an Interval – Exam Style Questions – MCQs

AP Calculus BC 8.1 Finding the Average Value of a Function on an Interval - Exam Style Questions - MCQs - New Syllabus

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Question

For \(x\ge0\), the function \(g\) is continuous and \(g(x)>0\). If the area bounded by \(y=g(x)\) and the \(x\)-axis from \(x=0\) to \(x=2\) is \(3\), what is the average value of \(g\) on \([0,2]\)?
(A) \( \dfrac{2}{3} \)
(B) \( \dfrac{3}{2} \)
(C) \( 2 \)
(D) \( 3 \)
▶️ Answer/Explanation
Detailed solution

Average value on \([a,b]\): \[ \text{Avg}=\frac{1}{b-a}\int_{a}^{b}g(x)\,dx. \] Here \(a=0\), \(b=2\), and \(\displaystyle \int_{0}^{2} g(x)\,dx=3\).

\[ \text{Avg}=\frac{1}{2-0}\cdot 3=\frac{3}{2}. \]
✅ Answer: (B)

Question

Bacteria weight (grams) is modeled by \[ w(t)=0.1\,(t^{2}+3)\,e^{-0.2t}, \] where \(t\) is hours since the start of an experiment. What is the average weight during the first \(3\) hours?

(A) \(0.120\)
(B) \(0.418\)
(C) \(0.479\)
(D) \(1.255\)
▶️ Answer/Explanation

Correct answer: (B) \(0.418\)

Average value on \([0,3]\): \[ \overline{w}=\frac{1}{3}\int_{0}^{3} 0.1\,(t^{2}+3)\,e^{-0.2t}\,dt. \] Set \(a=0.2\). Using \[ \int t^{2}e^{-at}\,dt = e^{-at}\!\left(-\frac{t^{2}}{a}-\frac{2t}{a^{2}}-\frac{2}{a^{3}}\right)+C, \qquad \int e^{-at}\,dt=-\frac{1}{a}e^{-at}+C, \] we get \[ \int (t^{2}+3)e^{-at}\,dt = e^{-at}\!\left(-\frac{t^{2}+3}{a}-\frac{2t}{a^{2}}-\frac{2}{a^{3}}\right)+C. \] Therefore \[ \overline{w} = \frac{0.1}{3}\left[ e^{-0.2t}\!\left(-\frac{t^{2}+3}{0.2}-\frac{2t}{0.2^{2}}-\frac{2}{0.2^{3}}\right) \right]_{0}^{3} \;\approx\; 0.418. \]

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