AP Calculus AB 8.1 Finding the Average Value of a Function on an Interval - MCQs - Exam Style Questions
Calc-Ok Question
Aayush’s heart rate during a workout is modeled by the differentiable function \(h\), where \(h(t)\) is measured in beats per minute and \(t\) is measured in minutes from the start of the workout. Which of the following expressions gives Aayush’s average heart rate from \(t=30\) to \(t=60\) ?
(A) \(\displaystyle \int_{30}^{60} h'(t)\,dt\)
(B) \(\displaystyle \frac{1}{30}\int_{30}^{60} h(t)\,dt\)
(C) \(\displaystyle \frac{1}{30}\int_{30}^{60} h'(t)\,dt\)
(D) \(\displaystyle \frac{h'(30)+h'(60)}{2}\)
(B) \(\displaystyle \frac{1}{30}\int_{30}^{60} h(t)\,dt\)
(C) \(\displaystyle \frac{1}{30}\int_{30}^{60} h'(t)\,dt\)
(D) \(\displaystyle \frac{h'(30)+h'(60)}{2}\)
▶️ Answer/Explanation
Average value on \([a,b]\) is \(\dfrac{1}{b-a}\int_{a}^{b} h(t)\,dt\).
Here \(a=30\), \(b=60\) so \(\dfrac{1}{60-30}\int_{30}^{60} h(t)\,dt=\dfrac{1}{30}\int_{30}^{60} h(t)\,dt\).
✅ Answer: (B)
Here \(a=30\), \(b=60\) so \(\dfrac{1}{60-30}\int_{30}^{60} h(t)\,dt=\dfrac{1}{30}\int_{30}^{60} h(t)\,dt\).
✅ Answer: (B)