Home / AP calculus AB and BC concise summary notes / AP Calculus BC 8.1 Finding the Average Value of a Function on an Interval Study Notes

AP Calculus BC 8.1 Finding the Average Value of a Function on an Interval Study Notes - New Syllabus

AP Calculus BC 8.1 Finding the Average Value of a Function on an Interval Study Notes- New syllabus

AP Calculus BC 8.1 Finding the Average Value of a Function on an Interval Study Notes – AP Calculus BC-  per latest AP Calculus BC Syllabus.

LEARNING OBJECTIVE

  • Definite integrals allow us to solve problems involving the accumulation of change over an interval.

Key Concepts: 

  •  Average Value of a Function on an Interval

AP Calculus BC-Concise Summary Notes- All Topics

AP Calculus BC Exam Questions MCQs and FRQs

 Finding the Average Value of a Function on an Interval

 Finding the Average Value of a Function on an Interval

The average value of a continuous function \( f(x) \) over the interval \([a, b]\) represents the “mean height” of the function on that interval.

It is given by the formula:

\(\text{Average Value} = \dfrac{1}{b-a} \int_{a}^{b} f(x) \, dx\)

This formula works similarly to finding the average of a set of numbers, except here the “sum” is replaced by the definite integral (representing the area under the curve), and we divide by the length of the interval \( b-a \).

Key Notes:

  • The function \( f(x) \) must be continuous on the closed interval \([a, b]\).
  • The average value can be interpreted as the constant height of a rectangle over \([a, b]\) that has the same area as the area under \( f(x) \) on that interval.
  • If the average value is multiplied by \( b-a \), it gives the total accumulated value (area) over that interval.

Step-by-Step Process:

  1. Identify the function \( f(x) \) and the interval \([a, b]\).
  2. Set up the integral \(\int_{a}^{b} f(x) \, dx\).
  3. Multiply the result of the integral by \(\dfrac{1}{b-a}\) to find the average value.

Example:

Find the average value of \( f(x) = x^2 \) on the interval \([0, 3]\).

▶️ Answer/Explanation

We use the formula: \( \text{Average Value} = \dfrac{1}{b-a} \int_{a}^{b} f(x) \, dx \)

Here, \( a = 0 \), \( b = 3 \), and \( f(x) = x^2 \).

\( \text{Average Value} = \dfrac{1}{3-0} \int_{0}^{3} x^2 \, dx \)

\( = \dfrac{1}{3} \left[ \dfrac{x^3}{3} \right]_{0}^{3} \)

\( = \dfrac{1}{3} \left( \dfrac{27}{3} – 0 \right) \) \( = \dfrac{1}{3} \times 9 = 3 \)

So, the average value is \( 3 \).

Example:

The velocity of a particle is given by \( v(t) = 4\sin(t) \) m/s.

Find its average velocity over the time interval \([0, \pi]\).

▶️ Answer/Explanation

Here, \( a = 0 \), \( b = \pi \), and \( v(t) = 4\sin(t) \).

\( \text{Average Velocity} = \dfrac{1}{\pi – 0} \int_{0}^{\pi} 4\sin(t) \, dt \)

\( = \dfrac{1}{\pi} \left[ -4\cos(t) \right]_{0}^{\pi} \) \( = \dfrac{1}{\pi} \left( -4\cos(\pi) + 4\cos(0) \right) \)

\( = \dfrac{1}{\pi} \left( -4(-1) + 4(1) \right) \)

\( = \dfrac{1}{\pi} (4 + 4) = \dfrac{8}{\pi} \)

So, the average velocity is \( \dfrac{8}{\pi} \) m/s.

Example:

The temperature of a cup of coffee is modeled by \( T(t) = 70 + 50e^{-0.1t} \) °F, where \( t \) is in minutes.

Find the average temperature from \( t = 0 \) to \( t = 10 \) minutes.

▶️ Answer/Explanation

We use: \( \text{Average Temperature} = \dfrac{1}{10 – 0} \int_{0}^{10} \left( 70 + 50e^{-0.1t} \right) dt \)

Split the integral: \( = \dfrac{1}{10} \left[ \int_{0}^{10} 70 \, dt + \int_{0}^{10} 50e^{-0.1t} \, dt \right] \)

First integral: \( \int_{0}^{10} 70 \, dt = 70t \big|_{0}^{10} = 700 \)

Second integral: \( \int_{0}^{10} 50e^{-0.1t} \, dt = \dfrac{50}{-0.1} e^{-0.1t} \big|_{0}^{10} = -500 \left( e^{-1} – 1 \right) \)

\( = -500 \left( \dfrac{1}{e} – 1 \right) = 500 \left( 1 – \dfrac{1}{e} \right) \)

Adding: \( \text{Total Area} = 700 + 500 \left( 1 – \dfrac{1}{e} \right) \)

Average value: \( \text{Average Temperature} = \dfrac{1}{10} \left[ 700 + 500 \left( 1 – \dfrac{1}{e} \right) \right] \approx 111.64 \ \text{°F} \)

Scroll to Top