Average Value

Introduction to Average Value of a Function

  • Basic Concept of Average: The average value can be illustrated using the ages of three friends:

    • Average = (Age1 + Age2 + Age3) / 3

    • In mathematical terms, this can be expressed as the sum from 1 to 3 of a_i (i = 1, 2, 3) divided by 3.

Transition to Functions

  • The average value of a function over an interval can be computed similarly:

    • Consider a graph of a function f(x) defined on the interval [a, b].

    • We divide this interval into n subintervals, each with width delta x.

Steps to Calculate Average Value

  • Defining Average Value of Function:

    • Average Value (f_avg) = (Sum of Function Values) / (Number of Function Values)

    • For n subintervals, the average value of the function can be expressed as:

      • f_avg ≈ (Sum from 1 to n of f(x_i)) / n.

  • Calculating delta x:

    • delta x = (b - a) / n

    • Therefore, the approximate average value becomes:

      • f_avg ≈ (Sum from 1 to n of f(x_i)) / ((b - a)/delta x).

Limit as n Approaches Infinity

  • As n approaches infinity:

    • The precise average value of the function is given by:

      • f_avg = limit as n->infinity of (Sum from 1 to n of f(x_i) * delta x) / (b - a)

    • This limit expression corresponds to the definition of a definite integral.

Definitive Integral Representation

  • The average value can thus be represented as:

    • f_avg = (1 / (b - a)) * ∫ from a to b f(x) dx

    • This means we calculate the definite integral of the function over the interval and normalize it by the interval's width.

Example with Cosine Function

  • Function to Analyze: f(x) = cos(x) over the interval [−π/2, π/2]

    • The function hits zero at both endpoints (cos(−π/2) = cos(π/2) = 0).

    • Maximum value of cosine on this interval is 1.

  • Checking for Reasonableness: Expect f_avg to lie between 0 and 1 due to the known max and min values.

  • Calculating Average Value:

    • f_avg = (1 / (π/2 - (−π/2))) * ∫ from −π/2 to π/2 cos(x) dx

    • Which simplifies to:

      • f_avg = (1 / π) * (sin(π/2) - sin(−π/2))

  • Evaluation of Integral:

    • sin(π/2) = 1 and sin(−π/2) = -1

    • Therefore, f_avg = (1 / π) * (1 - (−1)) = (2/π)

Conclusion from Example

  • Average value of f(x) = cos(x) over [−π/2, π/2] is f_avg = 2/π.

  • This is approximately 0.6366, which confirms it lies between 0 and 1, consistent with the earlier expectations.