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Section 5.5 Skills: Average Value of a Function

Section 5.5: Average Value of a Function

This section covers computing the average value of a function over an interval and the Mean Value Theorem for Integrals.

Learning Path

graph LR
    subgraph Foundation
        A["Average Value<br/>Formula"]
    end

    subgraph Theorem
        B["Mean Value Theorem<br/>for Integrals"]
    end

    A --> B

    style A fill:#d1fae5,stroke:#a565f0,stroke-width:3px
    style B fill:#d1fae5,stroke:#a565f0,stroke-width:2px

    click A "average-value-formula.html"
    click B "mean-value-theorem-integrals.html"

Skills in This Section

Skill	Description	Difficulty	Time
Average Value Formula	Computing $f_{\text{avg}}$ over $[a,b]$	Beginner	~15 min
Mean Value Theorem for Integrals	Existence of $c$ where $f(c) = f_{\text{avg}}$	Intermediate	~20 min

Key Formulas

Formula	Meaning
$f_{\text{avg}} = \dfrac{1}{b-a} \displaystyle\int_a^b f(x)\,dx$	Average value of $f$ on $[a,b]$
$\displaystyle\int_a^b f(x)\,dx = f_{\text{avg}} \cdot (b-a)$	Integral equals average height × base
$f(c) = f_{\text{avg}}$ for some $c \in [a,b]$	MVT for Integrals (requires continuity)

Prerequisites

Before starting this section, ensure you're comfortable with:

Definite integrals and the Fundamental Theorem of Calculus
Riemann sums (conceptual understanding of how integrals arise)
Continuity (required for the Mean Value Theorem)

Tip: Each skill page has a "Before You Start" self-check. If you struggle with those diagnostics, follow the links to review prerequisites before proceeding.

Recommended Study Order

Start with Average Value Formula: learn the formula and its derivation
Then Mean Value Theorem for Integrals: understand why the average is achieved

The Big Picture

Why Average Value Matters

When you compute the average of test scores, you add them up and divide by the number of tests. But what if you have infinitely many "scores," like temperature readings every instant of the day? You can't literally add infinitely many numbers.

The average value formula solves this by using integration: the integral "adds up" all the function values, and dividing by $(b-a)$ "counts" how many values there are (in a continuous sense).

The Geometric Interpretation

The average value $f_{\text{avg}}$ is the height of a rectangle with base $[a, b]$ that has the same area as the region under the curve $y = f(x)$.

    y
    │        ╭───╮
    │      ╱│     │╲
f_avg├─────┼─────────┼──────  ← rectangle height
    │    ╱ │         │ ╲
    │   ╱  │         │  ╲
    │  ╱   │         │   ╲
    └──────┼─────────┼──────→ x
           a         b

    Area under curve = Area of rectangle = f_avg × (b-a)

The Mean Value Theorem for Integrals

If $f$ is continuous on $[a, b]$, then there exists at least one number $c$ in $[a, b]$ such that:

$$f(c) = f_{\text{avg}} = \frac{1}{b-a} \int_a^b f(x)\,dx$$

What this says: A continuous function must actually equal its average value somewhere in the interval. It can't "skip over" its average.

← Section 5.4 Skills	Section 5.5 Overview	Chapter 6 →

Last updated: 2026-01-23