← MathScape

Average Value of a Function

MATH162
Reference: Stewart 5.5  •  Chapter: 5  •  Section: 5

Navigation: Wiki Home > Skills > Average Value of a Function

Average Value of a Function

Quick Navigation


Before You Start

🔍 Prerequisite Check (2 minutes) — Do this first!

Can you evaluate a definite integral?

Quick test: Compute $\displaystyle\int_1^3 (2x + 1)\,dx$

Check Your Answer

$$\int_1^3 (2x + 1)\,dx = \left[x^2 + x\right]_1^3 = (9 + 3) - (1 + 1) = 12 - 2 = 10$$

Got it? You're ready for this page!

Stuck on the antiderivative? Review Antiderivatives first (~10 min)

Stuck on the evaluation step? Review FTC Part 2 first (~15 min)

Do you remember what a Riemann sum represents?

A Riemann sum $\sum_{i=1}^n f(x_i^*)\Delta x$ approximates the area under a curve by adding up rectangles.

Need a refresher on Riemann sums?

Key idea: We divide $[a,b]$ into $n$ pieces of width $\Delta x = \frac{b-a}{n}$, pick sample points $x_i^*$ in each piece, and add up $f(x_i^*) \cdot \Delta x$ (height × width of each rectangle).

As $n \to \infty$, this sum becomes the integral: $\displaystyle\lim_{n\to\infty} \sum_{i=1}^n f(x_i^*)\Delta x = \int_a^b f(x)\,dx$

If this feels unfamiliar, review Riemann Sums (~10 min)


From Discrete to Continuous Averaging

How do you find the "average temperature" over an entire day when temperature changes continuously? You can't just add up infinitely many values and divide. Yet averaging is one of the most natural things we do with data—so there must be a way to extend it to continuous functions.

The key insight: averaging finitely many values becomes an integral when we have infinitely many. This connection between sums and integrals is exactly what the definite integral was designed for.

Prerequisite Map

graph LR
    subgraph Foundation["Foundation (Earlier Chapters)"]
        AA["Limits"]
        AB["Antiderivatives"]
    end

    subgraph DirectPrereqs["Direct Prerequisites"]
        A["Definite Integrals<br/>(min level 3)"]
        B["FTC Part 2<br/>(min level 2)"]
        AC["Riemann Sums<br/>(conceptual)"]
    end

    subgraph ThisSkill["This Skill"]
        C["Average Value<br/>Formula"]
    end

    subgraph Unlocks["What This Unlocks"]
        D["Mean Value Theorem<br/>for Integrals"]
        E["Probability<br/>Density Functions"]
        F["Expected Value<br/>(Statistics)"]
    end

    AA --> AC
    AB --> B
    AC --> A
    A --> C
    B --> C
    AC -.->|"intuition"| C
    C --> D
    C --> E
    D --> F

    style C fill:#d1fae5,stroke:#a565f0,stroke-width:3px
    style A fill:#fef3c7,stroke:#f59e0b
    style B fill:#fef3c7,stroke:#f59e0b

    click A "../ch4-sec2/definite-integral-evaluation.html"
    click B "../ch4-sec3/ftc-part2.html"
    click AC "../ch4-sec2/riemann-sums.html"
    click D "mean-value-theorem-integrals.html"
    click E "../ch8-sec5/probability-density.html"

    click C "average-value-formula.html"
    click F "../ch8-sec5/expected-value.html"

Legend: Yellow nodes = direct prerequisites you need. Green = this skill. Dashed arrows = helpful but not required.

Quick Reference

Property Value
Concept Applications of Integration
Chapter 5, Section 5
Difficulty Beginner
Time ~15 minutes

Key Concepts

The Average Value Formula

For a continuous function $f$ on the interval $[a, b]$, the average value is:

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

Breaking this down:

Where This Comes From

For $n$ discrete values $y_1, y_2, \ldots, y_n$, the average is:

$$y_{\text{avg}} = \frac{y_1 + y_2 + \cdots + y_n}{n}$$

For a function sampled at $n$ points with spacing $\Delta x = \frac{b-a}{n}$:

$$\frac{f(x_1^*) + f(x_2^*) + \cdots + f(x_n^*)}{n} = \frac{1}{b-a} \sum_{i=1}^n f(x_i^*)\Delta x$$

As $n \to \infty$, the Riemann sum becomes the integral:

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

Visualization

    y
    │        ╭───╮
    │      ╱│     │╲
f_avg├─────┼─────────┼──────  ← horizontal line at average height
    │    ╱ │         │ ╲
    │   ╱  │         │  ╲
    │  ╱   │         │   ╲
    └──────┼─────────┼──────→ x
           a         b

    Area under curve = Area of rectangle with height f_avg

The Rectangle Interpretation

Key insight: 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.

$$\text{Area under curve} = \int_a^b f(x)\,dx = f_{\text{avg}} \cdot (b-a) = \text{Area of rectangle}$$

Physical Interpretation

The average value appears naturally in many contexts:

💡 Alternative Way to Think About It

The "Redistribution" View: Imagine the area under the curve is made of water. The average value is what happens if you redistribute that water evenly across the interval—all the peaks fill in the valleys until you have a flat surface at height $f_{\text{avg}}$.

The "Sampling" View: If you took a huge number of random samples from the function and averaged them, you'd get $f_{\text{avg}}$. The integral is doing this sampling "infinitely densely."


Common Pitfalls

💡 Don't worry if you make these mistakes—almost everyone does at first!

Mistake Why It Happens How to Avoid It
Forgetting to divide by $(b-a)$ You compute $\int_a^b f(x)\,dx$ and stop Always write the full formula first: $f_{\text{avg}} = \frac{1}{b-a}\int_a^b f(x)\,dx$
Using $(b+a)$ instead of $(b-a)$ Interval notation confusion Remember: the length of $[a,b]$ is $b - a$ (subtraction!)
Getting the sign wrong when $a < 0$ E.g., interval $[-2, 3]$ has length $3-(-2)=5$, not $1$ Be extra careful: subtracting a negative adds
Confusing average value with average rate of change Both use "average" Average value = $\frac{1}{b-a}\int f$. Average rate = $\frac{f(b)-f(a)}{b-a}$
Thinking average value must be at the midpoint Intuition from symmetric functions Average value location depends on the function's shape, not just the interval
🔧 Quick Self-Check

Before submitting any average value problem, verify:

  • [ ] Did I compute the integral correctly?
  • [ ] Did I divide by the interval length $(b-a)$?
  • [ ] Is my answer between the min and max of $f$ on $[a,b]$?

If your answer is outside the range of $f$, something went wrong!


Worked Example

Problem: Find the average value of $f(x) = x^2 - 2x + 3$ on the interval $[1, 4]$.

📋 What's the game plan?

We need to:

  1. Identify the interval endpoints ($a$ and $b$)
  2. Set up the average value formula
  3. Compute the definite integral
  4. Divide by the interval length

Solution:

Step 1: Identify the interval.

Step 2: Write down the formula with our values plugged in: $$f_{\text{avg}} = \frac{1}{b-a}\int_a^b f(x)\,dx = \frac{1}{3}\int_1^4 (x^2 - 2x + 3)\,dx$$

Step 3: Find the antiderivative: $$\int (x^2 - 2x + 3)\,dx = \frac{x^3}{3} - x^2 + 3x + C$$

How did we get this antiderivative?

Using the power rule for each term:

  • $\int x^2\,dx = \frac{x^3}{3}$
  • $\int (-2x)\,dx = -x^2$
  • $\int 3\,dx = 3x$

Step 4: Evaluate at the bounds: $$\left[\frac{x^3}{3} - x^2 + 3x\right]_1^4 = \underbrace{\left(\frac{64}{3} - 16 + 12\right)}_{\text{at } x=4} - \underbrace{\left(\frac{1}{3} - 1 + 3\right)}_{\text{at } x=1}$$

$$= \left(\frac{64}{3} - 4\right) - \left(\frac{1}{3} + 2\right) = \frac{64}{3} - 4 - \frac{1}{3} - 2 = \frac{63}{3} - 6 = 21 - 6 = 15$$

Step 5: Apply the averaging factor: $$f_{\text{avg}} = \frac{1}{3} \cdot 15 = 5$$

✅ Sanity Check

Is this answer reasonable? Let's check:

  • At $x = 1$: $f(1) = 1 - 2 + 3 = 2$
  • At $x = 4$: $f(4) = 16 - 8 + 3 = 11$
  • Our answer $f_{\text{avg}} = 5$ is between 2 and 11 ✓

Since $f(x) = x^2 - 2x + 3 = (x-1)^2 + 2$ is an upward parabola with minimum value 2 at $x=1$, the function increases on $[1,4]$. The average of 5 lies between the minimum (2) and maximum (11), closer to the minimum—this makes sense because the function stays near its minimum for a while before climbing steeply near $x=4$.


😕 Still confused about average value?

If the formula doesn't make sense: Go back to Where This Comes From and trace through the discrete-to-continuous derivation.

If you're struggling with the integral: Review Definite Integrals and practice a few basic examples.

If the concept feels abstract: Think about averaging test scores. If you took tests continuously throughout a semester, the average value formula computes your "semester average."


Practice Problems

How to use these problems:

Level 1 Direct Formula Application

Find the average value of $f(x) = 6x$ on the interval $[0, 3]$.

Thought Process

This is a direct application of the formula. Identify $a$, $b$, set up the integral, evaluate it, then divide by the interval length.

The integral of $6x$ is $3x^2$. The interval length is $3 - 0 = 3$.

Show Answer

$$f_{\text{avg}} = \frac{1}{3-0}\int_0^3 6x\,dx = \frac{1}{3}\left[3x^2\right]_0^3 = \frac{1}{3}(27 - 0) = 9$$

Level 2 Polynomial Function

Find the average value of $g(x) = x^3 - x$ on $[-1, 2]$.

Thought Process

The interval length is $2 - (-1) = 3$. Integrate each term separately using the power rule. Be careful with the lower limit being negative.

Show Answer

$$g_{\text{avg}} = \frac{1}{3}\int_{-1}^2 (x^3 - x)\,dx = \frac{1}{3}\left[\frac{x^4}{4} - \frac{x^2}{2}\right]_{-1}^2$$

$$= \frac{1}{3}\left[\left(4 - 2\right) - \left(\frac{1}{4} - \frac{1}{2}\right)\right] = \frac{1}{3}\left[2 - \left(-\frac{1}{4}\right)\right] = \frac{1}{3} \cdot \frac{9}{4} = \frac{3}{4}$$

Level 3 Trigonometric Function

Find the average value of $h(x) = \sin(2x)$ on $[0, \pi/4]$.

Thought Process

The interval length is $\pi/4$. For $\int \sin(2x)\,dx$, use substitution or recall that the antiderivative is $-\frac{1}{2}\cos(2x)$.

Evaluate at the bounds: at $x = \pi/4$, we have $2x = \pi/2$, so $\cos(\pi/2) = 0$. At $x = 0$, $\cos(0) = 1$.

Show Answer

$$h_{\text{avg}} = \frac{1}{\pi/4}\int_0^{\pi/4} \sin(2x)\,dx = \frac{4}{\pi}\left[-\frac{1}{2}\cos(2x)\right]_0^{\pi/4}$$

$$= \frac{4}{\pi} \cdot \left(-\frac{1}{2}\right)\left[\cos\left(\frac{\pi}{2}\right) - \cos(0)\right] = \frac{4}{\pi} \cdot \left(-\frac{1}{2}\right)(0 - 1) = \frac{2}{\pi}$$

Level 4 Average Velocity Application

A particle moves along a line with velocity $v(t) = t^2 - 4t + 3$ m/s for $0 \leq t \leq 5$ seconds.

  1. Find the average velocity over the interval $[0, 5]$.
  2. Is the average velocity the same as the average speed? Explain.
Thought Process

🎯 Exam Strategy: This is a classic "trap" problem! Part (b) tests whether you understand the difference between velocity and speed.

For (a): Average velocity uses the signed velocity function directly. This is straightforward—just apply the average value formula.

For (b): This is where many students lose points. Average speed requires $\vert v(t)\vert $, which means:

  1. Find where $v(t) = 0$ (factor: $v(t) = (t-1)(t-3)$)
  2. Check the sign of $v(t)$ on each subinterval
  3. The question doesn't ask you to compute average speed—just explain why it differs
Show Answer

(a) Average velocity: $$v_{\text{avg}} = \frac{1}{5}\int_0^5 (t^2 - 4t + 3)\,dt = \frac{1}{5}\left[\frac{t^3}{3} - 2t^2 + 3t\right]_0^5$$

$$= \frac{1}{5}\left(\frac{125}{3} - 50 + 15\right) = \frac{1}{5} \cdot \frac{125 - 150 + 45}{3} = \frac{1}{5} \cdot \frac{20}{3} = \frac{4}{3} \text{ m/s}$$

(b) No, average velocity and average speed are different.

Quantity Formula What it measures
Average velocity $\frac{1}{b-a}\int_a^b v(t)\,dt$ Net displacement per unit time (signed)
Average speed $\frac{1}{b-a}\int_a^b \vert v(t)\vert \,dt$ Total distance per unit time (unsigned)

Since $v(t) = (t-1)(t-3)$ is negative on $(1,3)$, the particle moves backward during that interval. The integral of $v(t)$ subtracts this backward motion, but average speed would add the absolute value of all motion.

Physical intuition: If you drive 10 km east then 10 km west, your average velocity is 0, but your average speed is positive (you were always moving!).

Level 5 Average Value Inequality

Let $f$ be a continuous function on $[a, b]$.

  1. Prove that if $f(x) \geq 0$ for all $x \in [a,b]$, then $f_{\text{avg}} \geq 0$.
  2. Prove that if $m \leq f(x) \leq M$ for all $x \in [a,b]$, then $m \leq f_{\text{avg}} \leq M$.
  3. Use part (b) to show that $\frac{1}{2} \leq \frac{1}{e-1}\int_1^e \frac{1}{x}\,dx \leq 1$.
Thought Process

🎯 Exam Strategy: This is a proof-based problem. Each part builds on the previous one—recognize this structure!

For (a): If $f(x) \geq 0$, then the integral (area under curve) is non-negative. Dividing by positive $(b-a)$ preserves the inequality. This is almost trivial but sets up part (b).

For (b): Use the comparison property of integrals: if $g(x) \leq h(x)$ on $[a,b]$, then $\int_a^b g \leq \int_a^b h$. Apply this with $g = m$ (constant) and $h = f$, then with $f$ and $M$.

For (c): This applies part (b) to a specific function. Find $m$ and $M$ for $f(x) = 1/x$ on $[1, e]$. Since $1/x$ is decreasing:

  • Maximum at left endpoint: $f(1) = 1$
  • Minimum at right endpoint: $f(e) = 1/e$

Then use part (b) to bound $f_{\text{avg}}$.

Show Answer

(a) If $f(x) \geq 0$ on $[a,b]$, then by properties of integrals: $$\int_a^b f(x)\,dx \geq 0$$ Dividing by $(b-a) > 0$: $$f_{\text{avg}} = \frac{1}{b-a}\int_a^b f(x)\,dx \geq 0$$

(b) If $m \leq f(x) \leq M$ for all $x \in [a,b]$, then: $$\int_a^b m\,dx \leq \int_a^b f(x)\,dx \leq \int_a^b M\,dx$$ $$m(b-a) \leq \int_a^b f(x)\,dx \leq M(b-a)$$

Dividing by $(b-a) > 0$: $$m \leq \frac{1}{b-a}\int_a^b f(x)\,dx \leq M$$ $$m \leq f_{\text{avg}} \leq M$$

(c) For $f(x) = 1/x$ on $[1, e]$:

Upper bound (using part (b)): Since $\frac{1}{x} \leq 1$ for all $x \geq 1$, part (b) gives $f_{\text{avg}} \leq 1$. ✓

Lower bound: Part (b) only gives $f_{\text{avg}} \geq \frac{1}{e} \approx 0.37$, which is weaker than $\frac{1}{2}$. To get the sharper bound, we compute the integral directly:

$$f_{\text{avg}} = \frac{1}{e-1}\int_1^e \frac{1}{x}\,dx = \frac{1}{e-1}[\ln x]_1^e = \frac{1 - 0}{e-1} = \frac{1}{e-1}$$

Now we show $\frac{1}{e-1} \geq \frac{1}{2}$.

This is equivalent to showing $e - 1 \leq 2$, i.e., $e \leq 3$.

From the definition $e = \sum_{n=0}^{\infty} \frac{1}{n!}$, we can bound: $$e = 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \cdots < 1 + 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots = 1 + 2 = 3$$

(The inequality holds because $\frac{1}{n!} < \frac{1}{2^{n-1}}$ for $n \geq 2$.)

Therefore $e < 3$, so $e - 1 < 2$, giving $\frac{1}{e-1} > \frac{1}{2}$. ✓

CCI-Style Conceptual Questions

These questions test understanding, not computation. They're the type that separate students who memorized from those who truly understand.

Question 1: If $\int_0^{10} f(x)\,dx = 50$, what is the average value of $f$ on $[0, 10]$?

Answer

$f_{\text{avg}} = \frac{50}{10} = 5$

Why this is useful: On exams, you might be given the integral value and asked for the average (or vice versa). Know how to go both directions!

Question 2: The average value of a function $f$ on $[2, 8]$ is 7. What is $\int_2^8 f(x)\,dx$?

Answer

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

$$\int_2^8 f(x)\,dx = f_{\text{avg}} \cdot (b-a) = 7 \cdot (8-2) = 42$$

Question 3: Can the average value of a function on an interval be larger than the maximum value of the function on that interval? Explain.

Answer

No. If $f(x) \leq M$ for all $x$ in $[a,b]$, then $f_{\text{avg}} \leq M$.

Proof sketch: $\displaystyle f_{\text{avg}} = \frac{1}{b-a}\int_a^b f(x)\,dx \leq \frac{1}{b-a}\int_a^b M\,dx = \frac{M(b-a)}{b-a} = M$

Intuition: If no sample exceeds $M$, no average of samples can exceed $M$.

Question 4: True or False: If two different functions have the same average value on $[a, b]$, they must have the same integral on $[a, b]$.

Answer

True. If $f_{\text{avg}} = g_{\text{avg}}$ on $[a,b]$, then: $$\frac{1}{b-a}\int_a^b f(x)\,dx = \frac{1}{b-a}\int_a^b g(x)\,dx$$

Multiplying both sides by $(b-a)$: $$\int_a^b f(x)\,dx = \int_a^b g(x)\,dx$$

Question 5: A function $f$ is positive on $[0, 5]$ with $f_{\text{avg}} = 4$. Sketch a possible graph of $f$, and shade the rectangle with the same area as the region under $f$.

Answer

Any curve that:

  • Stays above the $x$-axis on $[0, 5]$
  • Has total area = $4 \times 5 = 20$

The rectangle has base $[0, 5]$ (width 5) and height 4.

    y
    │     ╭──╮
  4 ├─────┼──┼─────  ← rectangle height
    │    ╱│  │╲
    │   ╱ │  │ ╲
    │  ╱  │  │  ╲
    └──────────────→ x
         0    5

The shaded area under the curve equals the area of the rectangle (both = 20).

Mastery Checklist

Level 1-2: Novice → Competent

Level 3: Competent → Proficient

Level 4-5: Proficient → Expert

Self-Assessment

If you can check all boxes in Levels 1-3, you're ready for exams on this topic. Levels 4-5 prepare you for harder problems and future courses.

Mental Model

The "Smoothing" Analogy:

Imagine pouring water into a container shaped like the region under a curve. The average value is the water level if the container had straight vertical walls—all the peaks and valleys have been "averaged out" to a single uniform height.


Key Takeaways

📌 What to remember for exams:

  1. The formula: $f_{\text{avg}} = \frac{1}{b-a}\int_a^b f(x)\,dx$
  2. Geometric meaning: Height of equal-area rectangle
  3. Bounds: $\min(f) \leq f_{\text{avg}} \leq \max(f)$
  4. Common trap: Don't forget to divide by $(b-a)$!

Connections

Looking back:

Looking ahead:

Real-world connections:


Previous Up Next
§5.4 Work Chapter 5 Mean Value Theorem for Integrals

Last updated: 2026-01-23