Navigation: Wiki Home > Skills > Average Value of a Function
Quick test: Compute $\displaystyle\int_1^3 (2x + 1)\,dx$
$$\int_1^3 (2x + 1)\,dx = \left[x^2 + x\right]_1^3 = (9 + 3) - (1 + 1) = 12 - 2 = 10$$
✅ 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)
A Riemann sum $\sum_{i=1}^n f(x_i^*)\Delta x$ approximates the area under a curve by adding up rectangles.
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)
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.
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D["Mean Value Theorem<br/>for Integrals"]
E["Probability<br/>Density Functions"]
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Legend: Yellow nodes = direct prerequisites you need. Green = this skill. Dashed arrows = helpful but not required.
| Property | Value |
|---|---|
| Concept | Applications of Integration |
| Chapter | 5, Section 5 |
| Difficulty | Beginner |
| Time | ~15 minutes |
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:
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:
$$f_{\text{avg}} = \frac{1}{b-a} \int_a^b f(x)\,dx$$
y
│ ╭───╮
│ ╱│ │╲
f_avg├─────┼─────────┼────── ← horizontal line at average height
│ ╱ │ │ ╲
│ ╱ │ │ ╲
│ ╱ │ │ ╲
└──────┼─────────┼──────→ x
a b
Area under curve = Area of rectangle with height f_avg
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.
$$\text{Area under curve} = \int_a^b f(x)\,dx = f_{\text{avg}} \cdot (b-a) = \text{Area of rectangle}$$
The average value appears naturally in many contexts:
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."
💡 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 |
Before submitting any average value problem, verify:
If your answer is outside the range of $f$, something went wrong!
Problem: Find the average value of $f(x) = x^2 - 2x + 3$ on the interval $[1, 4]$.
We need to:
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: $$\int (x^2 - 2x + 3)\,dx = \frac{x^3}{3} - x^2 + 3x + C$$
Using the power rule for each term:
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: $$f_{\text{avg}} = \frac{1}{3} \cdot 15 = 5$$
Is this answer reasonable? Let's check:
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 $x=4$.
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."
How to use these problems:
- Levels 1-2: Build confidence with the basic formula
- Level 3: Add complexity (trig, substitution)
- Levels 4-5: Applications and proofs—exam-level difficulty
Find the average value of $f(x) = 6x$ on the interval $[0, 3]$.
Find the average value of $g(x) = x^3 - x$ on $[-1, 2]$.
Find the average value of $h(x) = \sin(2x)$ on $[0, \pi/4]$.
A particle moves along a line with velocity $v(t) = t^2 - 4t + 3$ m/s for $0 \leq t \leq 5$ seconds.
Let $f$ be a continuous function on $[a, b]$.
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 $[0, 10]$?
$f_{\text{avg}} = \frac{50}{10} = 5$
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 $\int_2^8 f(x)\,dx$?
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.
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 $[a, b]$.
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 $f$.
Any curve that:
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).
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.
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.
📌 What to remember for exams:
- The formula: $f_{\text{avg}} = \frac{1}{b-a}\int_a^b f(x)\,dx$
- Geometric meaning: Height of equal-area rectangle
- Bounds: $\min(f) \leq f_{\text{avg}} \leq \max(f)$
- Common trap: Don't forget to divide by $(b-a)$!
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