Average Value of a Function Explained — Derivation, Examples & Quizzes

Q: What is the average value of a function?

f_avg = 1/(b−a) · ∫ₐᵇ f(x) dx: integrate over [a,b], then divide by the interval's width. It's the continuous version of (sum)/(count).

Q: What does the Mean Value Theorem for integrals add?

It guarantees some c in (a,b) with f(c) = f_avg: a continuous function actually attains its average value somewhere in the interval.

Where it comes from

The problem: you know the average of finitely many numbers — add them up, divide by how many. But a function on [a, b] has infinitely many values. What should “average” even mean? The translation: the “sum” becomes the integral (total accumulation), and the “count” becomes the interval’s width (b−a).

The tempting endpoint average:

Before reading on: the wrong guess averages the endpoints: (f(0)+f(3))/2 = 4.5 for f(x) = x². But the curve bends upward, spending most of its time low. Before reading: should the true average be above or below 4.5 — and why?

f avg = (f(a) + f(b))/2 ?? the tempting — and wrong (in general) — guess

Kill it with f(x) = x² on [0, 2]. The guess says (f(0) + f(2))/2 = (0 + 4)/2 = 2. But the true average weights every point, not just the ends:

f_avg

=

(1/(2−0)) · ∫₀² x² dx = (1/2) · (8/3) = 4/3

The true average is 4/3 ≈ 1.33, not 2. The endpoint guess overshoots because x² curves upward — it spends most of the interval below the chord joining the endpoints.

Intuition: f_avg is the height of the rectangle with the same area. A rectangle of height f_avg over [a, b] has area f_avg·(b−a); setting that equal to the true area ∫_a^b f and solving gives the formula. The endpoint average only matches for straight lines (where the area is exactly a trapezoid).

Derivation

Start from the discrete average and take the limit — or argue geometrically with the equal-area rectangle. Both roads reach the same formula.

average of n values

=

(f(x₁) + ⋯ + f(x_n)) / n

Step 1 — the discrete template. Sum of values divided by count. We generalize each piece.

“sum”: Σ f(x_i) Δx

→

integral: ∫_a^b f(x) dx

Step 2 — continuous total. Sampling n evenly spaced points with spacing Δx = (b−a)/n, the Riemann sum Σf(x_i)Δx tends to the integral — the “sum” of all values.

“count”

→

width — n·Δx = b − a

Step 3 — continuous count. Dividing the Riemann sum by n: [Σf(x_i)Δx]/n = [Σf(x_i)/n]·Δx·n/n… precisely, (1/n)Σf(x_i)Δx = [1/(b−a)]Σf(x_i)Δx ·(b−a)/n · n/(b−a)… the limit is [∫_a^bf]/(b−a).

f_avg

=

(1/(b−a)) · ∫_a^b f(x) dx

Step 4 — the formula. Equivalently: the constant height whose rectangle f_avg·(b−a) has the same area as ∫_a^bf. ∎

Bonus — Mean Value Theorem for integrals: if f is continuous on [a, b], some c in (a, b) has f(c) = f_avg. The function actually attains its average value somewhere inside — the equal-area rectangle’s top edge always touches the curve.

How to use it

Before reading on: the average of n sampled heights is (sum)/n. As the samples get infinitely dense, the sum becomes an integral. Before reading: what continuous thing replaces the “divide by n”?

The procedure, every time:

Integrate f over [a, b] — FTC Part 2, as usual.
Divide by the width (b−a). Not by b, not by n — by the length of the interval.
Sanity-check against the shape: the average must lie between the function’s min and max on [a, b]. If it doesn’t, recheck.

Shortcuts worth knowing

Constants: the average of f(x) = c is c (collapse check: (1/(b−a))·c(b−a) = c). Lines: the average of a linear function is its midpoint value f((a+b)/2) — the only case where the endpoint average (f(a)+f(b))/2 works, because both equal the midpoint value. Symmetry: odd functions on symmetric intervals average to 0.

Common mistake: dividing by b instead of (b−a). For f(x) = x on [2, 4]: correct is (1/2)∫₂⁴ x dx = (1/2)(8−2) = 3, but ÷b gives (1/4)(6) = 1.5 — not even between f’s min (2) and max (4). The width is 2, not 4.

Worked examples

Four problems, easiest first. In each one, read every step — the why of each move is the lesson.

Example 1 — the basic move: average of x² on [0, 3]

Integrate: ∫₀³ x² dx = [x³/3]₀³ = 9.
Divide by the width (3 − 0 = 3): 9/3 = 3.
Sanity check: x² runs from 0 to 9 on [0, 3]; the average 3 sits between. Also plausible: the curve bends upward, so the average should exceed the midpoint value f(1.5) = 2.25 — and 3 > 2.25. ✓

Your turn: Find the average value of x² on [0, 2].

Answer: 4/3

Integrate: ∫₀² x² dx = [x³/3]₀² = 8/3. Divide by the width 2: (8/3)/2 = 4/3. Sanity: x² runs 0→4; the upward bend pushes the average above the midpoint value f(1) = 1, and 4/3 ≈ 1.33 fits. ✓

Example 2 — trig: average of sin x on [0, π]

Integrate: ∫₀^π sin x dx = 2 (the area-2 anchor).
Divide by the width π: 2/π ≈ 0.637.
Sanity check: sin x peaks at 1 and the hump is rounded, so the average should be well under 1 but well above 0. 0.637 fits. ✓

Common mistake: answering 2 — the integral without the division. The integral is the total; the average is the total per unit width. Forgetting ÷(b−a) is the #1 error on this page.

Your turn: Find the average value of cos x on [−π/2, π/2].

Answer: 2/π ≈ 0.637

Integrate: ∫_−π/2^π/2 cos x dx = [sin x]_−π/2^π/2 = 1 − (−1) = 2. Divide by the width π: 2/π. Sanity: cos x peaks at 1 with a rounded hump, so the average sits well under 1 but well above 0. ✓

Example 3 — constant: average of 5 on [2, 7]

Integrate: ∫₂⁷ 5 dx = 5·(7−2) = 25.
Divide by the width 5: 25/5 = 5.
The lesson: the average of a constant is itself — the collapse check. If your method ever gives anything else for a constant, the method is broken.

Your turn: Find the average value of −3 on [1, 4].

Answer: −3

Integrate: ∫₁⁴ (−3) dx = −3·3 = −9. Divide by the width 3: −3. The lesson: the average of a constant is itself — the collapse check. ✓

Example 4 — motion: average velocity for v(t) = 3t² on [0, 2]

Integrate (total displacement): ∫₀² 3t² dt = [t³]₀² = 8.
Divide by the time width 2: 8/2 = 4.
Interpret: average velocity 4 over 2 seconds → displacement 8 — consistent. (Also: v runs 0→12, curving upward, so average 4 < midpoint v(1) = 3? No: 4 > 3 ✓ — upward curve pushes the average above the midpoint value.)

Common mistake: averaging the endpoint velocities: (v(0)+v(2))/2 = (0+12)/2 = 6 ≠ 4. Endpoint averaging fails for curves — only the integral sees the whole motion.

Your turn: Average velocity for v(t) = 2t on [0, 4].

Answer: 4

Integrate (total displacement): ∫₀⁴ 2t dt = [t²]₀⁴ = 16. Divide by the time width 4: 4. Note: endpoint averaging (v(0)+v(4))/2 = 4 agrees here only because v is linear — no bend, no trap. ✓

Memorization tips

Say it aloud: “total over width.” The integral is the total; (b−a) is the width. Two words, the whole formula.
The rectangle picture: f_avg is the height of the equal-area rectangle. Draw it once — the formula becomes geometry, not algebra.
Width, not b: on [2, 4] the width is 2. Whenever a ≠ 0, pause and compute b−a explicitly.
The min-max cage: the average must sit between the function’s minimum and maximum on the interval. Cage every answer.
Collapse check: constants average to themselves. Run it whenever a problem looks suspicious.
Endpoints lie (usually): (f(a)+f(b))/2 works only for straight lines. For anything curved, integrate.

Final challenge

Five mixed questions — the width trap, the endpoint trap, and solving backwards. Score 5/5 and averages are yours.

How to learn a formula here

Read each section in order — every section ends with a short quiz. Take it before moving on; the questions test exactly what you just read.
Work the examples with the answers covered, then uncover one step at a time and compare.
Finish with the final challenge — five mixed questions including the classic traps.
Retake what you miss — every quiz reshuffles each attempt, and every answer explains itself.

Frequently asked questions

What is the average value of a function?

f_avg = 1/(b−a) · ∫_a^b f(x) dx: integrate over [a, b], then divide by the interval’s width. It’s the continuous version of (sum)/(count).

Why not just average f(a) and f(b)?

That only works for linear functions. Counterexample: f(x) = x² on [0, 2] gives (0+4)/2 = 2, but the true average is (1/2)∫₀²x² dx = 4/3. Endpoints ignore the curve’s shape.

Do I divide by b or by (b−a)?

By (b−a), the width of the interval. Dividing by b only works when a = 0 — otherwise it answers a different question.

What is the average value of sin x on [0, π]?

(1/π)∫₀^π sin x dx = 2/π ≈ 0.637. Plausible: the max is 1, and the hump spends most of its time below it.

What does the Mean Value Theorem for integrals add?

It guarantees some c in (a, b) with f(c) = f_avg: a continuous function actually attains its average value somewhere in the interval.

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