Source: Calc AB Lecture Notes, UCF Advanced Calculus
Difficulty: Intermediate–Advanced | Prerequisites: Antiderivatives, definite integrals, FTC (Parts 1 and 2 of these notes), chain rule for derivatives
Tags: mean value theorem for integrals, MVT integrals, average value of a function, u-substitution, displacement, distance, net change theorem, Calc AB Unit 4
This section ties together everything from the first two parts of Unit 4 and adds the techniques you will actually use most often on exams. The Mean Value Theorem for Integrals connects average value to definite integrals. U-substitution is the integration equivalent of the chain rule and is the single most-used integration technique going forward. The displacement and distance distinction rounds out the motion applications that AP Calculus AB tests heavily.
The average value of a function on [a, b] is the integral divided by the interval width. U-substitution lets you integrate compositions of functions by identifying an inner function and its derivative. Displacement is the net integral of velocity; total distance is the integral of the absolute value of velocity.
Average value of a function
On [a, b], the average value of f is: f_avg = [1 / (b − a)] · ∫ₐᵇ f(x) dx. This gives the constant height of a rectangle with the same area as the region under f(x). Think of it as: "if f(x) were perfectly flat across the interval, what height would produce the same total area?"
Mean Value Theorem for Integrals (MVT for Integrals)
If f is continuous on [a, b], there exists at least one value c in [a, b] such that ∫ₐᵇ f(x) dx = f(c) · (b − a). In other words, the function actually hits its average value somewhere on the interval. In simple terms: "somewhere on the interval, the function equals its own average."
U-substitution
An integration technique for reversing the chain rule. You identify an inner function u = g(x), compute du = g'(x) dx, rewrite the integral in terms of u, integrate, and substitute back. Think of it as: "finding a hidden chain rule inside the integral and undoing it."
Displacement
The net change in position: ∫ v(t) dt. Positive and negative motion cancel each other out. In simple terms: "how far you end up from where you started, accounting for direction."
Total distance
The total length of path travelled: ∫ |v(t)| dt. All motion counts as positive, regardless of direction. Think of it as: "what the odometer reads."
Net Change Theorem
Another name for the Fundamental Theorem of Calculus applied to rates: ∫ₐᵇ f'(x) dx = f(b) − f(a). The integral of a rate of change gives the net change in the original quantity.
Constant multiple rule (for integrals)
∫ k · f(x) dx = k · ∫ f(x) dx. You can pull a constant factor outside the integral.
Average value formula: f_avg = [1 / (b − a)] · ∫ₐᵇ f(x) dx.
Geometrically, this is the height of the rectangle whose base is (b − a) and whose area equals the area under f(x).
The MVT for integrals guarantees that f(c) = f_avg for some c in [a, b], provided f is continuous.
Worked example: average value of √(x + 1) on [1, 3]
f_avg = [1 / (3 − 1)] · ∫₁³ √(x + 1) dx = (1/2) · ∫₁³ (x + 1)^(1/2) dx
Antiderivative of (x + 1)^(1/2): (2/3)(x + 1)^(3/2)
Evaluate: (1/2) · [(2/3)(4)^(3/2) − (2/3)(2)^(3/2)]
= (1/2) · [(2/3)(8) − (2/3)(2√2)]
= (1/2) · [16/3 − 4√2/3]
The lecture notes simplify to: √3 + 2/3
The foundation of u-substitution is recognising an inner function and its derivative within the integrand.
The method, step by step:
Identify the inner function: set u equal to it.
Differentiate to find du.
Rewrite the entire integral in terms of u and du (no x should remain).
Integrate with respect to u.
Substitute back to express the answer in terms of x.
Worked example 1: ∫ (4x² + 1)² · 8x dx
Inner function: u = 4x² + 1, so du = 8x dx.
Outer function: u².
Rewrite: ∫ u² du = u³/3 + C = (4x² + 1)³ / 3 + C.
Worked example 2: ∫ (1 + 6x)⁴ · 6 dx
u = 1 + 6x, du = 6 dx.
∫ u⁴ du = u⁵/5 + C = (1 + 6x)⁵ / 5 + C.
Worked example 3 (with a constant adjustment): (1/10) ∫₀¹⁰ x(5x² + 4)³ dx
u = 5x² + 4, du = 10x dx. The existing x dx is (1/10) du.
(1/10) ∫ u³ · (1/10) du would double-count the factor, so track carefully:
The original integral already has the 1/10 in front. ∫ x(5x² + 4)³ dx becomes (1/10) ∫ u³ du = u⁴/40 + C = (5x² + 4)⁴ / 40 + C.
When u-substitution works:
You can spot a function and its derivative (or a constant multiple of its derivative) in the integrand.
After substitution, the integral simplifies to a standard form you can integrate directly.
When it does not work:
No part of the integrand is the derivative of another part.
After substitution, x terms remain that you cannot express in terms of u.
Both use the velocity function v(t), but they treat direction differently.
Displacement (net change in position):
∫ v(t) dt = A₁ + A₃ − A₂ (where A₂ is the area below the t-axis).
Positive and negative regions cancel.
Total distance (total path length):
∫ |v(t)| dt = A₁ + A₂ + A₃ (all regions count as positive).
Take the absolute value of v(t) before integrating, which in practice means: find where v(t) changes sign, split the integral at those points, and add the absolute values of each piece.
You can split ∫ [f(x) + g(x)] dx = ∫ f(x) dx + ∫ g(x) dx.
You can split ∫ [f(x) − g(x)] dx = ∫ f(x) dx − ∫ g(x) dx.
You cannot split products or quotients this way: ∫ f(x)·g(x) dx ≠ ∫ f(x) dx · ∫ g(x) dx.
Formula | Expression |
|---|---|
Average value | f_avg = [1 / (b − a)] · ∫ₐᵇ f(x) dx |
MVT for Integrals | ∫ₐᵇ f(x) dx = f(c) · (b − a), for some c in [a, b] |
Displacement | ∫ₐᵇ v(t) dt (signed) |
Total distance | ∫ₐᵇ |
Constant multiple | ∫ k·f(x) dx = k · ∫ f(x) dx |
Average value shows up whenever you need a single representative number for a varying quantity: the average temperature over a day, the average voltage in an AC circuit, or the average concentration of a drug in the bloodstream over a dosing interval. Displacement versus distance is the difference between "how far are you from home?" and "how far did you drive today?", which matters in everything from navigation to fuel-consumption estimates.
Confusing displacement with total distance. Displacement allows cancellation (you walk 3 metres forward and 3 metres back, displacement is zero). Total distance does not (you walked 6 metres). The integral of v(t) gives displacement; the integral of |v(t)| gives distance.
Forgetting to substitute back after u-substitution. Your final answer must be in terms of x (or the original variable), not u.
Leaving stray x terms after substituting. If you cannot rewrite every part of the integrand in terms of u and du, the substitution you chose is wrong.
Thinking you can split integrals of products. ∫ f(x)·g(x) dx is not ∫ f(x) dx times ∫ g(x) dx. This error is surprisingly common under time pressure.
⚠️ U-substitution is the single most-tested integration technique in Calc AB. Expect it in both multiple-choice and free-response.
⚠️ Average value questions are a near-certainty on the AP exam. Know the formula cold: (1/(b − a)) times the integral.
⚠️ Displacement vs. distance is tested in motion problems, often with a velocity graph. You need to identify which quantity the question asks for and handle the signs accordingly.
⚠️ The constant multiple rule is frequently combined with u-substitution. When du does not match exactly, check whether the mismatch is just a constant factor you can pull out.
Fill in the blank: The average value of f on [a, b] equals ______.
True or false: Displacement can be negative.
True or false: Total distance can be negative.
Fill in the blank: In u-substitution, if u = 3x + 7, then du = ______.
True or false: ∫ f(x)·g(x) dx = ∫ f(x) dx · ∫ g(x) dx.
Answers: 1. [1/(b − a)] · ∫ₐᵇ f(x) dx. 2. True. 3. False (it is always non-negative). 4. 3 dx. 5. False.
Q: Find the average value of f(x) = 4x on [0, 2].
A: f_avg = [1/(2 − 0)] · ∫₀² 4x dx = (1/2) · [2x²]₀² = (1/2)(8) = 4.
Q: Evaluate ∫ (2x + 3)⁵ · 2 dx using u-substitution.
A: Let u = 2x + 3, du = 2 dx. ∫ u⁵ du = u⁶/6 + C = (2x + 3)⁶ / 6 + C.
Q: A particle has velocity v(t) = t − 3 on [0, 5]. Find displacement and total distance.
A: Displacement = ∫₀⁵ (t − 3) dt = [t²/2 − 3t]₀⁵ = (25/2 − 15) − 0 = −5/2. v(t) = 0 at t = 3. Total distance = |∫₀³ (t − 3) dt| + |∫₃⁵ (t − 3) dt| = |−9/2| + |2| = 9/2 + 2 = 13/2.
Q: Why does the MVT for Integrals require continuity?
A: A continuous function on a closed interval is guaranteed to take every value between its minimum and maximum (Intermediate Value Theorem). Since the average value lies between these extremes, f must actually equal the average value at some point.
U-substitution is the gateway to more advanced integration techniques (integration by parts, partial fractions) in Calc BC. The net change theorem connects to accumulation problems and differential equations from Part 1 of these notes. Displacement and distance link directly to position-velocity-acceleration relationships tested throughout AP Calculus.
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