How does substitution undo the chain rule?
You will be able to: Choose an inner function, transform its differential and verify an indefinite integral.
How does substitution undo the chain rule?
If a machine first squares an input and then exponentiates it, differentiating includes both stages. To undo that derivative, the matching inner-rate factor must be present.
A useful starting point: How does one known value choose an antiderivative? →
Words and symbols before equations
- Substitution u=g(x)
- A new name for an inner expression.
- du=g′(x)dx
- The differential replacement carrying the inner derivative.
- Composite function
- One function applied inside another.
- Back-substitution
- Returning an indefinite antiderivative to the original variable.
What this picture assumes
Original model; readouts are rounded. u=x²+1, du=2x dx. f=2x(x²+1)²; F=(x²+1)³/3. Lower x bound 0 becomes u=1; changing bounds gives ∫₁ᵘ v²dv. All formulas valid on this displayed range.
Read the picture in three steps
- Read the axes and labels first. Identify what each symbol and line represents. Read the units and fixed conditions before comparing quantities.
- x=1 → u=2; lower x=0 → u=1. F′=8; definite integral=((u³−1)/3)=2.33333; F(x)=2.66667. Initial F(0)=1/3 is subtracted.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the mathematics
To integrate 2x(x²+1)³, let u=x²+1. Then du=2x dx, so the whole integral becomes ∫u³du. The factor 2x is exactly what matches the inner derivative.
Integrate in u: u⁴/4+C. Replace u to get (x²+1)⁴/4+C. Differentiating gives 4(x²+1)³(2x)/4, recovering the original integrand.
If the integrand is x(x²+1)³ instead, x dx=du/2. A constant adjustment is allowed; inventing a variable factor is not.
Finish the variable change consistently. An expression containing leftover x factors is not yet a basic u-integral unless those factors are also rewritten in u.
A worked example, step by step
Find ∫3x√(x²+4)dx.
- Choose u=x²+4, so du=2x dx and 3x dx=(3/2)du.
- Rewrite as (3/2)∫u^(1/2)du.
- The power rule gives (3/2)(2/3)u^(3/2)+C=u^(3/2)+C.
- Back-substitute: (x²+4)^(3/2)+C. Its derivative is 3x√(x²+4).
Choosing u does not remove dx automatically. Replace the matching differential factor, including any constant multiplier.
May you multiply an integrand by x to create a missing derivative without compensating?
Compare with an explanation
No. That changes the integral. Only valid algebraic replacements preserve the original expression.
Predict. Change one thing. Explain.
Compare f(x)=2x(x²+1)² with F(x)=(x²+1)³/3. Move x across zero and explain how the inner derivative 2x controls the sign of F′.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
x=1 → u=2; lower x=0 → u=1. F′=8; definite integral=((u³−1)/3)=2.33333; F(x)=2.66667. Initial F(0)=1/3 is subtracted.
Original model; readouts are rounded. u=x²+1, du=2x dx. f=2x(x²+1)²; F=(x²+1)³/3. Lower x bound 0 becomes u=1; changing bounds gives ∫₁ᵘ v²dv. All formulas valid on this displayed range.
Explain what you noticed: Answer the investigation prompt above. State one observation and explain it using the rate units, signed contributions, theorem conditions, domain or antiderivative check. Identify what the representation cannot tell you.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionIntegrate x cos(x²+1) dx and verify the coefficient.
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Compare with the answer and four-point rubric
- 1 point: Set u=x²+1 and du=2x dx.
- 1 point: Then x dx=du/2, giving (1/2)∫cos u du.
- 1 point: The result is (1/2)sin(x²+1)+C.
- 1 point: Differentiating gives (1/2)cos(x²+1)×2x=x cos(x²+1).
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1Which derivative is needed?
The derivative of the chosen inner expression.
RECALL 2Where do constant corrections go?
Outside the transformed integral.
RECALL 3How do you validate a substitution answer?
Differentiate it and recover the original integrand.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
How does substitution undo the chain rule?
- u=g(x), du=g′(x)dx.
- Transform every factor, integrate in u, then return to x for an indefinite integral.
- Verify with the chain rule.
Remember: Choosing u does not remove dx automatically. Replace the matching differential factor, including any constant multiplier.
Conditions: Original model; readouts are rounded. u=x²+1, du=2x dx. f=2x(x²+1)²; F=(x²+1)³/3. Lower x bound 0 becomes u=1; changing bounds gives ∫₁ᵘ v²dv. All formulas valid on this displayed range.
Refresh Kid · AP Calculus BC Unit 6 · Objectives FUN-6.D · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 6.9, FUN-6.D. CED effective Fall 2020 and Fall 2026 clarifications checked September 17, 2026. BC scope includes all fourteen topics, including integration by parts (6.11), nonrepeating linear partial fractions (6.12) and improper integrals (6.13). Topic 6.14 integrates the BC toolkit and Skill 1.C. Focused lesson titles, examples and questions are original Refresh Kid material.
Distinguish signed accumulation, unsigned area and initial amount. Error direction needs interval-wide shape information. FTC hypotheses, variable-bound chain factors, integration constants, transformed bounds and domain restrictions are explicit. Bounded jumps are treated separately from differentiability of accumulation. The BC extension teaches parts from the product rule, decomposition with distinct linear factors and independent improper limits. Finite truncations do not certify convergence. Shared foundation lessons are maintained with AB; BC extensions and method selection are authored for this course.
The Organic Chemistry Tutor Fundamental Theorem of Calculus Part 1 video title, creator and description were checked; the full video was not reviewed. Khan Academy’s unit destination was checked, but its JavaScript lesson contents were not fully readable by the research tool. OpenStax Sections 5.2, 5.3 and 5.5 were consulted for conceptual cross-checking. No provider scripts, questions, diagrams or artwork were copied. Refresh Kid is not affiliated with or endorsed by these providers.
GitHub’s 3D website collection informed optional spatial inspection and camera controls. The original tank geometry uses existing self-hosted Three.js with its MIT license. Its 2×2 dm base and water depth use the same volume as the rate calculation: 1 dm³=1 L. The initial water is blue and newly accumulated inflow is teal. The rate graph is an abstract amount calculation; it is not the physical shape of water. Camera rotation changes only the view. Full 2D graphs, readouts and explanations remain available without WebGL.
Independent teacher review and observation of students remain pending. Technical checks do not certify mathematical accuracy, accessibility or learning effectiveness. This is a review edition.
Optional official resource: Released AP Calculus BC questions and scoring guides. The archive spans multiple units; no entire exam question is assigned here.
Learn → Explore → Practice → Review is informed by the IES learning guide; this exact implementation has not been evaluated with learners.
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