When changing variables, what happens to the endpoints?
You will be able to: Evaluate definite integrals using transformed bounds or back-substituted antiderivatives consistently.
When changing variables, what happens to the endpoints?
A trip endpoint labeled in miles must be converted if the calculation switches to kilometers. Integral endpoints likewise belong to the variable in use.
A useful starting point: How does substitution undo the chain rule? →
Words and symbols before equations
- x-bound
- An endpoint in the original variable.
- u-bound
- Its transformed value u=g(x).
- Orientation
- The order from lower to upper bound, retained after substitution.
- Equivalent method
- Back-substitute first and then use original bounds.
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
For ∫₀¹ 2x(x²+1)²dx, choose u=x²+1 and du=2x dx. The original endpoints x=0,1 become u=1,2.
The transformed integral is ∫₁² u²du=[u³/3]₁²=7/3. Do not use 0 and 1 as bounds for u³/3.
Alternatively find (x²+1)³/3 first and evaluate at x=1 and x=0. This produces the same answer. The variable of the expression and the endpoint labels must agree.
Substitution can reverse bound order: u=1−x changes x=0,1 to u=1,0 and dx=−du. Track both effects consistently instead of silently sorting the bounds.
A worked example, step by step
Evaluate ∫₀¹ 2x/(x²+1)dx by changing variables.
- Set u=x²+1, du=2x dx.
- Convert x=0 to u=1 and x=1 to u=2.
- The new integral is ∫₁² (1/u)du.
- Evaluate ln2−ln1=ln2; the result is positive, consistent with the nonnegative integrand.
Never combine a u-antiderivative with unchanged x-endpoints. Either transform the bounds or return the formula to x before evaluation.
Must a definite integral’s final answer be written in x?
Compare with an explanation
No. After correct endpoint evaluation it is a number; either consistent method gives it.
Predict. Change one thing. Explain.
On the model set the upper x endpoint to 1. Identify the new u endpoint 2 and compare F(1)−F(0) with ∫₁² u²du. Explain why the lower u endpoint is 1.
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 questionEvaluate ∫₀² (x+1)/(x²+2x+3)dx with substitution; explain the bounds and sign.
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Compare with the answer and four-point rubric
- 1 point: Let u=x²+2x+3, so du=2(x+1)dx.
- 1 point: The bounds become u=3 and u=11.
- 1 point: The integral becomes (1/2)∫₃¹¹ du/u=(1/2)ln(11/3).
- 1 point: It is positive, agreeing with the positive original numerator and denominator on [0,2].
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 1How do you transform each bound?
Insert its x value into u=g(x).
RECALL 2What is the alternative to changing bounds?
Return the antiderivative to x before using the original endpoints.
RECALL 3Should reversed transformed bounds be silently reordered?
No; a reversal requires the corresponding minus sign.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
When changing variables, what happens to the endpoints?
- The integral from a to b of f(g(x))g′(x)dx equals the integral from g(a) to g(b) of f(u)du under the stated continuous/differentiable conditions.
- Keep endpoint order; handle minus signs explicitly.
Remember: Never combine a u-antiderivative with unchanged x-endpoints. Either transform the bounds or return the formula to x before evaluation.
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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