How does one known value choose an antiderivative?
You will be able to: Determine a constant from a known function value and distinguish initial value from initial rate.
How does one known value choose an antiderivative?
Knowing every moment’s velocity tells how position changes, but not where a cart began. One measured position chooses which member of the antiderivative family fits the cart.
A useful starting point: Which familiar derivative pairs become basic integrals? →
Words and symbols before equations
- Initial condition
- A known value F(a)=b.
- Particular antiderivative
- One member selected from the family F+C.
- Initial rate
- The derivative value at the start, not the position.
- Consistency check
- Verifying both the derivative and the given value.
What this picture assumes
Original model; readouts are rounded. F=x²+C and F′=2x on all real x. Display window −2≤x≤2. A known value F(1)=5 selects C=4; the family otherwise has arbitrary C.
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.
- C=0; F(1)=1; F′(x)=2x for every C. The condition F(1)=5 requires C=4.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the mathematics
Suppose F′=2x. The family is F=x²+C. If F(1)=5, substitute both the input and output: 5=1+C, so C=4.
The value F′(1)=2 alone would not choose C because every member has that slope. Function values and derivative values provide different information.
For a quantity Q with rate r and Q(a)=Q₀, the same relationship can be written Q(x)=Q₀+∫ₐˣ r(t)dt. The integral vanishes at a, so the initial condition is automatically met.
After finding C, differentiate the answer and substitute the known input. Both checks matter; a correct derivative can coexist with a wrong starting amount.
A worked example, step by step
A cart has velocity v(t)=3t² m/s and position s(1)=7 m. Find its position formula.
- Position is an antiderivative of velocity, so s(t)=t³+C meters.
- Use s(1)=7: 1+C=7.
- Thus C=6 and s(t)=t³+6.
- Check s′=3t² and s(1)=7; at t=2, position is 14 m and the change since t=1 is 7 m.
Do not insert the initial rate as the initial amount. C is determined from a function value, not a derivative value already built into the family.
Does knowing F′(0)=0 determine C in F=x²+C?
Compare with an explanation
No. Every constant choice gives derivative zero at 0.
Predict. Change one thing. Explain.
For F=x²+C, move C until F(1)=5. Explain why only one vertical shift fits this point while every shift still has derivative 2x.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
C=0; F(1)=1; F′(x)=2x for every C. The condition F(1)=5 requires C=4.
Original model; readouts are rounded. F=x²+C and F′=2x on all real x. Display window −2≤x≤2. A known value F(1)=5 selects C=4; the family otherwise has arbitrary C.
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 questionGiven F′(x)=6x−2 and F(2)=9, find F and then F(0). Verify the condition.
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Compare with the answer and four-point rubric
- 1 point: Integrate: F=3x²−2x+C.
- 1 point: At 2, 9=12−4+C, so C=1.
- 1 point: F=3x²−2x+1 and F(0)=1.
- 1 point: Its derivative is 6x−2 and F(2)=9.
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 1What chooses C?
A known function value.
RECALL 2Can derivative data alone choose C?
No; constant shifts have identical derivatives.
RECALL 3Which two checks are needed?
Recover the derivative and satisfy the given value.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
How does one known value choose an antiderivative?
- Integrate, substitute F(a)=b, solve C, verify.
- Q(x)=Q(a)+∫ₐˣ Q′(t)dt.
Remember: Do not insert the initial rate as the initial amount. C is determined from a function value, not a derivative value already built into the family.
Conditions: Original model; readouts are rounded. F=x²+C and F′=2x on all real x. Display window −2≤x≤2. A known value F(1)=5 selects C=4; the family otherwise has arbitrary C.
Refresh Kid · AP Calculus AB Unit 6 · Objectives FUN-6.C · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 6.8, FUN-6.C. CED effective Fall 2020 and Fall 2026 clarifications checked September 17, 2026. AB scope includes 6.1–6.10 and 6.14. Topics 6.11–6.13 are BC-only; the numbering gap is intentional. Topic 6.14 consolidates the AB antidifferentiation objectives 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. No improper-integral evaluation, integration by parts or partial-fraction decomposition is taught in this AB unit.
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 AB 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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