Why do antiderivative endpoint values give the integral?
You will be able to: Use FTC to evaluate a definite integral and check the result’s meaning.
Why do antiderivative endpoint values give the integral?
A trip’s odometer reading at the finish minus its reading at the start gives the distance added. An antiderivative plays a similar bookkeeping role for signed accumulation.
A useful starting point: Can geometric area still work when a graph has a jump? →
Words and symbols before equations
- Antiderivative F
- A function whose derivative F′ equals the integrand f.
- Endpoint evaluation
- Subtracting F(a) from F(b).
- [F(x)]ₐᵇ
- Notation for F(b)−F(a).
- Constant C
- A vertical shift that disappears in the subtraction.
What this picture assumes
Original model; readouts are rounded. f=2x+1; lower bound −1; F=x²+x. ∫₋₁ᵇ f=F(b)−F(−1)=b²+b. Negative contributions are orange.
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.
- b=2; F(b)=6; F(−1)=0. Definite integral=6. Positive and negative regions contribute with signs.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the mathematics
If f is continuous on [a,b] and F′=f, the Fundamental Theorem gives ∫ₐᵇ f(x)dx=F(b)−F(a). The accumulated-area function and F have the same derivative, so differ only by a constant.
For f(x)=3x², use F=x³ because differentiating x³ recovers 3x². Then ∫₁² 3x²dx=8−1=7.
Any antiderivative x³+C gives the same difference: (8+C)−(1+C)=7. A definite integral needs no final +C.
Evaluate both endpoint values in parentheses, especially for a negative lower endpoint. Finally check sign, approximate size and contextual units.
A worked example, step by step
Evaluate ∫₋₁² (2x+1)dx.
- The integrand is continuous, so FTC applies.
- An antiderivative is F=x²+x; differentiating confirms F′=2x+1.
- F(2)=6 and F(−1)=0.
- The signed integral is 6−0=6. This is net area, not necessarily total geometric area because the integrand changes sign.
Subtract antiderivative values, not integrand values. Write parentheses around the full lower-bound expression.
Could F=x³+100 be used instead of x³ to integrate 3x²?
Compare with an explanation
Yes. The 100 cancels between endpoint evaluations.
Predict. Change one thing. Explain.
For f=2x+1, move the upper bound from −1 to 3 while keeping the lower bound −1. Compare the signed shaded area with F(b)−F(−1). Explain why the value first decreases.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
b=2; F(b)=6; F(−1)=0. Definite integral=6. Positive and negative regions contribute with signs.
Original model; readouts are rounded. f=2x+1; lower bound −1; F=x²+x. ∫₋₁ᵇ f=F(b)−F(−1)=b²+b. Negative contributions are orange.
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 ∫₁³ (x²−2x)dx, verify the antiderivative, and explain why a positive final result does not mean the integrand is always positive.
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Compare with the answer and four-point rubric
- 1 point: Choose F=x³/3−x².
- 1 point: Differentiate to get x²−2x.
- 1 point: F(3)=0 and F(1)=−2/3, so the integral is 2/3.
- 1 point: The integrand is negative on (1,2) and positive on (2,3); the positive contribution outweighs the negative one.
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 is the endpoint order?
Upper antiderivative value minus lower.
RECALL 2Why can we choose any C?
It cancels in the difference.
RECALL 3How can an antiderivative be checked?
Differentiate it and recover the integrand.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Why do antiderivative endpoint values give the integral?
- If F′=f and f is continuous, ∫ₐᵇ f=F(b)−F(a).
- Differentiate F to check it.
- The constant cancels.
Remember: Subtract antiderivative values, not integrand values. Write parentheses around the full lower-bound expression.
Conditions: Original model; readouts are rounded. f=2x+1; lower bound −1; F=x²+x. ∫₋₁ᵇ f=F(b)−F(−1)=b²+b. Negative contributions are orange.
Refresh Kid · AP Calculus BC Unit 6 · Objectives FUN-6.B · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 6.7, FUN-6.B. 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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