How can known integrals be combined without new antiderivatives?
You will be able to: Apply additivity, reversal, scaling and linearity of definite integrals.
How can known integrals be combined without new antiderivatives?
If you know how much water entered before lunch and after lunch, you can add the two amounts. The matching boundary splits one accumulation into adjacent pieces.
A useful starting point: How can a rate graph reveal an accumulation function’s shape? →
Words and symbols before equations
- Adjacent intervals
- Intervals sharing a boundary with no gap.
- Linearity
- Integrals respect addition and constant multiplication.
- Reversed bounds
- Traversing the interval in the opposite direction.
- Equal bounds
- An interval of zero width.
What this picture assumes
Original model; readouts are rounded. f=2−x on [0,4]. Adjacent signed pieces add; reversal negates. The full integral is zero, so reversal leaves its value zero but reverses each nonzero piece.
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=1. Forward pieces: integral 0→c is 1.5, integral c→4 is -1.5. Chosen orientation: 0→c→4, contributions 1.5 and -1.5, total 0.
- 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 integrable f, the integral from a to c plus the integral from c to b equals the integral from a to b. This is signed interval bookkeeping, valid even when c lies outside [a,b] if the relevant integrals exist.
Reversal gives: the integral from b to a is the negative of the integral from a to b, and equal bounds give ∫ₐᵃ f=0. Reversing orientation changes sign, not geometric area size.
For constants k,m, ∫ₐᵇ[kf+mg]=k∫ₐᵇ f+m∫ₐᵇ g. But integrals do not generally distribute over products or quotients.
A constant c has integral c(b−a). Distinguish this from multiplying two already accumulated totals.
A worked example, step by step
Given ∫₀² f=5, ∫₂⁵ f=−1 and ∫₀⁵ g=4, find ∫₅⁰(2f−3g).
- Combine adjacent intervals: ∫₀⁵ f=5−1=4.
- Use linearity on [0,5]: ∫₀⁵(2f−3g)=2×4−3×4=−4.
- Reverse the bounds to multiply by −1.
- The requested integral is 4; no formula for f or g was needed.
In general ∫fg is not (∫f)(∫g). Linearity applies to sums and constant factors.
If ∫₀² f=3, is ∫₂⁰ f also 3?
Compare with an explanation
No. It equals −3 because the interval orientation is reversed.
Predict. Change one thing. Explain.
Move the split point c in ∫₀⁴(2−x)dx. Watch the two oriented contributions change while their sum stays zero. Reverse the overall orientation and explain what changes.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
c=1. Forward pieces: integral 0→c is 1.5, integral c→4 is -1.5. Chosen orientation: 0→c→4, contributions 1.5 and -1.5, total 0.
Original model; readouts are rounded. f=2−x on [0,4]. Adjacent signed pieces add; reversal negates. The full integral is zero, so reversal leaves its value zero but reverses each nonzero piece.
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 ∫₁⁴ f=7 and ∫₁² f=3, find ∫₂⁴ f, ∫₄² f and ∫₂⁴(2f−1).
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Compare with the answer and four-point rubric
- 1 point: Subtract adjacent pieces: ∫₂⁴ f=7−3=4.
- 1 point: Reverse: ∫₄² f=−4.
- 1 point: Scale and separate the constant: ∫₂⁴(2f−1)=2×4−(4−2).
- 1 point: The last result is 6; the constant integrates over width 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 1What happens under reversal?
The sign changes.
RECALL 2How do adjacent intervals combine?
Their signed integrals add to the whole.
RECALL 3Can you distribute integration over multiplication?
Not in general.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
How can known integrals be combined without new antiderivatives?
- Adjacent intervals add.
- Reversing bounds negates an integral.
- Constants factor out; sums integrate term by term.
Remember: In general ∫fg is not (∫f)(∫g). Linearity applies to sums and constant factors.
Conditions: Original model; readouts are rounded. f=2−x on [0,4]. Adjacent signed pieces add; reversal negates. The full integral is zero, so reversal leaves its value zero but reverses each nonzero piece.
Refresh Kid · AP Calculus BC Unit 6 · Objectives FUN-6.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 6.6, FUN-6.A. 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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