Can geometric area still work when a graph has a jump?
You will be able to: Evaluate signed integrals geometrically and distinguish integrability from FTC differentiability.
Can geometric area still work when a graph has a jump?
A faucet switches abruptly from 2 L/min to 4 L/min. The switch creates a jump in its rate, yet the total water added still makes sense as the sum of two rectangles.
A useful starting point: How can known integrals be combined without new antiderivatives? →
Words and symbols before equations
- Integrable
- Having a well-defined finite definite integral on the interval.
- Jump discontinuity
- A mismatch between finite one-sided limits.
- Isolated point
- One input with no horizontal width by itself.
- Piecewise function
- A rule specified separately on parts of its domain.
What this picture assumes
Original model; readouts are rounded. f=2 for 0≤x<1; f=4 for 1<x≤3; f(1) is adjustable. Integral always 10. Accumulation is continuous with left slope 2 and right slope 4 at 1.
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.
- Assigned f(1)=5; integral=2×1+4×2=10. Left accumulation slope at 1 is 2; right slope is 4. A′(1) does not exist.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the mathematics
A bounded function with finitely many jumps or removable discontinuities on a finite closed interval is integrable. Compute each continuous piece and add the results.
If f=2 on [0,1) and f=4 on (1,3], then ∫₀³ f=2×1+4×2=10. Assigning any finite value to f(1) changes no area because a single input has zero width.
However A(x)=∫₀ˣ f(t)dt has left slope 2 and right slope 4 at x=1. A is continuous there but not differentiable. Integrability alone does not justify A′(1)=f(1).
Use familiar shapes when possible: rectangles, triangles and semicircles. A semicircle below the axis contributes minus half the area of its full circle.
A worked example, step by step
A graph on [0,2] is an upper semicircle of radius 1 centered at (1,0); on [2,4] it is the constant −1. Evaluate its integral.
- The semicircle lies above the axis and contributes π(1)²/2=π/2.
- The rectangle lies below the axis, with width 2 and signed height −1.
- Add the signed contributions: π/2−2.
- Changing values at the joining point alone leaves this integral unchanged; bounded jumps do not prevent this piecewise calculation.
Integrability, continuity and differentiability are different properties. A well-defined integral does not guarantee an accumulation derivative at a jump.
Does the point f(1)=9 add nine square units to the integral?
Compare with an explanation
No. One point has zero width, so its finite height contributes no area.
Predict. Change one thing. Explain.
Change the displayed value at the single jump input x=1. Explain why the point marker moves while the integral and the two one-sided accumulation slopes stay fixed.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Assigned f(1)=5; integral=2×1+4×2=10. Left accumulation slope at 1 is 2; right slope is 4. A′(1) does not exist.
Original model; readouts are rounded. f=2 for 0≤x<1; f=4 for 1<x≤3; f(1) is adjustable. Integral always 10. Accumulation is continuous with left slope 2 and right slope 4 at 1.
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 questionLet f=3 on [0,2), f=−1 on (2,5], and f(2)=8. Find ∫₀⁵ f and explain whether A(x)=∫₀ˣ f is differentiable at 2.
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Compare with the answer and four-point rubric
- 1 point: First rectangle contributes 3×2=6.
- 1 point: Second rectangle contributes −1×3=−3, so the integral is 3.
- 1 point: The finite isolated value 8 changes no area.
- 1 point: A has left derivative 3 and right derivative −1 at 2, so it is not differentiable there.
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 1Can a bounded jump function be integrable?
Yes; finitely many jumps on a finite interval cause no problem.
RECALL 2Does one finite point change the integral?
No.
RECALL 3What happens to accumulation slope at a rate jump?
The one-sided slopes can disagree, preventing differentiability.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Can geometric area still work when a graph has a jump?
- Signed geometric contributions add.
- Finite changes at isolated points do not change a bounded integrable function’s integral.
- At a jump, inspect one-sided accumulation slopes.
Remember: Integrability, continuity and differentiability are different properties. A well-defined integral does not guarantee an accumulation derivative at a jump.
Conditions: Original model; readouts are rounded. f=2 for 0≤x<1; f=4 for 1<x≤3; f(1) is adjustable. Integral always 10. Accumulation is continuous with left slope 2 and right slope 4 at 1.
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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