How can a rate graph reveal an accumulation function’s shape?
You will be able to: Infer increase, extrema and concavity of accumulation from an integrand graph.
How can a rate graph reveal an accumulation function’s shape?
A tank can keep filling while its tap slows down. Its water amount is increasing, but its graph becomes less steep. The sign of a rate and the direction the rate changes tell different stories.
A useful starting point: What changes when an integral’s boundary is x squared? →
Words and symbols before equations
- A′=f
- The integrand is the slope of the accumulation function.
- A″=f′
- Where f is differentiable, its slope controls accumulation concavity.
- Local maximum
- An accumulated value larger than nearby values.
- Base value
- A known starting accumulation used to recover heights.
What this picture assumes
Original model; readouts are rounded. f(t)=2−t; A(x)=∫₀ˣ f(t)dt. Positive regions are teal, negative regions orange. The lower graph is accumulation, not the rate. Abstract coordinates are dimensionless.
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=3; current rate -1; signed accumulation 1.5; unsigned area 2.5. A′=f, A″=−1.
- 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 A(x)=∫₀ˣ(2−t)dt, f=2−x is positive before 2 and negative after 2. Therefore A increases on (0,2) and decreases on (2,4), giving a local maximum at 2.
The integrand slopes downward throughout: f′=−1. Thus A″=−1 and A is concave down throughout (0,4), including the portion where A is still increasing.
A zero of f is a stationary candidate for A, not automatically an extremum. For f=x², the accumulated function has a horizontal tangent at 0 but continues increasing through it.
To recover a height, add signed areas from a known base value. Derivative signs alone tell shape, not the vertical placement of an arbitrary antiderivative.
A worked example, step by step
For A(x)=∫₀ˣ(t−1)dt on [0,3], determine its local extremum and concavity.
- A′(x)=x−1 is negative on (0,1) and positive on (1,3).
- The sign change from negative to positive gives a local minimum at x=1.
- A(1) is the negative triangle area −1/2.
- A″=1>0, so A is concave up throughout; both signs and the accumulated value have distinct roles.
A graph labeled f is a graph of A′ in this setting. Its maxima are not automatically maxima of A.
If f has a local maximum, must A also have a local maximum there?
Compare with an explanation
No. A needs f to change from positive to negative; an extremum of f concerns possible concavity changes in A.
Predict. Change one thing. Explain.
Use the rate graph 2−x and the accumulation graph together. Identify the interval where A is positive, increasing and concave down at once. Explain each property separately.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
x=3; current rate -1; signed accumulation 1.5; unsigned area 2.5. A′=f, A″=−1.
Original model; readouts are rounded. f(t)=2−t; A(x)=∫₀ˣ f(t)dt. Positive regions are teal, negative regions orange. The lower graph is accumulation, not the rate. Abstract coordinates are dimensionless.
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 A(x)=∫₀ˣ (t²−1)dt on [−2,2]. Identify increase/decrease intervals, local extrema and concavity.
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Compare with the answer and four-point rubric
- 1 point: A′=x²−1 gives increase on (−2,−1),(1,2) and decrease on (−1,1).
- 1 point: At −1 the + to − change gives a local maximum; at 1 the − to + change gives a local minimum.
- 1 point: A=x³/3−x, so the extrema values are 2/3 and −2/3.
- 1 point: A″=2x gives concave down on (−2,0), concave up on (0,2), and an inflection at 0.
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 1Which graph gives the slope of A?
The integrand f.
RECALL 2Which integrand feature suggests an extremum of A?
A zero with a sign change.
RECALL 3How do you recover A values?
Start from a known value and add signed areas.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
How can a rate graph reveal an accumulation function’s shape?
- f>0 ⇒ A increasing; f<0 ⇒ A decreasing.
- A extrema need sign changes in f.
- f increasing ⇒ A concave up; f decreasing ⇒ A concave down.
Remember: A graph labeled f is a graph of A′ in this setting. Its maxima are not automatically maxima of A.
Conditions: Original model; readouts are rounded. f(t)=2−t; A(x)=∫₀ˣ f(t)dt. Positive regions are teal, negative regions orange. The lower graph is accumulation, not the rate. Abstract coordinates are dimensionless.
Refresh Kid · AP Calculus BC Unit 6 · Objectives FUN-5.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 6.5, FUN-5.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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