Why does the derivative of accumulated change recover the rate?
You will be able to: Apply the Fundamental Theorem to an accumulation function and explain why it works.
Why does the derivative of accumulated change recover the rate?
A tank’s water total grows by roughly the current flow rate times the next short time interval. Dividing that small addition by the short interval recovers the flow rate.
A useful starting point: How do we translate between a sum limit and an integral? →
Words and symbols before equations
- A(x)
- An accumulation function with a variable upper bound.
- Dummy variable t
- The input summed over inside the integral.
- FTC
- Fundamental Theorem of Calculus, connecting integration and differentiation.
- Continuous
- Having no breaks at the inputs where the theorem is applied.
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
Define A(x)=∫₀ˣ (2−t) dt. Unlike an integral with two fixed bounds, A varies as its upper bound moves. At x=0 it is zero because no interval has been accumulated.
A(x+h)−A(x) is the signed accumulation over the short interval from x to x+h. For small h, continuity makes its average height close to 2−x.
Taking the limit of [A(x+h)−A(x)]/h gives A′(x)=2−x. In general, if f is continuous on an interval containing the fixed base a and x, then d/dx[∫ₐˣ f(t) dt]=f(x).
Here geometry also gives A(x)=2x−x²/2. Its derivative is 2−x, confirming the theorem. But the theorem also works when finding a simple antiderivative is difficult.
A worked example, step by step
Let H(x)=∫₁ˣ (t³+2) dt. Find H(1) and H′(2).
- The integrand is a polynomial, hence continuous.
- H(1)=0 because its two bounds coincide.
- FTC gives H′(x)=x³+2.
- H′(2)=10. This is the slope of accumulation at 2, not the accumulated value H(2).
FTC gives the integrand evaluated at the upper bound, not the derivative of the integrand.
If A′(3)=−1, must A(3)=−1?
Compare with an explanation
No. The derivative is the current rate; A(3) includes all earlier signed contributions.
Predict. Change one thing. Explain.
For A(x)=∫₀ˣ(2−t)dt, move x through 2. Match the current rate to the slope of the accumulation graph. Explain the horizontal tangent at x=2.
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)=∫₂ˣ (t²−1)dt. Find A(2), A′(3), and A″(3), stating the theorem’s condition.
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Compare with the answer and four-point rubric
- 1 point: The polynomial integrand is continuous, so FTC applies.
- 1 point: A(2)=0.
- 1 point: A′(x)=x²−1, hence A′(3)=8.
- 1 point: A″(x)=2x, so A″(3)=6; differentiate only after applying FTC.
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 1Why is continuity useful?
It makes the average height over a shrinking interval approach the value at the endpoint.
RECALL 2Must an elementary antiderivative be found first?
No; FTC differentiates the accumulation directly.
RECALL 3What is A(a)?
Zero when A(x)=∫ₐˣ f(t)dt.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Why does the derivative of accumulated change recover the rate?
- A(x)=∫ₐˣ f(t) dt ⇒ A′(x)=f(x), for continuous f.
- A(a)=0.
- Distinguish accumulated height A from its slope f.
Remember: FTC gives the integrand evaluated at the upper bound, not the derivative of the integrand.
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 6.4, 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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