How can termwise integration create a new function?
You will be able to: Integrate a power series, determine its constant and examine endpoints.
How can termwise integration create a new function?
Accumulating a rate from zero can produce a function that is harder to expand directly. Integrating a familiar series gives both the accumulated function and a new coefficient pattern.
A useful starting point: What changes when a power series is differentiated? →
Words and symbols before equations
- Termwise integration
- Integrate each power-series term inside its radius.
- Constant of integration C
- The value determined by an initial condition or a definite lower limit.
- Dummy variable t
- The integration variable, distinct from the endpoint x.
- Endpoint convergence
- Convergence checked after integration changes the coefficients.
What this picture assumes
Controls stay strictly inside radius 1, where termwise rules hold. Substitution and derivative examples exclude both endpoints. The integrated arctangent series converges at both endpoints; equality there needs a limiting argument.
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.
- Substitute: 1/(1+x²). At x=0.5, 4 nonzero terms give 0.796875. Function value: 0.8. The termwise rule holds inside abs(x)<1; endpoint tests remain separate.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the mathematics
Integrate 1/(1+t²)=Σ(−1)ⁿt²ⁿ from 0 to x, initially for abs(x)<1. The result is arctan x=Σ(−1)ⁿx²ⁿ⁺¹/(2n+1). Both sides are zero at x=0, fixing the constant.
The radius remains 1. At x=1 and −1, the new series converge by the alternating test even though the original geometric series diverge there. The endpoint values agree with the continuous arctangent limit; endpoint equality needs this additional limiting justification.
For an indefinite integral write C explicitly. For a definite integral from the center, substituting the lower limit often determines C automatically. Never extend an interior termwise rule to endpoints without checking.
A worked example, step by step
Integrate the geometric series from 0 to x to represent −ln(1−x).
- For abs(t)<1, 1/(1−t)=Σₙ₌₀∞tⁿ.
- Integrating from 0 to x gives −ln(1−x)=Σₙ₌₀∞xⁿ⁺¹/(n+1).
- At x=0 both sides are 0, fixing the integration constant.
- At x=−1 the series is alternating harmonic and converges; at x=1 it is harmonic and diverges. Its interval is [−1,1).
Integration preserves radius, but it can gain endpoints. Include the constant or show why it is zero.
Why is there no arbitrary constant in the integral from 0 to x?
Compare with an explanation
The definite lower limit fixes the value at x=0 as zero.
Predict. Change one thing. Explain.
Select the integration case for arctan x. Inspect its powers and denominators, then reason separately about x=±1 using the alternating test. The plotted interior alone does not establish endpoint equality.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Substitute: 1/(1+x²). At x=0.5, 4 nonzero terms give 0.796875. Function value: 0.8. The termwise rule holds inside abs(x)<1; endpoint tests remain separate.
Controls stay strictly inside radius 1, where termwise rules hold. Substitution and derivative examples exclude both endpoints. The integrated arctangent series converges at both endpoints; equality there needs a limiting argument.
Explain what you noticed: Answer the investigation prompt above. State one observation and explain it using the term formula, partial sums, test conditions, remainder bound or interval of convergence. 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 questionStarting from 1/(1+t²), obtain the first three nonzero terms of arctan x, determine the integration constant and test both endpoints.
This response is not submitted or saved. Copy it before leaving.
Compare with the answer and four-point rubric
- 1 point: Integrate 1−t²+t⁴−… term by term from 0 to x.
- 1 point: arctan x=x−x³/3+x⁵/5−… and the value at zero fixes C=0.
- 1 point: The radius is 1; at x=1 the alternating odd-reciprocal series converges.
- 1 point: At x=−1 it is the negative of that convergent series, so the interval is [−1,1].
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 1How is an integration constant fixed?
By an initial value or the definite integral’s lower limit.
RECALL 2Can integration gain convergent endpoints?
Yes; arctangent is an example.
RECALL 3Does endpoint convergence alone justify every endpoint function equality?
No; a separate limiting or continuity argument is needed.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
How can termwise integration create a new function?
- ∫Σcₙ(x−a)ⁿdx=C+Σcₙ(x−a)ⁿ⁺¹/(n+1), inside the radius.
- Integrated series has the same radius; endpoints can change.
- Use an initial value or definite integral to fix C.
Remember: Integration preserves radius, but it can gain endpoints. Include the constant or show why it is zero.
Conditions: Controls stay strictly inside radius 1, where termwise rules hold. Substitution and derivative examples exclude both endpoints. The integrated arctangent series converges at both endpoints; equality there needs a limiting argument.
Refresh Kid · AP Calculus BC Unit 10 · Objectives LIM-8.G · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 10.15, LIM-8.G. CED effective Fall 2020 and Fall 2026 clarifications checked September 17, 2026. This unit covers BC topics 10.1–10.15, Infinite Sequences and Series. Required convergence methods are the nth-term, integral, comparison, alternating and ratio tests; the root test is not assigned as required AP content.
Finite plots illustrate terms and partial sums but do not establish infinite convergence. Positive-series comparisons retain their hypotheses and inequality directions. Alternating bounds use the next omitted magnitude. Taylor bounds use the next derivative over the full interval. Power-series endpoints are checked separately, including after differentiation and integration. Taylor representation requires a remainder tending to zero.
All focused explanations, examples, practice and models are original Refresh Kid work. OpenStax was consulted for mathematical cross-checking. Khan Academy’s destination was checked, but JavaScript lesson content was not fully readable by the research tool. Organic Chemistry Tutor video titles, creator and destinations were checked; full videos were not reviewed. No provider questions, diagrams or scripts were copied. Resources are optional, and no paid resource is needed. Refresh Kid is not affiliated with or endorsed by these providers.
GitHub’s 3D website collection informed optional camera and spatial inspection. The original geometric slab model uses self-hosted Three.js with its MIT license. A unit cube receives slabs of widths 1/2, 1/4, 1/8 and so on, each spanning unit height and depth. The complete 2D strip and numerical volume readout remain available. No autoplay or WebGL is required to learn the mathematics.
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 implementation has not been evaluated with learners.
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