How do bounds and repeated steps work in integration by parts?
You will be able to: Evaluate definite parts integrals and repeat the method when the remainder still needs it.
How do bounds and repeated steps work in integration by parts?
Suppose a changing rate has the numerical form xeˣ. To total that rate from 0 to 1, both the product contribution and the correction must use the same endpoints.
A useful starting point: How does reversing the product rule help integrate a product? →
Words and symbols before equations
- Boundary term [uv]ₐᵇ
- u(b)v(b)−u(a)v(a).
- Remainder integral
- The simpler integral subtracted after a parts step.
- Repeated parts
- Applying the identity again to a remaining product.
- Definite value
- A number obtained by evaluating the full expression at both bounds.
What this picture assumes
Original BC model; rounded readouts. Integrand xeˣ on [0,b]. Product boundary term b eᵇ minus remainder eᵇ−1 equals (b−1)eᵇ+1. All expressions are dimensionless. Shading is the finite integral.
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.
- b=1: boundary 2.71828 minus remainder 1.71828 gives 1. F=(x−1)eˣ differentiates to xeˣ.
- 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 continuously differentiable u and v on [a,b], ∫ₐᵇu dv=[uv]ₐᵇ−∫ₐᵇv du. Do not evaluate just the first term or drop the lower endpoint.
For ∫₀¹xeˣdx, the boundary term is e and the remainder is e−1. Their difference is 1. This matches the model at b=1.
For ∫x²eˣdx, take u=x², dv=eˣdx. Then x²eˣ−2∫xeˣdx still contains a product. A second parts step gives eˣ(x²−2x+2)+C.
Differentiate eˣ(x²−2x+2): the bracket plus its derivative simplifies to x². Some problems need a different technique or an algebraic rearrangement instead of blindly repeating parts.
A worked example, step by step
Evaluate ∫₀¹x²eˣdx.
- A first parts step gives [x²eˣ]₀¹−2∫₀¹xeˣdx.
- The boundary term is e−0=e.
- The remaining integral is 1, so the value is e−2.
- Equivalently evaluate F=eˣ(x²−2x+2): F(1)−F(0)=e−2≈0.718; the positive sign agrees with the integrand.
Keep parentheses around subtracted antiderivatives. Constants cancel in definite evaluation; do not append +C to a numerical definite value.
Is [xeˣ]₀¹ alone the integral of xeˣ?
Compare with an explanation
No. Subtract ∫₀¹eˣdx=e−1 to obtain 1.
Predict. Change one thing. Explain.
At b=0,1 and2, identify both terms in ∫₀ᵇ xeˣdx=b eᵇ−(eᵇ−1). Explain why the lower-bound contribution supplies the +1.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
b=1: boundary 2.71828 minus remainder 1.71828 gives 1. F=(x−1)eˣ differentiates to xeˣ.
Original BC model; rounded readouts. Integrand xeˣ on [0,b]. Product boundary term b eᵇ minus remainder eᵇ−1 equals (b−1)eᵇ+1. All expressions are dimensionless. Shading is the finite integral.
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 questionEvaluate ∫₀¹x sin(x)dx, in radians, and justify each endpoint contribution.
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Compare with the answer and four-point rubric
- 1 point: Choose u=x, dv=sin(x)dx; v=−cos(x).
- 1 point: An antiderivative is −xcos(x)+sin(x).
- 1 point: At 1 it is −cos(1)+sin(1), and at 0 it is 0.
- 1 point: The value sin(1)−cos(1)≈0.301 is positive, consistent with the integrand on (0,1).
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 1What does [uv]ₐᵇ mean?
The upper product value minus the lower product value.
RECALL 2Why repeat parts for x²eˣ?
The first remainder still contains xeˣ.
RECALL 3Does a definite answer need +C?
No; constants cancel when endpoint values are subtracted.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
How do bounds and repeated steps work in integration by parts?
- ∫ₐᵇu dv=[uv]ₐᵇ−∫ₐᵇv du.
- Repeated parts can reduce a polynomial degree at each step.
Remember: Keep parentheses around subtracted antiderivatives. Constants cancel in definite evaluation; do not append +C to a numerical definite value.
Conditions: Original BC model; rounded readouts. Integrand xeˣ on [0,b]. Product boundary term b eᵇ minus remainder eᵇ−1 equals (b−1)eᵇ+1. All expressions are dimensionless. Shading is the finite integral.
Refresh Kid · AP Calculus BC Unit 6 · Objectives FUN-6.E · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 6.11, FUN-6.E. 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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