How does reversing the product rule help integrate a product?
You will be able to: Derive integration by parts and choose factors that simplify the remaining integral.
How does reversing the product rule help integrate a product?
A rectangle with sides u and v can grow in two ways: one side changes, or the other does. The product rule accounts for both contributions. Reversing that rule lets us trade one difficult integral for a simpler one.
A useful starting point: How can completing the square uncover an inverse-tangent integral? →
Words and symbols before equations
- u and v
- Differentiable functions of the same input x.
- du
- The expression u′(x)dx.
- dv
- The expression v′(x)dx; integrating it gives v.
- Integration by parts
- The identity ∫u dv=uv−∫v du, with +C for an indefinite result.
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
The product rule says (uv)′=u′v+uv′. Integrate and rearrange: ∫uv′dx=uv−∫vu′dx. That is why the remaining integral is subtracted.
For ∫xeˣdx, choose u=x so du=dx, and dv=eˣdx so v=eˣ. The remaining integral ∫eˣdx is simpler than the original product.
The result is xeˣ−eˣ+C=(x−1)eˣ+C. Differentiating gives eˣ+(x−1)eˣ=xeˣ. An integral of a product is not generally the product of two integrals.
A useful choice makes u simpler when differentiated and dv manageable when integrated. This is a strategy, not a universal rule. For ∫ln(x)dx on x>0, choose u=ln(x), dv=dx; the hidden second factor is 1.
A worked example, step by step
Find ∫ln(x)dx on x>0.
- Write the integrand as ln(x)×1 and choose u=ln(x), dv=dx.
- Then du=dx/x and v=x.
- Apply the identity: xln(x)−∫x(1/x)dx=xln(x)−x+C.
- Differentiate: ln(x)+1−1=ln(x); the stated domain x>0 is preserved.
Do not multiply separate antiderivatives. Keep the minus sign and verify the whole answer using the product rule.
Why choose u=x rather than u=eˣ here?
Compare with an explanation
Differentiating x removes that factor, leaving an elementary exponential integral.
Predict. Change one thing. Explain.
Move the upper endpoint b. Compare the product boundary term with the subtracted remainder for ∫₀ᵇ xeˣdx. Explain why the product term alone overcounts.
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 questionFind ∫x cos(x)dx, using radians, and verify your result.
This response is not submitted or saved. Copy it before leaving.
Compare with the answer and four-point rubric
- 1 point: Choose u=x and dv=cos(x)dx.
- 1 point: Then du=dx and v=sin(x).
- 1 point: Obtain xsin(x)−∫sin(x)dx=xsin(x)+cos(x)+C.
- 1 point: Differentiate to sin(x)+xcos(x)−sin(x)=xcos(x).
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 1Where does the minus sign come from?
Rearranging the integrated product rule.
RECALL 2Must u be a polynomial?
No; it should simplify the remaining work when possible.
RECALL 3What checks a parts answer?
Differentiate the complete expression on its domain.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
How does reversing the product rule help integrate a product?
- ∫u dv=uv−∫v du.
- Choose u, compute du, integrate dv to v, then simplify.
- Add +C to the final indefinite family.
Remember: Do not multiply separate antiderivatives. Keep the minus sign and verify the whole answer using the product rule.
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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