C26 · Unit C6 Integration and accumulation
Integration by parts
BC extensionReview editionNot yet human-reviewed
- Time
- About 45 minutes, untimed
- Course framework reference
- AP 6.11, 6.14 · BC
- Part of
Progress on this lesson counts in every pathway listed.
Understand the idea
Integrating the product rule yields ∫u dv equals uv minus ∫v du. Choose u so that differentiating it simplifies the remaining integral, and choose dv that can be integrated. For definite integrals use [uv]ₐᵇ minus ∫ₐᵇv du. Integration by parts is a BC technique here; it is not required for the AB route.
Worked example
For ∫x eˣdx, choose u equals x and dv equals eˣdx. Then du equals dx,v equals eˣ, giving xeˣ minus eˣ+C equals eˣ(x minus 1)+C. Differentiation gives eˣ(x minus 1)+eˣ equals xeˣ.
Explore the mathematics
Integration by parts
Transfer a derivative from one factor to simplify an integral.
Scroll sideways to see the full diagram.
Predict before continuing
Which factor becomes simpler when differentiated?
Read the complete explanation
- Which factor becomes simpler when differentiated?
- Choose u=x and dv=eˣ dx; then du=dx and v=eˣ.
- Integration by parts gives [xeˣ]₀¹−∫₀¹eˣ dx.
- The result is e−(e−1)=1. The subtraction term is required.
For ∫x eˣdx, choose u=x and dv=eˣdx. Then du=dx,v=eˣ, giving xeˣ−eˣ+C=eˣ(x−1)+C. Differentiation gives eˣ(x−1)+eˣ=xeˣ.
Discuss: explain what each label represents, then draw a related example with different values.
A common misconception
The second term is an integral with a minus sign. Choosing u and dv is a strategy decision, not an arbitrary letter assignment.
Investigate and explain
Compare parts with substitution on x eˣ and x e^(x squared). Explain why different structures favor different methods.
Identify one step in the worked example, name the mathematical fact that permits it, and state its domain conditions.