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AP Calculus AB/BC · Mathematics · Lesson 6.11

6.11 Integration by parts (BC)

Level
Advanced Placement (AP)
Reading time
8 min reading
Errata

Abstract

For a product of two different kinds of function, trade the integral for an easier one using the reverse of the product rule.

Concept preview

What you will be able to do

  1. 01State the integration-by-parts formula ∫u dv=uv−∫v du\int u\,dv = uv - \int v\,du and explain how it is the product rule for derivatives read in reverse.
  2. 02Choose uu and dvdv wisely using the LIATE guideline so that the new integral ∫v du\int v\,du is genuinely simpler, and carry out the method on the classics ∫xex dx\int xe^x\,dx, ∫xcos⁡x dx\int x\cos x\,dx, and ∫ln⁡x dx\int \ln x\,dx.
  3. 03Apply integration by parts more than once when needed (repeated parts), and evaluate a definite integral by parts while tracking the crucial minus sign in −∫v du-\int v\,du.

Start from something you already own — the product rule for derivatives. If uu and vv are functions of xx, then ddx(uv)=udvdx+vdudx\dfrac{d}{dx}\left(uv\right)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}. Now integrate both sides with respect to xx. The left side unwinds directly, because integrating a derivative returns the original product: ∫ddx(uv) dx=uv\int \dfrac{d}{dx}(uv)\,dx = uv. The right side splits into two integrals: ∫u dv+∫v du\int u\,dv + \int v\,du. Putting the pieces together gives uv=∫u dv+∫v duuv = \int u\,dv + \int v\,du.

All-Access Plan

6.11 Integration by parts (BC)

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