CalculusLesson 2 of 2 in the course Techniques of integration

Integration by Parts: Formulas, Strategies, and Applications

5 flashcards · 5 quiz questions · 5 match pairs · 4 fill-in sentences · English

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Summary

Integration by parts transforms an integral of a product into another integral that may be easier to evaluate. It is the integral counterpart of the product rule for differentiation.

∫u(x)v′(x)dx=u(x)v(x)−∫u′(x)v(x)dx.∫u(x)v'(x) dx=u(x)v(x)-∫u'(x)v(x) dx.

The constant of integration is understood in an indefinite integral.

Definition: Integration by parts is a method that replaces an integral of a product with a product term minus an integral in which differentiation has shifted from one factor to the other.

All flashcards

Indefinite integration-by-parts formula
∫u dv=uv−∫v du\int u\,dv=uv-\int v\,du; here du=u′(x) dxdu=u'(x)\,dx and vv is an antiderivative of dvdv.
Definite integration-by-parts formula
∫abu(x)v′(x) dx=[uv]ab−∫abu′(x)v(x) dx\int_a^b u(x)v'(x)\,dx=[uv]_a^b-\int_a^b u'(x)v(x)\,dx, where [uv]ab=u(b)v(b)−u(a)v(a)[uv]_a^b=u(b)v(b)-u(a)v(a).
LIATE
A rule of thumb for choosing uu: logarithmic, inverse trigonometric, algebraic, trigonometric, then exponential.
Cyclic integral
An integral that returns after repeated integration by parts and can then be isolated algebraically.
Multivariable integration by parts
For a scalar uu and vector field VV, ∫Ωu(∇⋅V) dΩ=∫∂ΩuV⋅n^ dΓ−∫Ω∇u⋅V dΩ\int_\Omega u(\nabla\cdot V)\,d\Omega=\int_{\partial\Omega}uV\cdot\hat n\,d\Gamma-\int_\Omega\nabla u\cdot V\,d\Omega.

Quiz questions

  1. Which differentiation rule gives rise to the integration-by-parts formula?Answer: The product rule, (uv)′=u′v+uv′(uv)'=u'v+uv'.
  2. How does LIATE order the categories used as a guide for choosing uu?Answer: Logarithmic, inverse trigonometric, algebraic, trigonometric, exponential.
  3. If two integrations by parts give a cyclic integral equation I=A−II=A-I, how can II be isolated?Answer: Add II to both sides, then divide by 2.
  4. Using Ff(ξ)=∫e−2πiyξf(y) dyFf(\xi)=\int e^{-2\pi iy\xi}f(y)\,dy, what is Ff′(ξ)Ff'(\xi) when the boundary term vanishes?Answer: 2πiξFf(ξ)2\pi i\xi Ff(\xi)
  5. For ∫xcos⁡x dx\int x\cos x\,dx, choose u=xu=x and dv=cos⁡x dxdv=\cos x\,dx. What are dudu and vv?Answer: du=dxdu=dx and v=sin⁡xv=\sin x.

Based on the Wikipedia article Integration by parts, licensed CC BY-SA 4.0. Adapted into a summary, flashcards and quizzes.

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