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Ch. 8 - Techniques of Integration
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Capitolo 8, Problema 8.GYR.2

2. When applying the formula for integration by parts, how do you choose the u and dv? How can you apply integration by parts to an integral of the form ∫ f(x) dx?

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Recall the integration by parts formula: \(\int u \, dv = uv - \int v \, du\). The goal is to rewrite the integral in a way that simplifies the problem.
To choose \(u\) and \(dv\), use the LIATE rule as a guideline, which prioritizes functions in this order: Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential. Choose \(u\) as the function that appears first in this list and \(dv\) as the remaining part of the integral.
Once \(u\) is chosen, compute its differential \(du\) by differentiating \(u\) with respect to \(x\): \(du = u' \, dx\).
Next, determine \(v\) by integrating \(dv\): \(v = \int dv\). This step requires you to integrate the part of the integral assigned to \(dv\).
Finally, substitute \(u\), \(v\), and \(du\) into the integration by parts formula: \(\int f(x) \, dx = uv - \int v \, du\). Then, simplify and evaluate the resulting integral.

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Integration by Parts Formula

Integration by parts is based on the product rule for differentiation and is given by ∫u dv = uv - ∫v du. It transforms the integral of a product of functions into a simpler form, often making the integral easier to solve.
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Percorso guidato
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Introduction to Integration by Parts

Choosing u and dv

Selecting u and dv wisely is crucial; typically, u is chosen as a function that simplifies when differentiated, and dv is chosen as a function that is easy to integrate. The LIATE rule (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential) helps prioritize the choice of u.
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Percorso guidato
07:51
Choosing a Convergence Test

Applying Integration by Parts to ∫ f(x) dx

To apply integration by parts to ∫ f(x) dx, rewrite f(x) as a product of two functions u and dv. Then differentiate u to find du, integrate dv to find v, and substitute into the formula ∫u dv = uv - ∫v du to evaluate the integral.
Video consigliato:
Percorso guidato
08:30
Introduction to Integration by Parts