Ch. 8 - Integration Techniques

Briggs - Calculus: Early Transcendentals 3rd Edition

Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook

Briggs 3rd Edition

Ch. 8 - Integration Techniques

Problem 8.2.1

Chapter 8, Problem 8.2.1

1. On which derivative rule is integration by parts based?

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Recall that integration by parts is a technique used to integrate products of functions.

Integration by parts is based on the product rule for differentiation, which states that the derivative of a product of two functions \(u(x)\) and \(v(x)\) is given by: \[\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)\]

By rearranging the product rule, we isolate one term to express an integral in terms of another: \[u(x)v'(x) = \frac{d}{dx}[u(x)v(x)] - u'(x)v(x)\]

Integrate both sides with respect to \(x\) to get: \[\int u(x)v'(x) \, dx = u(x)v(x) - \int u'(x)v(x) \, dx\]

This last equation is the formula for integration by parts, showing that it directly follows from the product rule for derivatives.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Product Rule for Derivatives

The product rule states that the derivative of the product of two functions is given by the derivative of the first times the second plus the first times the derivative of the second. It is expressed as (fg)' = f'g + fg'. This rule is fundamental in deriving the integration by parts formula.

Integration by Parts Formula

Integration by parts is a technique used to integrate products of functions. It is derived from the product rule and states that ∫u dv = uv - ∫v du, where u and v are functions of a variable. This formula helps transform complex integrals into simpler ones.

Relationship Between Differentiation and Integration

Differentiation and integration are inverse processes. Understanding how differentiation rules, like the product rule, relate to integration techniques allows one to manipulate integrals effectively. Integration by parts leverages this inverse relationship to simplify integration of products.

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The following figure shows the discharge rate r(t) of the Snoqualmie River near Carnation, Washington, starting on a February day when the air temperature was rising. The variable t is the number of hours after midnight, r(t) is given in millions of cubic feet per hour, and ∫(0 to 24) r(t) dt equals the total amount of water that flows by the town of Carnation over a 24-hour period. Estimate ∫(0 to 24) r(t) dt using the Trapezoidal Rule and Simpson's Rule with the following values of n.

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