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Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550

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BuyFindarrow_forward

Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550
Textbook Problem

The double integral 0 1 0 1 1 1 x y d x d y is an improper integral and could be defined as the limit of double integrals over the rectangle [0, t] × [0, t] as t → 1. But if we expand the integrand as a geometric series, we can express the integral as the sum of an infinite series. Show that 0 1 0 1 1 1 x y d x d y = n 1 1 n 2

To determine

To show: 010111xydxdy=n=11n2.

Explanation

Given:

The double integral 010111xydxdy is an improper integral and can be defined as the limit of double integrals over the rectangle [0,t]×[0,t] as t1.

Calculation:

As t1, it is observed that the value of |xy|<1. So, the sum of the geometric series is given by, 11xy=n=0(xy)n except at the point (1,1).

010111xydxdy=0101(n=0(xy)n)dxdy=n=0(0101(xy)ndxdy)=n=0(01yndy01xndx)

Integrate it and apply the limit

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Chapter 15 Solutions

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Sect-15.1 P-11ESect-15.1 P-12ESect-15.1 P-13ESect-15.1 P-14ESect-15.1 P-15ESect-15.1 P-16ESect-15.1 P-17ESect-15.1 P-18ESect-15.1 P-19ESect-15.1 P-20ESect-15.1 P-21ESect-15.1 P-22ESect-15.1 P-23ESect-15.1 P-24ESect-15.1 P-25ESect-15.1 P-26ESect-15.1 P-27ESect-15.1 P-28ESect-15.1 P-29ESect-15.1 P-30ESect-15.1 P-31ESect-15.1 P-32ESect-15.1 P-33ESect-15.1 P-34ESect-15.1 P-35ESect-15.1 P-36ESect-15.1 P-37ESect-15.1 P-38ESect-15.1 P-39ESect-15.1 P-40ESect-15.1 P-41ESect-15.1 P-42ESect-15.1 P-43ESect-15.1 P-44ESect-15.1 P-47ESect-15.1 P-48ESect-15.1 P-49ESect-15.1 P-50ESect-15.1 P-52ESect-15.2 P-1ESect-15.2 P-2ESect-15.2 P-3ESect-15.2 P-4ESect-15.2 P-5ESect-15.2 P-6ESect-15.2 P-7ESect-15.2 P-8ESect-15.2 P-9ESect-15.2 P-10ESect-15.2 P-11ESect-15.2 P-12ESect-15.2 P-13ESect-15.2 P-14ESect-15.2 P-15ESect-15.2 P-16ESect-15.2 P-17ESect-15.2 P-18ESect-15.2 P-19ESect-15.2 P-20ESect-15.2 P-21ESect-15.2 P-22ESect-15.2 P-23ESect-15.2 P-24ESect-15.2 P-25ESect-15.2 P-26ESect-15.2 P-27ESect-15.2 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