Calculus for Business, Economics, Life Sciences, and Social Sciences (13th Edition)

13th Edition

ISBN: 9780321869838

Author: Raymond A. Barnett, Michael R. Ziegler, Karl E. Byleen

Publisher: PEARSON

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Textbook Question

Chapter 7.7, Problem 19E

In Problems 21–26, use the description of the region R to evaluate the indicated integral.

25. ∬ R e x + y d A ; R = { ( x , y ) | − x ≤ y ≤ x , 0 ≤ x ≤ 2 }

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Chapter 7.7, Problem 18E

Chapter 7.7, Problem 20E

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For each real-valued nonprincipal character x mod k, let A(n) = x(d) and F(x) = Σ : dn * Prove that F(x) = L(1,x) log x + O(1). n

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Proof: LN⎯⎯⎯⎯⎯LN¯ divides quadrilateral KLMN into two triangles. The sum of the angle measures in each triangle is ˚, so the sum of the angle measures for both triangles is ˚. So, m∠K+m∠L+m∠M+m∠N=m∠K+m∠L+m∠M+m∠N=˚. Because ∠K≅∠M∠K≅∠M and ∠N≅∠L, m∠K=m∠M∠N≅∠L, m∠K=m∠M and m∠N=m∠Lm∠N=m∠L by the definition of congruence. By the Substitution Property of Equality, m∠K+m∠L+m∠K+m∠L=m∠K+m∠L+m∠K+m∠L=°,°, so (m∠K)+ m∠K+ (m∠L)= m∠L= ˚. Dividing each side by gives m∠K+m∠L=m∠K+m∠L= °.°. The consecutive angles are supplementary, so KN⎯⎯⎯⎯⎯⎯∥LM⎯⎯⎯⎯⎯⎯KN¯∥LM¯ by the Converse of the Consecutive Interior Angles Theorem. Likewise, (m∠K)+m∠K+ (m∠N)=m∠N= ˚, or m∠K+m∠N=m∠K+m∠N= ˚. So these consecutive angles are supplementary and KL⎯⎯⎯⎯⎯∥NM⎯⎯⎯⎯⎯⎯KL¯∥NM¯ by the Converse of the Consecutive Interior Angles Theorem. Opposite sides are parallel, so quadrilateral KLMN is a parallelogram.

Chapter 7 Solutions

Calculus for Business, Economics, Life Sciences, and Social Sciences (13th Edition)

Ch. 7.1 - Prob. 1MP Ch. 7.1 - Matched Problem 2 Find f(2, 2, 3) for f in Example...Ch. 7.1 - Prob. 3MP Ch. 7.1 - Matched Problem 4 For the box in Example 4, find...Ch. 7.1 - Matched Problem 5 Refer to Example 5. Find the...Ch. 7.1 - Matched Problem 6 Find the coordinates of the...Ch. 7.1 - Matched Problem 7 (A) Describe the cross sections...Ch. 7.1 - Explore and Discuss 1 Imagine that you are facing...Ch. 7.1 - In Problems 18, find the indicated value of the...Ch. 7.1 - Prob. 2E

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In Problems 21–26, use the description of the region R to evaluate the indicated integral. 25. ∬ R e x + y d A ; R = { ( x , y ) | − x ≤ y ≤ x , 0 ≤ x ≤ 2 }

Solution Summary: The author explains the value of the iterated integral, which is e4-52.

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