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H. The integral 1 -dz = --- --(28). Then, the improper integral 10+ 2z L 1 -dz = lim (-----(29)) | = 10 + 2z 1 -dz is 10 + 2z --(30). Therefore, ---(31). %3D -5 -5 2 1 In |10 + 2z| + C f. 2 ln |10 + 2z| +C a. b. 10 1 с. g. -10 5 In |10 + 2z| h. -5+ i. divergent d. convergent e. +∞ j. -00

H. The integral 1 -dz = --- --(28). Then, the improper integral 10+ 2z L 1 -dz = lim (-----(29)) | = 10 + 2z 1 -dz is 10 + 2z --(30). Therefore, ---(31). %3D -5 -5 2 1 In |10 + 2z| + C f. 2 ln |10 + 2z| +C a. b. 10 1 с. g. -10 5 In |10 + 2z| h. -5+ i. divergent d. convergent e. +∞ j. -00

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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H.
H. The integral 1 -dz = 10+ 2z -(28). Then, the improper integral 1 dz = lim --(29)) |; 1 -dz is 10 + 2z --(30). Therefore, --(31). %3D 10 + 2z -5 In |10 + 2z| + C f. 2 ln |10 + 2z| +C а. b. 10 g. –10 5 In |10 + 2z| h. -5+ с. d. convergent i. divergent e. +∞ j. -00
Transcribed Image Text:H. The integral 1 -dz = 10+ 2z -(28). Then, the improper integral 1 dz = lim --(29)) |; 1 -dz is 10 + 2z --(30). Therefore, --(31). %3D 10 + 2z -5 In |10 + 2z| + C f. 2 ln |10 + 2z| +C а. b. 10 g. –10 5 In |10 + 2z| h. -5+ с. d. convergent i. divergent e. +∞ j. -00
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